DSC1520/001/4/2020
SG001/4/2020
Quantitative Modelling 1
DSC1520
Semesters 1 and 2
Department of De ision S ien es
This is the only study guide for DSC1520.
Bar code
Contents
1 Introdu tion
1.1
1.2
1
The module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1.1
Knowledge assumed to be in pla e
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1.2
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.1.3
Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.1.4
A tivities
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Development of a mathemati al model . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Modelling
1.2.1
I Linear fun tions
5
2 Appli ations of linear fun tions
6
2.1
2.2
2.3
Revenue,
ost and prot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.1.1
Cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.1.2
Revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.1.3
Prot
7
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Demand and supply
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.2.1
Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
2.2.2
Supply . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
Pri e elasti ity of demand and supply
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1
Pri e elasti ity of demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.3.2
Pri e elasti ity of supply . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
3 Appli ations of simultaneous fun tions
3.1
13
17
Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.1.1
Equilibrium in the goods market
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.1.2
Equilibrium in the labour market . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
3.1.3
Pri e
. . . . . . . . . . . . . . . . . .
19
3.1.4
Complementary and substitute goods . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
3.1.5
Taxes and subsidies
ontrols and government intervention in markets
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
3.2
Break-even analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.3
Consumer and produ er surplus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
3.3.1
Consumer surplus
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
3.3.2
Produ er surplus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.3.3
Total surplus
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
Linear programming
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
3.4
ii
DSC1520/SG001
II Non-linear fun tions
38
4 Appli ations of non-linear fun tions
39
4.1
Non-linear fun tions in e onomi s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
4.2
Quadrati
40
fun tions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1
Optimising quadrati
. . . . . . . . . . . . . . . . .
41
4.2.2
Market equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
4.2.3
Break-even analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
4.3
Cubi
fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
4.4
Exponential and logarithmi
4.5
revenue, prot and
fun tions
ost fun tions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
4.4.1
Unlimited growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
4.4.2
Limited growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
4.4.3
Logisti
Hyperboli
growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
III Dierentiation
54
5 Dierentiation theory
55
5.1
Slope of a
urve and dierentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2
Dierentiation rules
55
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
5.2.1
The power rule for dierentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
5.2.2
Exponential and natural logarithmi
fun tions . . . . . . . . . . . . . . . . . . . . . . .
60
5.2.3
The
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
5.2.4
The produ t rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
5.2.5
The quotient rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
5.3
Higher derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
5.4
Optimisation of fun tions in one variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
5.4.1
The nature of stationary points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
5.4.2
Intervals along whi h a fun tion is in reasing or de reasing . . . . . . . . . . . . . . . .
71
5.4.3
Sket hing fun tions using dierentiation
72
hain rule
. . . . . . . . . . . . . . . . . . . . . . . . . .
6 Appli ations of dierentiation
6.1
75
Marginal and average fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
6.1.1
Marginal revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
6.1.2
Marginal
76
6.1.3
Average revenue and
ost fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
6.1.4
Produ tion fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
ost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2
Optimisation of e onomi
fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
6.3
Elasti ity of demand non-linear demand fun tions . . . . . . . . . . . . . . . . . . . . . . . .
87
IV Integration
90
7 Integration theory
91
7.1
Integration as the reverse of dierentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
7.2
Integration rules
91
7.3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.1
The power rule for integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
7.2.2
Integration by substitution
94
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The denite integral and the area under a
urve . . . . . . . . . . . . . . . . . . . . . . . . . .
8 Appli ations of integration
97
102
iii
DSC1520/SG001
8.1
8.2
From marginal to total fun tions
ost from marginal
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.1.1
Total
ost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.1.2
Total revenue from marginal revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.1.3
Rate of
hange to total quantity
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Consumer and produ er surplus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
8.2.1
Consumer surplus
8.2.2
Produ er surplus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
8.2.3
Total surplus
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
9 Solutions to a tivities
112
9.1
Se tion 2.1 (Revenue,
9.2
Se tion 2.2 (Appli ations: demand, supply,
9.3
Se tion 2.3 (Elasti ity of linear demand and supply fun tions) . . . . . . . . . . . . . . . . . . 115
9.4
Se tion 3.1 (Equilibrium and break-even)
9.5
Se tion 3.1.4 (Complementary and substitute goods)
9.6
Se tion 3.1.5 (Taxes and subsidies)
9.7
Se tion 3.2 (Break-even analysis)
9.8
Se tion 3.3 (Consumer and produ er surplus)
9.9
Se tion 3.4 (Linear programming) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
9.10 Se tion 4.1 (Quadrati
ost and prot fun tions) . . . . . . . . . . . . . . . . . . . . . . . . . . 112
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
. . . . . . . . . . . . . . . . . . . . . . . 116
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
. . . . . . . . . . . . . . . . . . . . . . . . . . . 118
fun tions in e onomi s)
fun tions)
. . . . . . . . . . . . . . . . . . . . . . . . . . 120
fun tions) . . . . . . . . . . . . . . . . . . . . . . . . 124
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
9.13 Se tion 5.2.1 (The power rule for dierentiation)
9.14 Se tion 5.2.3 (The
. . . . . . . . . . . . . . . . . . . . 113
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
9.11 Se tion 4.4 (Exponential and logarithmi
9.12 Se tion 4.5 (Hyperboli
ost, revenue)
. . . . . . . . . . . . . . . . . . . . . . . . . 126
hain rule) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
9.15 Se tion 5.2.4 (The produ t rule) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
9.16 Se tion 5.2.5 (The quotient rule)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
9.17 Se tion 5.3 (Higher derivatives) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
9.18 Se tion 5.3 (Optimisation of fun tions in one variable)
9.19 Se tion 6.1 (Marginal and average fun tions)
9.20 Se tion 6.1.4 (Produ tion fun tions)
9.21 Se tion 6.2 (E onomi
. . . . . . . . . . . . . . . . . . . . . . 130
. . . . . . . . . . . . . . . . . . . . . . . . . . . 131
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
appli ations of optimisation) . . . . . . . . . . . . . . . . . . . . . . . . 132
9.22 Se tion 6.3 (Elasti ity of demand non-linear demand fun tions)
9.23 Se tion 7.2.1 (The power rule for integration)
. . . . . . . . . . . . . . . . 134
. . . . . . . . . . . . . . . . . . . . . . . . . . . 135
9.24 Se tion 7.2.2 (Integration by substitution) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
9.25 Se tion 7.3 (The denite integral and the area under a
urve)
. . . . . . . . . . . . . . . . . . 136
9.26 Se tion 8.1 (From marginal to total fun tions) . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
9.27 Se tion 8.2 (Consumer and produ er surplus)
. . . . . . . . . . . . . . . . . . . . . . . . . . . 138
V Mathemati al ba kground assumed to be in pla e
140
A Numbers and variables
141
A.1
Priorities
A.2
Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
A.3
Laws of operations
A.4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
A.3.1
Addition and subtra tion
A.3.2
Multipli ation and division
Fra tions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
A.4.1
Multipli ation and division
A.4.2
Addition and subtra tion
A.5
Fa torisation
A.6
Fun tions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
iv
DSC1520/SG001
A.7
Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
A.8
Per entages
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
B Equations and inequalities
B.1
B.2
152
Solving equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
B.1.1
Linear equations
B.1.2
Quadrati
Inequalities
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
equations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
C Linear fun tions
157
C.1
Graphs of linear fun tions
C.2
Plotting linear fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
C.3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
y
C.2.1
Using the slope and
C.2.2
Using the inter epts on the
axis inter ept
x
and
Determining formulæ of linear fun tions
y
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
axes . . . . . . . . . . . . . . . . . . . . . . . . . . 159
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
C.3.1
Given two points on the line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
C.3.2
Given the slope of the line and any point on the line
. . . . . . . . . . . . . . . . . . . 161
C.4
Plotting linear inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
C.5
Simultaneous equations/inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
C.5.1
Solving simultaneous linear equations
. . . . . . . . . . . . . . . . . . . . . . . . . . . 163
C.5.2
Solving simultaneous equations graphi ally
C.5.3
Systems of inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
. . . . . . . . . . . . . . . . . . . . . . . . 165
D Non-linear fun tions
D.1
Quadrati
D.1.1
fun tions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
Properties and graphs
D.2
Cubi
D.3
Exponential fun tions
D.4
168
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
D.3.1
Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
D.3.2
Working with exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
D.3.3
Solving equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
Logarithmi
fun tions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
D.4.1
Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
D.4.2
Graphs and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
D.4.3
Working with logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
D.4.4
Solving equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
D.5
Hyperboli
D.6
Composite fun tions
fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
E Solutions to a tivities (Appendix)
186
E.1
Se tion A.1 (Priorities) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
E.2
Se tion A.2 (Variables) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
E.3
Se tion A.3 (Laws of operations)
E.4
Se tion A.4 (Fra tions) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
E.5
Se tion A.5 (Fa torisation)
E.6
Se tion A.6 (Fun tions)
E.7
Se tion A.7 (Polynomials) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
E.8
Se tion A.8 (Per entages)
E.9
Se tion B.1 (Solving equations) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
E.10 Se tion B.2 (Inequalities)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
E.11 Se tion C.2 (Plotting linear fun tions)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
E.12 Se tion C.3 (Determining formulæ of linear fun tions)
v
. . . . . . . . . . . . . . . . . . . . . . 190
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E.13 Se tion C.5 (Simultaneous equations/inequalities) . . . . . . . . . . . . . . . . . . . . . . . . . 191
E.14 Se tion D.1 (Quadrati
E.15 Se tion D.2 (Cubi
fun tions) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
fun tions) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
E.16 Se tion D.3 (Exponential fun tions)
E.17 Se tion D.4 (Logarithmi
E.18 Se tion D.5 (Hyperboli
fun tions)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
fun tions)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
E.19 Se tion D.6 (Composite fun tions)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
vi
Chapter 1: Introdu tion
Students who enrol for a business-related degree su h as De ision S ien es (or E onomi s) often nd it
surprising that mathemati s form su h a huge part of the study.
However, in order to make informed
business de isions, it very often is ne essary to set up and solve some mathemati al formulæ representing
real-life problems.
modelling
(or
This setting up of mathemati al formulæ to represent problems is
quantitative modelling).
alled
mathemati al
1.1 The module
This module, Quantitative Modelling, provides introdu tory notes on elementary mathemati al te hniques
and illustrates the appli ation thereof on de ision-making situations.
The study guide is divided into four main parts that in lude this introdu tion, seven hapters and an appendix
with details of the mathemati al ba kground that you should have a quired at s hool or somewhere else.
Ea h
hapter starts with the learning obje tives that highlight the key
hapter.
When you have nished a
thoroughly familiar with all the relevant
All
hapters
also
he klist to ensure that you are
on epts.
ontain worked examples to show how questions on a topi
ontain a tivities that are intended as self-assessment tests. If you
making any signi ant errors, then you have apparently mastered the
reveal the areas in whi h you require further study. In this
the
on epts you will en ounter in the
hapter, use these obje tives as a
should be answered. The
hapters
an manage the a tivities without
hapter. If not, the exer ise should
ase you should go ba k to the relevant parts of
hapter.
Should you, at this stage, after an earnest attempt to understand troublesome passages, nd that you
make progress, you should
annot
onta t your le turer or tutor at on e and ask for his or her help. Setting aside
too many unresolved problems for another day is bound to lead to
onfusion and disillusionment.
1.1.1 Knowledge assumed to be in pla e
You were allowed to enter this module sin e you had either passed Grade 12 with good marks in mathemati s or
ompleted a higher
erti ate su
essfully. We therefore assume that you are pro ient in basi
mathemati s, whi h in ludes the following:
⊲
Numbers and working with numbers
priorities
variables
laws of operations
fra tions
fa torisation
fun tions
polynomials
1
DSC1520/SG001
⊲
Fun tions
⊲
powers and roots
per entages
linear fun tions
linear systems
evaluating and transforming fun tions
Equations and inequalities
solving linear and non-linear equations
solving simple inequalities
⊲
Linear fun tions
⊲
Simultaneous linear fun tions
⊲
Non-linear fun tions
quadrati
ubi
fun tions
fun tions
exponential and logarithmi
fun tions
omposite fun tions
If you should nd that you la k knowledge and/or understanding of any one of these
on epts, you will nd
notes and exer ises on them in Appendix V.
1.1.2 Notation
Throughout the module, we use
fun tions.
variables
p
ursive lower ase letters for variables and upper ase
For example, the fun tions for
and
q
ost, revenue and prot are denoted by
C, R
ursive letters for
and
P,
while the
denote pri e and quantity.
1.1.3 Software
In this module we introdu e you to the mathemati al pa kage Maxima. This is free software that
an be
downloaded from the internet. Notes on the installation and use of Maxima are available at the following
https://my.unisa.a .za/portal/site/DSC1520-20-S1
https://my.unisa.a .za/portal/site/DSC1520-20-S2.
links:
and
Maxima is really easy to use for mathemati al operations su h as plotting fun tions or solving equations.
When you observe the graphs of dierent fun tions regularly, you may develop an intuitive sense of the
properties of
ertain fun tions. You may also use Maxima to
he k your answers to assignment questions.
1.1.4 A tivities
Throughout ea h
hapter, you will nd a tivities. Do these a tivities by hand (with pen on paper) before
you look at the solutions. This will give you an indi ation of how well you understand the relevant study
material and provide the opportunity to exer ise your mathemati al writing skills.
Solutions to a tivity questions are available in Chapter 9. Please try to do the a tivities before looking at
the solutions.
2
DSC1520/SG001
1.2 Modelling
In many areas of life, models are used to represent aspe ts of reality. In this se tion, we attempt to put the
notion of mathemati al modelling into
Models may be
⊲
lassied as either physi al or abstra t, and stati
or dynami . Examples are the following:
Physi al models
Stati : fashion models; miniature
Dynami : the aerodynami s of a
⊲
ontext with referen e to other kinds of models.
repli as of buildings/trains/ ars; et
ar in a wind tunnel; et
Abstra t models
Stati
: a
hart showing interrelations in an organisation; the design of a
ar engine; the relationship
between the pri e of a good and the quantity demanded; et
Dynami
: the ow of information in an organisation; how a
how a population
Abstra t models may also be
⊲ Deterministi
ar engine works when it is swit hed on;
hanges over time; et
lassied as deterministi
: The out ome of the model
or sto hasti .
an be determined exa tly when all the inputs are given.
Example:
If
p
is the pri e of an item and
q
the number of items sold, then the equation
q = 20 − 2p
des ribes
how the number of items demanded depends on the pri e per item. With this equation available, the
number of items demanded
⊲ Sto hasti
: The out ome is
an be determined exa tly if the pri e is known.
al ulated with a
ertain probability.
Example:
When a fair di e is tossed, any one of the numbers 1, 2, . . . , 6 may turn up. The probability of getting
any one of these numbers is
x
is a
sto hasti variable.
1
6 . If
x
represents the number turning up when the di e is tossed, we say
In this module we are mainly
We do not in lude any dynami
on erned with stati , deterministi , abstra t models.
or sto hasti
models.
1.2.1 Development of a mathemati al model
The pro ess that needs to be followed to develop a mathemati al model for a real-life situation,
an be
summed up in the following four steps:
1. Study and understand the situation ask pertinent questions.
2. Formulate the mathemati al model.
(a) Identify variables.
(b) Set up mathemati al formulæ.
( ) Solve the equations and interpret solutions.
3. Che k the validity of the model.
4. If the model passes the validity test, it may be used, subje t to the
onditions under whi h it has been
set up. If not, start again.
Consider the following real-life example:
Judy provides fudge to be sold at a railway station kiosk. She needs to de ide how many blo ks of fudge she
should make per day.
3
DSC1520/SG001
To formulate a model for Judy's problem, we follow the steps explained above:
1. We assume that only the pri e harged per blo k of fudge determines the number that will be demanded,
sin e the fudge is very tasty and Judy is the only person providing fudge at the station. Past experien e
shows that 50 blo ks of fudge are sold per day if the pri e is R1 per blo k, while only 20 blo ks are
demanded if the pri e is R4 per blo k.
2. To formulate the model, we go through the following steps:
(a) First of all, we de ide to let
q
denote the number of blo ks of fudge demanded and
p
the pri e per
blo k. These are the variables that are measurable.
q = a − bp.
(p; q), are given through whi h the line representing our problem should pass, namely
(1; 50) and (4; 20). Using these points, it is found1 that our problem an be represented by the
line q = 60 − 10p.
(b) The simplest mathemati al formula is the linear fun tion, in general represented by
Two points,
( ) The solution and interpretation are straightforward in this
ase, sin e we
an simply
he k whether
the mathemati al formula represents the given data. However, this might not always be the
3. To
he k the validity of this model,
he k whether the demand at a
ertain pri e
ase.
orresponds reasonably
well to the out ome of the model. If the model passes the validity test, it may be used. If not, something
is wrong and you will have to start again.
Throughout the study guide, modelling is used for su h real-life examples. So, let's get started.
1
Details of how to solve this, are provided in Se tion 2.2.1.
4
PART I
LINEAR FUNCTIONS
5
Chapter 2: Appli ations of linear fun tions
Learning obje tives
After you have
ompleted this
⊲
explain the
⊲
explain the
⊲
ost and prot
ost and prot fun tions from given information
on epts of demand and supply
plot demand and supply fun tions and distinguish between the two types of graphs
determine equations for demand and supply from given information
explain the
These
on epts of revenue,
set up revenue,
hapter, you should be able to
on ept of pri e elasti ity of demand and supply
al ulate and interpret point and ar
elasti ity of demand and supply
on epts form the basis of quantitative modelling.
moving on to more
You should understand them really well before
ompli ated topi s.
2.1 Revenue, ost and prot
The aim of any business is to generate prot. In order to
ne essary to determine what the operation
al ulate the prot earned from an operation, it is
osts and what in ome
an be generated
2.1.1 Cost
In general,
ost is what one needs to pay (or give up) in order to get something.
In business,
ost is an amout of money that represents the (i) eort, (ii) material, (iii) resour es, (iv) time
onsumed, (v) utilities
onsumed, (vi) risks in urred and (vii) opportunities that might have been missed in
the pro ess of produ ing something or delivering a servi e.
Total ost (T C )
is broken down into
xed osts (F C )
and
variable osts (V C ),
with
T C = F C + V C.
(a) Fixed
sold)
ost is the expense or
hanges. Fixed
ost that does not
hange when the number of items that are produ ed (or
osts are expenses that have to be paid by a
ompany, independent of any business
a tivity.
(b) Variable
osts depend on produ tion output it is a
may in lude the
and sales
ommissions to be paid.
multiplied by the
onstant amount per unit produ ed. These
ost of raw materials used in produ tion, the
Variable
ost is
osts
osts
al ulated as the quantity of goods produ ed (q )
ost per unit (c), that is
V C = q × c.
6
ost of utilities like ele tri ity, labour
DSC1520/SG001
linear
The
total
ost fun tion
TC = FC + V C
an be graphed by using either the slope and the verti al
inter ept (or another point on the line), or any two points on the line (e.g. the
Consider, for example, the total
oordinates on the axes).
ost fun tion
T C = 20 + 4q,
F C = 20 and V C = 4q . To
(when q = 0) a the slope is four.
with
1
graph the fun tion , we note that the inter ept on the verti al axis is 20
The graph is shown in Figure 2.1.
TC
(5; 40)
40
30
20
Fixed
ost
10
1
2
Figure 2.1:
3
4
q
5
T C = 20 + 4q .
2.1.2 Revenue
In retail stores, the pri es of produ ts are usually xed. The amount of money that a storekeeper will re eive
from selling a
ertain produ t is
total revenue2 (T R) from the produ
the number of units sold (q ), that is
alled the
by multiplying the pri e per unit, (p) by
t. Total revenue is
al ulated
T R = q × p.
If, for example, a produ t is sold for R10 per unit, total revenue is given by
T R = 10q.
This fun tion is plotted as shown in Figure 2.2. Note that the slope is 10 and the verti al inter ept is 0 the line goes through the origin
(0; 0).
2.1.3 Prot
Prot (P ) is the surplus that remains after total
su
osts have been dedu ted from total revenue. A
ompany's
ess is usually measured by the prot it makes.
Prot is
al ulated by subtra ting total
osts from total revenue, that is
P = T R − T C = T R − (F C + V C).
Note:
1
2
⊲
When
T R = T C,
no prot is made (P
= 0)
⊲
When
TR > TC
(i.e.
P > 0),
the
ompany
makes a prot.
⊲
When
TR < TC
(i.e.
P < 0),
the
ompany
makes a loss.
and we say the
ompany
breaks even.
If you nd it di ult to plot su h a graph, please onsult Appendix C (page 157).
Total revenue is the in ome generated from the sale of goods or servi es before any osts or expenses are dedu ted.
7
DSC1520/SG001
TR
(4; 40)
40
30
20
10
1
2
3
Figure 2.2:
4
q
T R = 10q
Example
A tu k shop at the station has xed
while the produ tion
The tu k shop's total
osts of R600 per week. They sell pies at a xed pri e of R25,00 ea h,
ost per pie is R10,00. We assume all the pies that are produ ed will be sold.
ost fun tion is
T C = F C + V C = 600 + 10q,
where
q
is the number of pies produ ed per week.
Their total revenue fun tion is
T R = price per pie × number of pies sold = 25q.
The prot fun tion is given by
P = TR − TC
= 25q − (600 + 10q)
= 15q − 600.
Figure 2.3 shows the graph of the prot fun tion as obtained from Maxima. From the graph it is
Figure 2.3:
the tu k shop needs to sell 40 pies to
pies, their prot is R900, whi h we
an
P = 15q − 600
over their overhead
al ulate as
lear that
osts and break even. Also, when they sell 100
P = 15(100) − 600 = 900.
8
DSC1520/SG001
A tivity
1. The surng
lub provides swimming lessons to boost their funds. The
day when oering these
lasses mostly for insuran e and it
(a) Write down the total daily
(b) Cal ulate the total
osts of R250 per
osts them R25 for ea h lesson given.
ost fun tion.
ost to provide 20 lessons.
( ) How many lessons did they provide if total
2. A rm's xed
lub has xed
osts to produ e a
osts amounted to R1 400?
ertain kind of lamp amount to R1 000 per week and it
osts R15 to
produ e ea h lamp. These lamps are sold for R35 ea h.
(a) Write down the total
(b) What is total
ost and total revenue fun tions.
ost if 400 lamps are produ ed per week?
( ) How many lamps are produ ed when total revenue is R1 750?
(d) Write down the prot fun tion for this kind of lamp and nd the prot if 100 lamps are produ ed
and sold.
3. Adam sells wat hes at the ea market on Saturdays. His xed
osts per week amount to R900. He
buys the wat hes wholesale at R30 ea h and he sells them for R60 ea h.
(a) Write down Adam's weekly total
ost, total revenue and prot fun tions.
(b) How many wat hes did Adam sell on Saturday if his total revenue was R4 200?
4. A rm produ es
al ulators for a
to produ e. The
ertain shop. Their xed
al ulator
osts R15
al ulators are sold for R35 ea h.
(a) Write down the equation for total weekly
(b) Graph the total
( ) What is the total
(d) How many
ost is R1 000 and ea h
ost fun tion for
ost.
q = 0 to 100.
ost to produ e 25
al ulators?
al ulators are produ ed if total
osts amount to R7 000?
(e) Find the total revenue fun tion.
(f ) Graph the
(g) How many
TR
fun tion for
q = 0 to 100.
al ulators are sold when
T R = 1 750?
(h) Does the rm's total revenue ex eed total
osts when 80
al ulators are produ ed? What does
this mean?
(i) Cal ulate the rm's prot when 80
al ulators are produ ed.
2.2 Demand and supply
The level of e onomi
a tivity in an e onomy is determined by the de isions about demand and supply that
onsumers, rms and the government make.
These de isions play a vital role in business and
onsumer
a tivity, and we need mathemati al tools to analyse them.
In this module we
onsider the simplest models of demand and supply where the demand for and the supply
of a produ t depend on pri e only. All other variables are
For example, demand may be ae ted by
9
onsidered to be
onstant.
DSC1520/SG001
(i)
onsumer in ome
3 or supplementary4 good
(ii) the pri e of a substitute
(iii) the level of advertising
Supply may be inuen ed by
(i) the
ost of produ tion
(ii) the pri e of other goods
(iii) available te hnology
(iv) taxes or subsidies
2.2.1 Demand
The term
demand
represents the quantity of a produ t or servi e (q ) that
onsumers would buy at a
ertain
pri e (p).
The demand for a produ t is negatively related to the pri e that is asked for the produ t, that is
the quantity demanded de reases when the pri e in reases.
Consumers will buy less of a produ t if its pri e in reases, while they will buy more of a produ t at a lower
pri e.
The demand fun tion
an be modelled by the linear equation
q = a − bp,
where
a and b are
onstants,
q
is the dependent variable (on the verti al axis) and
p the independent variable
(on the horizontal axis). Comparing this with the linear fun tion, we see that the verti al inter ept is given
by
a.
q = 0,
We nd the inter ept on the horizontal axis by setting
that is
0 = a − bp
or
p=
a
b.
The general linear demand fun tion is graphed in Figure 2.4.
q
a
Slope
= −b
a
b
Figure 2.4: Linear demand fun tion
p
q = a − bp
Note that the demand fun tion falls in the rst quadrant of the graph, sin e
q
be less than zero.
Example
Consider the demand fun tion
q = 200 − 2p.
3
4
A substitute good an be used instead of another, like trains and buses.
A supplementary good is onsumed in onjun tion with another, like petrol and ars.
10
(quantity) and
p (pri
e)
annot
DSC1520/SG001
To plot the linear demand fun tion, we
For example, by setting
0 = 200 − 2p
or
p = 100
p = 0 we
an either nd two points on the line, or use the slope and inter ept.
q = 200 − 2(0) = 200
(100; 0)).
nd
(the point
(the point
(0; 200))
and by setting
q=0
we nd
The graph of this demand fun tion is shown in Figure 2.5.
q
200
150
100
50
20 40 60 80 100
Figure 2.5: Demand fun tion
p
q = 200 − 2p
lear that the slope of the demand line is negative (−2). When the pri e is zero,
It is
q = 200
items are
demanded. On the other hand, when the pri e is R100, the quantity demanded is zero, whi h means that
the pri e is too high and nobody will buy the produ t.
2.2.2 Supply
The term
supply
represents the quantity of a produ t or servi e (q ) that is made available in the market,
depending on the pri e (p) of the produ t or servi e. There is a
positive relationship
between the supply of
a produ t and its pri e, that is
the quantity supplied in reases when the pri e in reases.
When the pri e of a good is high, suppliers want to sell more in order to make more prot, thus in reasing
supply. When the pri e of a good de reases, however, suppliers will also de rease the quantity supplied.
As is the
ase with the demand fun tion, we
onsider supply (q ) as the dependent variable and pri e (p) as
the independent variable. The general linear supply fun tion is given by
q = c + dp,
where
q
is the number of units supplied,
p
is the pri e per unit, and
c and d
are
onstants. The graph of this
supply fun tion is shown in Figure 2.6.
Example
John supplies hand-printed T-shirts to a
raft market.
He will only supply T-shirts when the pri e per
T-shirt is more than R50 and will in rease his output by 20 units for every R10 in rease in pri e.
To nd the supply fun tion, we need either a point on the line and the slope of it, or two points on the line.
⊲
It is given that
q = 0 if p < 50, whi
c
this to Figure 2.6 above, we see that −
d
⊲
⊲
= 50,
giving the verti al inter
The slope of the line is found from the fa t that
the slope of the line as
d=
20
10
(p; q) = (50; 0).
ept as c = −50d.
h means the line goes through the point
q
Comparing
will in rease by 20 for every R10 in rease in
p, giving
= 2.
The verti al inter ept is therefore at
c = −50(2) = −100,
so that the supply fun tion is given by
q = −100 + 2p.
11
DSC1520/SG001
q
Slope
c
=d
p
− dc
Figure 2.6: Linear supply fun tion
q = c + dp
We plot this supply fun tion as shown in Figure 2.7.
q
200
100
0
50
−100
100
p
150
Figure 2.7: Linear supply fun tion
q = −100 + 2p
Note:
* The slope of the line is positive, indi ating a positive relationship between pri e and supply.
* The line under the horizontal line is dashed to indi ate that it is only in luded to draw the graph.
* If we need the supply fun tion with
p
as the dependent variable, we transform the fun tion to nd that
2p = 100 + q
or p = 50 + 0,5q.
A tivity
1. Suppose the demand fun tion for joyrides at a merry-go-round is given by
number of rides per hour and
(a) Find the demand when
(b) Use these
p
is the pri e per ride in rand.
p=0
and the pri e when
q = 0,
oordinates to plot the demand fun tion with
( ) What is the
q = 64 − 4p,
and write these results as
q
where
q
is the
oordinates.
on the verti al axis.
hange in demand (q ) if pri e (p) in reases by one unit?
(d) Transform the demand fun tion to have pri e (p) as dependent variable.
2. The demand and supply fun tions for baby marrows are
tively, with
p
the pri e in rand and
q
q = 210 − 3,5P
the quantity in boxes.
(a) Transform the supply fun tion to have
q
as the dependent variable.
12
and
p = 0,25q + 22,5,
respe -
DSC1520/SG001
(b) By using the inter epts on the verti al and horizontal axes, graph the demand and supply fun tions
on the same axes with
q
as dependent variable.
( ) Find the point where the demand and supply fun tions interse t. What does this point tell us?
3. A supplier supplies 50 T-shirts when the pri e is R60 per T-shirt and 90 T-shirts when the pri e is
R110 per T-shirt.
(a) Determine the equation of the supply fun tion as a fun tion of
(b) How many additional T-shirts are supplied for ea h su
q.
essive R1 in rease in pri e?
( ) How many T-shirts are supplied when the pri e is R85?
(d) What is the pri e when 120 T-shirts are supplied?
2.3 Pri e elasti ity of demand and supply
It is ne essary for a rm to know how qui kly and ee tively it
espe ially to pri e
an respond to
hanging market
onditions,
hanges. For example, if the pri e of a produ t is R900 per unit, will a 10% in rease in
pri e result in a 10% de rease in demand, or will it have a larger per entage de rease in demand?
Su h information is very important to rms as their total revenue from a produ t depends on both the pri e
and the quantity sold.
They need to know whether the additional revenue that will be generated by the
pri e in rease will make up for the loss in revenue due to the de rease in demand.
This is where elasti ity plays an important role, sin e it measures the responsiveness of quantity demanded/supplied to
The
oe ient of
hange in an e onomi
elasti ty
variable su h as pri e or in ome.
is determined as the ratio of the per entage
supplied) to the per entage
hange in quantity demanded (or
hange in pri e.
In general, the numeri al value of elasti ty (ε) is given without the sign. This is
alled the
absolute value,
whi h is dened as
|ε| =
The
oe ient of elasti ty
1. When
|ε| > 1
(i.e.
|ε|
ε < −1
or
ε > 1),
(i.e.
−1 < ε < 1),
supply is weakly responsive to
|ε| = 1
per entage
(i.e.
ε = −1
or
ategories:
we say demand or supply is
elasti
. This means that demand or
hange in demand or
hange in pri e.
we say demand or supply is
inelasti
. This means that demand or
hange in pri e, in other words the per entage
supply is less than the per entage
3. When
−ε if ε < 0.
hange in pri e, in other words the per entage
supply is greater than the per entage
|ε| < 1
ε if ε ≥ 0,
an fall in one of the following three
supply is strongly responsive to
2. When
(
hange in demand or
hange in pri e.
ε = 1),
we say demand or supply is
hange in demand or supply is equal to the per entage
unit elasti
. This means that the
hange in pri e.
2.3.1 Pri e elasti ity of demand
Pri e elasti ity of demand (εd ) is a measure of how demand responds to a rise (or fall) in
the pri e of a good or servi e.
13
DSC1520/SG001
Pri e elasti ty of demand is
al ulated as
% change in quantity demanded
% change in price
%∆q
=
%∆p
εd =
The numeri al value of
εd
=
∆q
q
∆p
p
=
∆q p
· .
∆p q
× 100
× 100
is negative, sin e the slope of the demand line given by
∆q
∆p is negative. We know
that the slope of a straight line is given by the number of units the variable on the verti al axis
unit
hange in the variable on the horizontal axis, whi h is
∆q
∆p
= −b
hanges per
for the demand fun tion
q = a − bp.
Therefore,
εd =
∆q
∆p
·
p
q
= −b ·
p
q is also negative.
This negative sign indi ates the dire tion of responsiveness of demand with regard to
hange in pri e, namely
an in rease in pri e will result in a de rease in demand, while a de rease in pri e will result in an in rease in
demand.
Note that pri e elasti ity
an also be written as a fun tion of only
εd = −b ·
This formula
We
ontains only
p
as a variable, sin e
a
p,
that is
p
p
= −b ·
.
q
a − bp
and
b
are
onstants in the demand fun tion.
onsider two approa hes to measuring pri e elasti ity, namely point elasti ity and ar
elasti ity.
Point pri e elasti ity of demand
Point elasti ity measures elasti ity at a point on the demand fun tion. For the demand fun tion
q = a − bp,
point elasti ity at any point
(p0 ; q0 )
is
∆q p0
∆p q0
p0
= −b × ,
q0
εd =
sin e the slope of the demand line is
∆q
∆p
= −b.
Example
The demand fun tion for a
ertain kind of
omputer is given by
q = 4 800 − 2p.
Let's determine the point elasti ity of demand when
p = 1 800, p = 1 200
14
and
p = 600.
DSC1520/SG001
⊲
When
p = 1 800,
then
q = 4 800 − 2(1 800) = 1 200,
εd = −b ×
|ε| = 3 > 1 and
This means that
(or de rease) in pri e will
⊲
When
p = 1 200,
then
⊲
When
p = 600,
then
and
an further say that a 1% in rease
ause a 3% de rease (or in rease) in demand.
that is
(p0 ; q0 ) = (1 200; 2 400)
and
p0
1 200
= −2 ×
= −1.
q0
2 400
and demand is unit elasti
at this pri e. This means that a 1% in rease (or
ause a 1% de rease (or in rease) in demand.
q = 4 800 − 2(600) = 3 600,
εd = −b ×
This means that
at this pri e. We
q = 4 800 − 2(1 200) = 2 400,
|ε| = 1
de rease) in pri e will
(p0 ; q0 ) = (1 800; 1 200)
1 800
p0
= −2 ×
= −3.
q0
1 200
demand is elasti
εd = −b ×
This means that
that is
|ε| = 0,33 < 1
(or de rease) in pri e will
that is
(p0 ; q0 ) = (600; 3 600)
and
p0
600
= −2 ×
= −0,33.
q0
3 600
and demand is inelasti
at this pri e. This means that a 1% in rease
ause a 0,33% de rease (or in rease) in demand.
Ar pri e elasti ity of demand
Ar
pri e elasti ity measures the elasti ity of demand over an interval. It uses the average of the pri es and
quantities at the beginning and end of the interval under
If the interval is from
(p1 ; q1 )
to
(p2 ; q2 )
εd =
onsideration to
on the demand line, ar
al ulate elasti ity.
pri e elasti ity is
∆q 12 (p1 + p2 )
p1 + p2
= −b ·
· 1
.
∆p 2 (q1 + q2 )
q1 + q2
Example
Consider the demand fun tion for a kind of
omputer as before, namely
q = 4 800 − 2p.
To determine ar
elasti ity if pri e in reases from R1 000 to R1 500, we need to nd the quantities demanded
at these pri es.
If
p = 1 000,
then
q = 4 800 − 2(1 000) = 2 800
to get
(p1 ; q1 ) = (1 000; 2 800).
If
p = 1 500,
then
q = 4 800 − 2(1 500) = 1 800
to get
(p2 ; q2 ) = (1 500; 1 800).
Then ar
pri e elasti ity is
εd = −b ·
Sin e
|εd | > 1,
demand is elasti
1 000 + 1 500
p1 + p2
= −2 ·
= −2 × 0,54 = −1,08.
q1 + q2
2 800 + 1 800
over the given interval.
15
DSC1520/SG001
2.3.2 Pri e elasti ity of supply
The
on ept of elasti ity
derived for point and ar
an also be applied to the analysis of the pri e elasti ity of supply. The formulae
sensitivity of demand,
an also be derived for supply, but sin e the supply fun tion
has a positive slope, elasti ity will also be positive.
Pri e elasti ity of supply (εs ) is a measure of the how supply responds to a rise (or fall) in
the pri e of a good or servi e.
Consider the supply fun tion
q = c + dp
with slope
d.
Point pri e elasti ity of supply is
al ulated as
% change in quantity supplied
% change in price
∆q p
·
=
∆p q
p
=d· ,
q
εs =
and ar
elasti ity of supply over the interval
εs =
(p1 ; q1 )
to
(p2 ; q2 )
on the supply line, is
∆q 12 (p1 + p2 )
p1 + p2
· 1
.
=d·
∆p 2 (q1 + q2 )
q1 + q2
Example
Consider the supply fun tion
p = 10 + 0,5q,
The point elasti ity of supply at pri e
whi h we
p = R25
(where
εs = d ·
Sin e
|εs | > 1,
supply is elasti
an transform to
q = 30)
q = −20 + 2p.
is
25
p
=2·
= 1,67.
q
30
at a pri e of R25. If the pri e in reases by 1%, supply will in rease by 1,67%.
A tivity
demand fun tion
demanded at pri e p.
1. The
for a
ertain kind of
al ulator is
q = 250 − 5p,
with
p
only.
(a) Find the expression for point elasti ity of demand in terms of
(b) Cal ulate point elasti ity of demand at pri es
( ) Cal ulate ar
p = 20
and
p = 30,
q
the number of
and explain ea h of the results.
elasti ity of demand if the pri e in reases from R25 to R35.
2. The supply fun tion for a
ertain produ t is given by
p = 90 + 0,05q ,
with
p
quantity, respe tively.
(a) Find the formula for pri e elasti ity of supply in terms of
p.
(b) Determine pri e elasti ity of supply when the pri e is R70.
( ) Cal ulate ar
al ulators
elasti ity of supply when the pri e in reases from R40 to R60.
16
and
q
the pri e and
Chapter 3: Appli ations of simultaneous fun tions
Learning obje tives
After you have
⊲
⊲
hapter, you should be able to
determine the point where two (or more) equations or inequalities interse t
explain the
⊲
ompleted this
determine pri e and quantity at market equilibrium in the goods and labour markets
determine and interpret the ee t of intervention through pri e
determine and interpret the ee t of
determine the break-even point after setting up
explain the
⊲
explain the
omplementary and substitute goods
on ept of break-even
determine the break-even point, given total revenue (T R) and total
⊲
eilings, pri e oors, taxes and
subsidies
explain the
on ept of equilibrium
on epts of
determine
T R and T C
ost (T C ) fun tions
fun tions frome available information
onsumer and produ er surplus
onsumer, produ er and total surplus from given demand and supply fun tions
on ept of linear programming (LP)
set up the obje tive fun tion and
plot the
onstraints for an LP model from given information
onstraints, nd the feasible area and solve an LP model graphi ally
3.1 Equilibrium
E onomi
equilibrium is dened as a state in whi h e onomi
for es su h as supply and demand are balan ed.
When in equilibrium, the values of e onomi
variables will not
In a perfe tly
equilibrium is the point where the supply of and demand for a
ompetitive market, e onomi
hange if there are no external inuen es.
produ t are equal. This point is found on a graph where the lines representing supply and demand interse t.
3.1.1 Equilibrium in the goods market
Market equilibrium
o
1 for and supply2 of a produ t (or good) are in balan e, that is
urs when the demand
when the number of units that
(qs ). Equilibrium also o
onsumers demand (qd ) is equal to the number of units that produ ers supply
the pri e that the produ er is willing to a
At market equilibrium, the following
ept (ps ).
onditions hold:
qd = qs
1
2
onsumers are willing to pay (pd ) for a good is equal to
urs when the pri e that
and
pd = ps .
Consumer demand is the number of units of a produ t that onsumers buy in stores.
Produ er supply is the number of units that a produ er makes available to be sold in retail stores.
17
DSC1520/SG001
Example
The demand and supply fun tions for a
ertain produ t are given as
qd = 210 − 3,5pd
and
qs = −90 + 4ps
To nd the equilibrium pri e and quantity, we need to solve the demand and supply fun tions simultaneously.
Sin e the fun tions are both given in the form
q(p),
we set them as equal and solve for
p
(the equilibrium
pri e), that is
qd = qs
210 − 3,5p = −90 + 4p
−3,5p − 4p = −90 − 210
7,5p = 300
p = 40.
To nd the equilibrium quantity, we substitute
p = 40
into either one of the given fun tions, to nd
q = 210 − 3,5(40) = 70
or
q = −90 + 4(40) = 70.
Figure 3.1 shows how market equilibrium is determined graphi ally by plotting the fun tions on the same
axes and nding the point of interse tion, that is at point
E0 .
q
210
180
150
120
90
b
60
E0 = (40; 70)
30
0
−30
10 20 30 40 50 60 70
p
−60
−90
Figure 3.1: Equilibrium if
qd = 210 − 3,5pd
and
qs = −90 + 4ps
3.1.2 Equilibrium in the labour market
Labour market equilibrium o
supply (ls ). Also, when the
workers are willing to a
urs when the labour that rms demand (ld ) is equal to the labour that workers
wage (or salary) that a rm is willing to pay (ws ) is equal to the wage that
ept (wd ), the labour market is in equilibrium.
At labour market equilibrium, the following
onditions hold:
ld = ls
and
18
wd = ws .
DSC1520/SG001
Example
At a small bottling
ompany, the demand fun tion is given as
ld = 15 − 1,67wd ,
and the supply fun tion as
ls = −5 + 2,5ws ,
where labour is measured in number of workers and wage in hourly salary. To nd the
we set ld
= ls
ompany's equilibrium,
to nd
15 − 1,67w = −5 + 2,5w
−1,67w − 2,5w = −5 − 15
4,17w = 20
w = 4,8.
Substituting this into either the demand or supply fun tions, we nd
l = −5 + 2,5(4,8) = 7.
Therefore, at equilibrium the
ompany should employ seven workers at a wage of R4,80 per hour.
Graphi ally, we represent the situation as shown in Figure 3.2, where
E0
is the labour market equilibrium the demand and supply lines interse t.
l
15
12
9
b
6
E0 = (4,8; 7)
3
0
−3
1
2
3
Figure 3.2: Equilibrium if ld
4
5
6
7
= 15 − 1,67wd
8
9
and ls
w
= −5 + 2,5ws
3.1.3 Pri e ontrols and government intervention in markets
It often happens that markets fail to a hieve market equilibrium owing to
ertain fa tors, su h as the existen e
of rms with monopoly power or intervention by the government. In this se tion we look at
government intervenes through pri e
ases where the
ontrols.
Pri e eilings
When government believes that the equilibrium pri e is too high for
pri e
pri e.
3
onsumers to pay, they may establish a
eiling3 below market equilibrium. The pri e is then not allowed to ex eed this
Maximum pri e ontrol.
19
eiling (or maximum)
DSC1520/SG001
Example
Suppose the demand and supply fun tions for a good are given by
pd = 100 − 0,5qd
These fun tions
an be transformed to have
q
pd = ps
or
qd = qs ,
and
qs = −20 + 2ps .
we nd the equilibrium pri e and quantity to be
qe = 55
Suppose a pri e
ps = 10 + 0,5qs .
as the dependent variable, that is
qd = 200 − 2pd
By setting either
and
eiling of R40 is introdu ed (p
and
= 40).
pe = 90.
The number of units demanded and supplied at this
pri e, is then
qd = 200 − 2(40) = 120
and
qs = −20 + 2(40) = 60.
4 at this pri e and a shortage of
This means that demand is higher than supply
the market.
This shortage
120 − 60 = 60
units exists in
reates an opportunity for someone to buy the 60 units at the lower pri e of R40 and sell them
on the bla k market for more than the equilibrium pri e, sin e there is a demand for them.
The pri e that
onsumers are willing to pay if these 60 units are made available, is
pd = 100 − 0,5(60) = R70
and the bla k marketeer
an potentially make a prot of
P = TR − TC
= number sold (demanded) × price per unit − number bought × cost per unit
= 60 × 70 − 60 × 40
= R1 800.
Figure 3.3 illustrates this situation graphi ally, with the potential prot to be made by the bla k marketeer
represented by the shaded area.
Pri e oors
When government believes that the equilibrium pri e is too low for produ ers to re eive, a minimum pri e
( alled a
pri e oor)
an be set to prote t produ ers. Su h a pri e operates above market equilibrium.
Example
Consider a labour market with demand and supply fun tions
wd = 9 − 0,6ld
These fun tions
an be transformed to have
l
and
ws = 2 + 0,4ls .
as dependent variable:
ld = 15 − 1,67wd
and
ls = −5 + 2,5ws .
On page 19, we found the equilibrium wage and number of labourers to be seven labourers at an hourly wage
of R4,80.
4
Produ ers will only make 60 units available to be sold in retail stores at the lower pri e.
20
DSC1520/SG001
p
100
90
80
70
60
b
50
E0 = (90; 55)
Pri e eiling
40
30
20
10
20
Figure 3.3: Pri e
40
60
80 100 120 140 160 180 200 220
eiling of R40 for market with
pd = 100 − 0,5qd
and
q
ps = 10 + 0,5qs
If the government makes a law that wages may not be lower than R6,00 per hour, whi h is higher than the
market equilibrium, employers will have to employ fewer people, de reasing employer demand to
ld = 15 − 1,67(6) = 4,98 ≈ 5 labourers.
More employees will be willing to work for this higher wage, in reasing the supply of employees to
ls = −5 + 2,5(6) = 10 labourers.
This means that there will be a surplus of ve employees in the market (unemployment will rise by ve).
Figure 3.4 depi ts the situation:
w
9
8
7
Pri e oor
6
5
b
E0 = (7; 4,8)
4
3
2
1
1
2
3
4
5
6
Figure 3.4: Equilibrium if ld
7
8
9 10 11 12 13 14 15 l
= 15 − 1,67wd
and ls
= −5 + 2,5ws
A tivity
1. Lindiwe designs and manufa tures fashion jewellery. The demand and supply fun tions for rings are
pd = 800 − 2qd
and
21
ps = −40 + 8qs ,
DSC1520/SG001
respe tively.
(a) Transform the supply and demand fun tions to have quantity as dependent variable.
(b) Cal ulate the equilibrium pri e and quantity.
( ) Find the ex ess of rings supplied if the pri e per ring is set as R720.
(d) Find the ex ess demand if the pri e is lowered to R560.
2. The demand and supply fun tions for labour are given as
wd = 70 − 4ld
respe tively, where
w
and
ws = 10 + 2ls
is the hourly wage payable by the employer and
l
is the number of workers
employed.
(a) Write the given demand and supply fun tions with labour
l
as dependent variable.
(b) Determine the equilibrium number of workers and wage per hour.
( ) Find the ex ess demand for labour when
(d) Find the ex ess supply of labour when
w = 20.
w = 40.
3.1.4 Complementary and substitute goods
Sin e produ ts are not always traded in isolation, we need to know how
omplementary and substitute goods
ae t market equilibrium.
Complementary goods
are goods that are
and software, et . The one
Substitute goods
onsumed together, like
ars and petrol; and
omputer hardware
annot fun tion without the other.
are goods that are
onsumed instead of ea h other, like
oee or tea and a train or a bus on
ertain routes.
Consider two goods,
⊲
If
X
and
Y
are
X
and
Y.
omplementary goods, then the demand fun tion for good
X
is given by
qX = a − bpX − cpY ,
where
a, b and c are
onstant
The negative sign before
cpY
oe ients, while
pX
and
pY
are the pri es of goods
indi ates that demand for good
in reases. For example, when the pri e of
ars in reases,
X
X
and
Y , respe
tively.
de reases as the pri e of good
onsumers will buy fewer
ars and
Y
onsequently
the demand for petrol will de rease.
⊲
If
X
and
Y
are substitute goods, then the demand for
X
is given by
qX = a − pX + cpY .
Adding
cpY
means that as the pri e of
Y
in reases, more of
X
will be demanded. For example, if train
fares in rease, people will rather take the bus, in reasing the demand for bus transport.
Example
Consider a two-goods market with demand and supply fun tions for good
qdX = 82 − 3pX + pY
and
22
X (Y
being a substitute good) are
qsX = −5 + 15pX .
DSC1520/SG001
The demand and supply fun tions for good
Y (X
being the substitute good) are
qdY = 92 + 2pX − 4pY
and
qsY = −6 + 32pY .
Then, at equilibrium,
qdX = qsX
82 − 3pX + pY = −5 + 15pX
−3pX − 15pX + pY = −5 − 82
−18pX + pY = −87
(3.1)
and
pdX = psX
92 + 2pX − 4pY = −6 + 32pY
2pX − 4pY − 32pY = −6 − 92
2pX − 36pY = −98.
(3.2)
We now need to solve this system of simultaneous equations. From Equation 3.1 we nd that
pY = −87 + 18pX .
(3.3)
Subtituting this into Equation 3.2 gives
2pX − 36(−87 + 18pX ) = −98
2pX + 3132 − 648pX = −98
−646pX = −3 230
pX = 5.
Substituting ba k into Equation 3.3 gives
pY = −87 + 18(5) = 3.
To nd the equilibrium quantities, we simply substitute
pX = 5
and
pY = 3
into the given equations:
qdX = 82 − 3pX + pY = 82 − 3(5) + 3 = 70
and
qdY = 92 + 2pX − 4pY = 92 + 2(5) − 4(3) = 90.
The equilibrium pri es and quantities in this two-goods market are
pX = 5, qx = 70, pY = 3, qY = 90.
A tivity
1. The demand and supply fun tions for two
omplementary produ ts, pit hing wedges (X ) and putters
(Y ), are given as
qdX = 190 − 2pX − 2pY
and
qsX = −10 + 2pX
qdY = 240 − 2pX − 4pY
and
qsY = −40 + pY ,
and
where
pX
and
pY
are measured in hundreds of rands.
(a) Determine the equilibrium pri es for pit hing wedges and putters.
(b) Find the equilibrium quantities for the two produ ts.
23
DSC1520/SG001
3.1.5 Taxes and subsidies
Governments
an intervene in the market by imposing taxes and/or subsidies.
When a tax is put on a
produ t, we say it is an indire t tax, whi h may be either of the following:
⊲
a xed amount per unit of output, for example the tax imposed on petrol, al ohol and toba
⊲
a per entage of the pri e of a good, for example value added tax (VAT)
In this module, we only
onsider xed tax per unit. We assume that the
o
onsumer always pays the equilibrium
pri e, while the supplier re eives the equilibrium pri e minus the tax.
Fixed tax per unit
When a xed tax per unit is imposed on a produ t, the produ er will re eive the pri e
is
p − t.
The supply fun tion
p = c + dq
p
minus tax
t,
that
will then be ome
p − t = c + dq
or
p = c + dq + t.
Example
The demand and supply fun tions for a good are given as
pd = 100 − 0,5qd
and
ps = 10 + 0,5qs .
The equilibrium pri e and quantity are found by equating the demand and supply fun tions, that is
100 − 0,5q = 10 + 0,5q,
whi h results in
q = 90 and p = 55.
If government now imposes a xed tax of R6 per unit sold, the supplier re eives
supply fun tion be omes
ps − 6 = 10 + 0,5qs
or
ps − 6
per unit, and the
ps = 16 + 0,5qs .
Now, we nd equilibrium by equating the new taxed supply fun tion and the demand fun tion, that is
100 − 0,5q = 16 + 0,5q,
whi h results in
q = 84 and p = 58.
In Figure 3.5, the supply fun tions before and after tax as well as the equilibrium points before (E0 ) and
after tax (Et ) are shown.
As stated before, the
After tax,
onsumer always pays the equilibrium pri e.
onsumers pay R58 per unit, whi h is R3 more than the equilibrium before tax (R55).
On the other hand, the produ er re eives the new equilibrium pri e, minus the R6 tax, that is
R58−R6 = R52,
whi h is R3 less than the equilibrium pri e before tax.
In this
ase, the tax is distributed evenly between the produ er and
onsumer ea h pays 50% of the tax.
Subsidies
When a produ t is subsidised by Rs per unit, the produ er will re eive the equilibrium pri e
subsidy, that is
p + s.
In this
ase the supply fun tion be omes
p + s = c + dq
or
24
p = c + dq − s.
p,
plus the
DSC1520/SG001
p
100
ps = 16 + 0,5qs
90
ps = 10 + 0,5qs
80
70
ET = (84; 58)
60
b
b
50
E0 = (90; 55)
40
pd = 100 − 0,5qd
30
20
10
50
100
150
200
q
Figure 3.5: Equilibrium before and after tax
Example
The demand and supply fun tions for
rates of pears are given as
pd = 450 − 2qd
The equilibrium pri e and quantity are
and
ps = 100 + 5qs .
al ulated by equating the demand and supply fun tions, that is
450 − 2q = 100 + 5q,
whi h results in
q = 50
that is 50
rates at a pri e of R350 per
If a subsidy of R70 per
and
rate.
rate is provided, the supply fun tion be omes
ps + 70 = 100 + 5qs
Now, the equilibrium is
p = 350,
or
ps = 30 + 5qs .
al ulated as
450 − 2q = 30 + 5q
−7q = −420
q = 60, with
p = 30 + 5(60) = 330.
Figure 3.6 shows the supply fun tions before and after subsidy as well as the equilibrium points before (E0 )
and after subsidy (Es ).
As before, the
onsumer pays the equilibrium pri e, that is R330 per
than before. The
onsumer therefore re eives
20
70
≈ 29%
rate after subsidy. That is R20 less
of the subsidy.
The produ er, on the other hand, re eives the equilibrium pri e plus the subsidy, that is
whi h is R50 more than before the subsidy. The produ er therefore re eives
25
50
70
≈ 71%
330 + 70 = 400,
of the subsidy.
DSC1520/SG001
p
ps = 100 + 5qs
500
Produ er pri e400
Consumer pri e
ps = 30 + 5qs
E0
b
ES
b
300
200
pd = 450 − 2qd
100
50
100
150
200
q
250
Figure 3.6: Equilibrium before and after subsidy
A tivity
The demand and supply fun tions for golf lessons at a
pd = 200 − 5qd
ertain golf
and
lub are given as
ps = 92 + 4qs .
1. Find the equilibrium pri e and quantity algebrai ally. Show this graphi ally.
2. Government imposes a tax of R9 per lesson.
(a) Write down the equation of the supply fun tion adjusted for tax and then graph it on the same
graph as in Question 1.
(b) Cal ulate the equilibrium pri e and quantity with tax taken into a
( ) How is the tax distributed between the
ount.
ustomer and the supplier (the
lub)?
3.2 Break-even analysis
In business, it is usually very important to know how many of a produ t should be sold to break even, that
is when the in ome from selling the produ t equals the
the
osts have been
overed and the business
ost to produ e or buy it. This is the point where
an start making prot.
At break-even the total revenue re eived (T R) is equal to the
osts asso iated with the pro esses needed to
generate the revenue (T C ). This means that at break-even
TR = TC
with
FC
xed
ost(s) and
VC
variable
or
T R = F C + V C,
ost.
Example
For a
ertain good, total revenue and total
T R = 3q
ost fun tions are given as
and
26
T C = 10 + 2q,
DSC1520/SG001
where
q
is the number of units produ ed and sold.
The break-even point is found by setting
TR = TC
3q = 10 + 2q
q = 10.
This means that when ten units of the good is sold, total revenue is equal to total
T R = 3 × 10 = 30
and
Figure 3.7 shows the
osts.
At this point,
T C = 10 + 2 × 10 = 30.
TR
and
TC
fun tions, and the break-even point at
TR
TC
T R = 3q
30
b
q = 10.
T C = 10 + 2q
Break-even point
20
10
1
2
3
4
5
6
7
8
9 10 11 12 13 14
q
Figure 3.7: Break-even point
A tivity
1. A rm sells their produ t for R30 per item. Their xed
osts amounts to R200 and the variable
ost
is R5 per unit produ ed.
(a) How many units should the rm produ e and sell to break even?
(b) Cal ulate the value of total revenue and total
2. A rm manufa tures
ost at break-even.
hildren's wat hes that are sold for R6,60 ea h. The rm's total
manufa turing these wat hes is
ost fun tion for
T C = 800 + 0,2q .
(a) Write down the total revenue fun tion.
(b) How many wat hes must be manufa tured and sold to break even?
( ) Suppose that at the rm's break-even point, they sell 160 wat hes. What is the pri e per wat h
in this situation?
3.3 Consumer and produ er surplus
3.3.1 Consumer surplus
Consumer surplus
CS
is the dieren e between what
onsumers are willing and able to spend on a produ t
and what they a tually spend (at market pri e).
The market pri e of a produ t is often lower than the pri e that
what we have learnt thus far, we know the following:
27
onsumers are willing to pay. In terms of
DSC1520/SG001
⊲
Market pri e is given by the pri e at equilibrium.
⊲
The pri e that
onsumers are willing to pay, is given by the demand fun tion.
For example, the demand and supply fun tions for a
p = 100 − 0,5q
ertain produ t are given by
and
p = 10 + 0,5q.
At equilibrium, we nd the market pri e and quantity of the produ t by equating the demand and supply
fun tions, that is
100 − 0,5q = 10 + 0,5q,
giving
q = 90,
and p = 100 − 0,5(90) = 55.
In Figure 3.8(a), the demand fun tion is graphed with the equilibrium point indi ated at
E0 = (p0 ; q0 ) = (90; 55).
The shaded area under the demand line, between
q=0
and
q = 90,
represents the amount that
onsumers
are willing to spend on the produ t.
At the market pri e of R55 per unit, 90 units of the produ t are sold. The amount that
spend on the produ t is
p0 × q0 = 55 × 90 =
onsumers a tually
R4 950. This amount is represented by the shaded area of the
re tangle in Figure 3.8(b).
As stated earlier, the
onsumer surplus is the amount that
onsumers are willing to spend over and above
expenditure at market pri e. This is represented by the area under the demand line from
q = 0 to q = 90 and
above the line representing market pri e. This is the shaded area in Figure 3.8( ) and is known as
onsumer
surplus.
p
p
A
100
p0 = 55
p
A
100
E0 (q0 ; p0 )
q0 = 90
200
100
E0
55
q
200
q
(b) Amount a tually spent
Figure 3.8: Cal ulating
Consumer surplus for our problem is
E0
55
90
(a) Amount willing to spend
A
90
200
q
( ) Consumer surplus
onsumer surplus
al ulated as follows:
CS = amount consumers are willing to spend − amount actually spent
= area under demand line from 0 to q0 − (p0 × q0 )
= area of triangle p0 E0 A
= 0,5 × 90 × (100 − 55)
= R2 025
(area of triangle = 1/2 × base × height)
3.3.2 Produ er surplus
Now
onsider the situation from produ ers' point of view.
minimum pri e at whi h produ ers are willing to produ e.
28
The market pri e might be higher than the
DSC1520/SG001
Again
onsider the demand and supply fun tions, whi h are
p = 100 − 0,5q
For the supply fun tion
p0 =
p = 10 + 0,5Q,
and
the produ er sells
R55 per unit. The resulting revenue is
Figure 3.9(a).
90 × 55 =
However, the supply line represents pri es that are a
ure 3.9(b) therefore represents the revenue that is a
supply line, between
q=0
and
p = 10 + 0,5q.
q0 = 90
units of the produ t at market pri e of
R4 950, whi h is represented by the shaded area in
eptable to the produ er.
The shaded area in Fig-
eptable to the produ er. This is the area under the
q = 90.
Produ er surplus is therefore given by the revenue at market pri e, minus the revenue that the produ er
would be willing to a
ept. This is shown as the shaded area in Figure 3.9( ).
p
p
100
p
100
p0 = 55
E0
100
10
10
q0 = 90
200
B = 10
q
90
(a) Revenue at market pri e
E0
p0 = 55
E0
55
200
q
q0 = 90
(b) A eptable revenue
200
q
( ) Produ er surplus
Figure 3.9: Cal ulating produ er surplus
Produ er surplus for our supply fun tion is
al ulated as follows:
P S = p0 × q0 − area under supply line to the left of q0
= area of triangle p0 E0 B
= 0,5 × 90 × (55 − 10)
(1/2 × base × height.)
= R2 025
3.3.3 Total surplus
The total surplus at market equilibrium is simply the sum of
onsumer surplus and produ er surplus at
market pri e.
Example
The demand and supply fun tions of shirts are given as
pd = 60 − 0,6qd
and
ps = 20 + 0,2qs .
Again, we nd the equilibrium pri e and quantity by setting demand equal to supply, that is
pd = ps
60 − 0,6q = 20 + 0,2q
−0,6q − 0,2q = 20 − 60
−40
= 50
q=
−0,8
29
DSC1520/SG001
and
p = 20 + 0,2(50) = 30.
Figure 3.10 shows the demand and supply fun tions, with the equilibrium point at
E0 .
p
A
60
50
40
CS
E0
p0 = 30
PS
20
B
10
25
q0 = 50
Figure 3.10: Market equilibrium;
Consumer surplus is represented by the triangle
75
100
q
onsumer surplus and produ er surplus
AE0 p0 .
This area is
al ulated as
CS = 0,5 × base × height = 0,5 × 50 × (60 − 30) = 750.
Produ er surplus is represented by the triangle
BE0 p0
and the area is
al ulated as
P S = 0,5 × base × height = 0,5 × 50 × 10× = 250.
Total surplus at market equilibrium is therefore
CS + P S = 750 + 250 = 1 000.
A tivity
1. The demand and supply fun tions for seats on a tour bus are given by
pd = 58 − 0,2qd
and
ps = 4 + 0,1qs .
(a) Find the equilibrium pri e and quantity. Plot the demand and supply fun tions, and show the
areas representing
onsumer surplus and produ er surplus at equilibrium.
(b) Cal ulate the following:
i. the amount
ii. the amount
iii.
onsumers pay (at equilibrium)
onsumers are willing to pay
onsumer surplus (CS )
( ) Cal ulate the following:
i. the amount the bus
ompany (the produ er) re eives for seats on the bus
ii. the amount that the bus
ompany is willing to a
iii. produ er surplus (P S )
30
ept
DSC1520/SG001
3.4 Linear programming
Linear programming (LP) is a method where a problem is modelled by representing the
exist in terms of the variables as linear fun tions. The
prot or minimising
ost, also
obje tive fun tion
onstraints
that
of the model, su h as maximising
onsists of a linear relationship between the variables of the problem.
When su h a model is solved, values for the variables are found that optimise the obje tive fun tion.
The important aspe ts of an LP model are the
are
onstraints and the obje tive fun tion. Also, the way variables
hosen to represent elements of the problem have an impa t on the su
variables should be
arefully
ess of su h a model, therefore
onsidered.
When given a real-life problem statement and one needs to formulate it as an LP model, the following steps
should be followed:
1. Dene the de ision variables. The following steps are all performed a
ording to these variables.
2. Analyse the given information and put it in some stru ture like a table.
3. Write down the obje tive fun tion, that is the fun tion that needs to be optimised.
4. Set up the
onstraints of the problem.
Now let us illustrate these steps based on the following problem statement:
A manufa turer of leather arti les produ es boots and ja kets. The manufa turing pro ess
sists of two a tivities, namely
making
( utting and stit hing) and
on-
nishing.
There are 800 labour hours available per month for making the arti les and 1 200 hours for
nishing them. It takes four hours to make a pair of boots and three hours to nish them. It
takes two hours to make a ja ket and four hours to nish it.
They sell a pair of boots for R1 200 and a ja ket for R900.
To formulate an LP model for this problem, we follow the steps above with the obje tive to maximise monthly
revenue.
1.
De ision variables
When we read this problem statement, we realise that the manufa turer needs to determine how many
pairs of boots and how many ja kets they should manufa ture to maximise revenue.
variables
2.
are therefore
The
de ision
hosen to represent this obje tive, namely letting
⊲ x
be the number of pairs of boots to manufa ture; and
⊲ y
be the number of ja kets to manufa ture.
Stru ture the information
To give stru ture to the given information, we set up the following table:
Ja kets
Hours
(x)
(y )
available
Making
4
2
800
Finishing
3
4
1 200
1 200
900
Pri e
3.
Boots
Obje tive fun tion
The revenue from selling boots is
Rx = 1 200x
and the revenue from selling ja kets is
31
Ry = 900y ,
with
DSC1520/SG001
total revenue
T R = Rx + Ry = 1 200x + 900y .
The obje tive fun tion is therefore
Maximise T R = 1 200x + 900y.
4.
Constraints
The
onstraints of the problem are represented by the following linear inequalities:
4x + 2y ≤
800
x, y ≥
0
(only 800 hours available for making)
3x + 4y ≤ 1 200
(only 1 200 hours available for finishing)
(impossible to produce a negative number of items)
The LP model for our problem is therefore the following:
Maximise T R = 1 200x + 900y
subject to
4x + 2y ≤
800 (Making)
3x + 4y ≤ 1 200 (Finishing)
x, y ≥
0
(Non-negativity)
To solve a model like this, we need to determine how many of ea h of the produ ts must be manufa tured
and sold to maximise total revenue. We use the graphi al method, whi h follows the following steps:
1. Plot the
onstraint inequalities on the same axes.
Treat them as equations and then determine on
whi h side of the line ea h inequality holds.
2. Find the feasible area. This is where the inequalities overlap and all the
onstraints are feasible.
3. Determine where the obje tive fun tion is optimised.
To plot the
we
onstraints in our LP model, we rst treat them as equations. For example, to graph
onsider the equation
4x+2y ≤ 800,
4x + 2y = 800.
x = 0 to nd the y inter ept
x = 200 (from 4x + 2(0) = 800).
Set
Draw the line through the
as
y = 400
(from
4(0) + 2y = 800)
and
y=0
to nd the
x
inter ept as
oordinates (0; 400) and (200; 0).
Determine whi h side of the line represents the inequality by using the origin (0; 0) and see whether it
satises the inequality. [Substitute
x=0
and
y=0
into the inequality to nd
4(0) + 2(0) = 0 < 800,
is true.℄ The origin therefore falls inside the feasible area of this inequality and we
whi h
an shade this area, as in
Figure 3.11(a).
The same pro edure is followed to draw the fun tion
3x + 4y = 1 200
and nd the feasible area as shown in
Figure 3.11(b).
The non-negative
onstraints
positive.
x≥0
and
y≥0
restri t us to the rst quadrant, where
x
and
y
are always
In Figure 3.11( ), all the inequalities are drawn on the same axes. The area where the feasible areas overlap
(the
he kered area) is the feasible area of the model. In this area all the
onstraints are satised.
Now, to nd the point in this feasible area where the total revenue fun tion
maximum, we
we
an either
al ulate the total revenue at ea h
orner point and
T R = 1 200x + 900y
an draw the isorevenue lines to nd the optimum.
Total revenue at the
orner points as shown in Figure 3.12(a) is as shown in Table 3.1.
32
is a
hoose the highest number, or
DSC1520/SG001
y
y
y
400
400
400
300
300
300
200
200
200
100
100
100
100 200 300 400
x
x
100 200 300 400
(a) 4x + 2y ≤ 800
100 200 300 400
(b) 3x + 4y ≤ 1 200
x
( ) Feasible area
Figure 3.11: Finding the feasible area
Point
Coordinates
Total Revenue
1 200x + 900y
A
(0, 300)
270 000
B
(80; 240)
312 000
C
(200; 0)
240 000
Table 3.1: Total revenue at
The isolines with slope
− 34
(from
T R = 1 200x + 900y )
←
Maximum
orner points
are shown in Figure 3.12(b) where it is
lear
that revenue is a maximum at the point (80; 240). (Find this point by solving the simultaneous equations
4x + 2y = 800
and
3x + 4y = 1 200.)
y
y
400
300
400
A(0; 300)
300
B(80; 240)
B(80; 240)
200
200
100
100
C(200; 0)
100
200
300
400
x
100
(a) Feasible area and orner points
200
300
400
x
(b) Feasible area and isorevenue lines
Figure 3.12: Maximise total revenue
The solution to the LP model is to manufa ture (and sell) 200 pairs of boots and 300 ja kets for maximum
revenue of R312 000.
Example
Consider the following problem statement:
33
DSC1520/SG001
A daily diet requires a minimum of 600 mg of vitamin C, 360 mg of vitamin D and 40 mg of vitamin E. Two
food mixes,
X
and
Y,
ontain these vitamins per portion as given in the following table:
Per portion of
Per portion of
Vit D
Vit E
Cost per mg
(mg)
(mg)
(mg)
(R)
20
10
4
5
30
20
1
4
600
360
40
X
Y
Minimum daily requirement
Find the number of portions of food mixes
minimum
Vit C
X
and
Y
that will satisfy the minimum dietary requirements at
ost.
To model this problem as an LP, we follow the steps spe ied before.
1.
De ision variables
x
Let
2.
be the number of portions of food mix
X
and
y
the number of portions of food mix
Y.
Obje tive fun tion
The obje tive is to minimise
in lude
X osts R5. So if we
Y osts 4y . Therefore,
ost. From the table, we see that a portion of mix
x portions per day, the
ost is
5x.
Likewise, to in lude
y
portions of mix
the total obje tive fun tion is
Minimise cost = 5x + 4y.
3.
Constraints
The
onstraints on the problem are the minimum daily requirements of ea h vitamin. The requirements
are not exa t one may take in more than the minimum therefore they are
A
ording to the table, food mix
and food mix
Y
X
inequality
onstraints.
ontains 20 mg, 10 mg and 4 mg of vitamins C, D and E, respe tively,
ontains 30 mg, 20 mg and 1 mg of vitamins C, D and E, respe tively.
It is also stated that the diet should in lude at least 600 mg of vitamin C, 360 mg of vitamin D and
40 mg of vitamin E.
Furthermore, the number of portions
annot be negative, so
x
and
y
must be greater than or equal to
zero.
The
onstraints are therefore
20x + 30y ≥ 600
(vit C)
10x + 20y ≥ 360
(vit D)
4x + y ≥ 40
(vit E)
x, y ≥ 0
4.
(non-negativity)
Plot the onstraints and nd the feasible area
To graph ea h
onstraint, the easiest is to determine the inter epts on the
x
and
y
axes. This is done
in the following table:
Equation
Vit C
Vit D
Vit E
20x + 30y = 600
10x + 20y = 360
4x + y = 40
x ≥ 0, y ≥ 0
x axis inter ept (y = 0)
20x + 0 = 600 → x = 30
10x + 0 = 360 → x = 36
4x + 0 = 40 → x = 10
First quadrant
34
y axis inter ept (x = 0)
0 + 30y = 600 → y = 20
0 + 20y = 360 → y = 18
0 + y = 40 → y = 40
DSC1520/SG001
Now, plot the lines and determine on whi h side of the line ea h inequality holds. (See Se tion C.4.)
Figure 3.13 shows the
onstraints and the area where the
onstraints overlap the feasible area.
x
40 P(0, 40)
Feasible area
20
18
Q(6, 16)
R(12, 12)
S(36, 0)
30
10
Vit E
36
y
Vit C Vit D
Figure 3.13: Inequalities and feasible area
5.
Optimise the obje tive fun tion
There are two ways of determining the optimal solution, namely by
value at ea h of the
whi h
al ulating the obje tive fun tion
orner points and sele ting the lowest, or by using isolines to determine graphi ally
orner point represents minimum
ost (the optimal solution).
35
DSC1520/SG001
⊲
The values of the obje tive fun tion at the
It is therefore
Point (x, y )
Cost
= 5x + 4y
P(0, 40)
Cost
5(0) + 4(40) = 160
5(6) + 4(16) = 94
5(12) + 4(12) = 108
5(36) + 4(0) = 180
Q(6, 16)
Cost
R(12, 12)
Cost
S(36, 0)
Cost
al ulated as follows:
← Minimum
lear that the optimum solution is at point Q where six portions of food mix
16 portions of food mix
at the minimum
⊲
orner points are
Y
are
onsumed. The minimum daily vitamin requirements are satised
We start by assuming that the diet
all have the same slope. The
ost fun tion at dierent values of
The line has a slope of
− 45
osts R200 per day to get the slope of the iso ost lines, whi h
5
y = − x + 50.
4
or
and is depi ted as a thi k, dashed line on the graph in Figure 3.14.
The lines parallel to this line that pass through the
orner points of the feasible area are shown as
thinner dashed lines. As the lines move lower in the graph, the
ost
ost de reases and is at it lowest
= R94.
This is the optimum point where six portions of
X
and 16 portions of
ost of R94.
x
40 P(0, 40)
Feasible area
20
18
ost.
ost fun tion at this value is
5x + 4y = 200
at the minimum
and
ost of R94.
Iso ost lines are lines representing the
at point Q, where
X
Q(6, 16)
R(12, 12)
S(36, 0)
30
10
Vit E
36
Vit C Vit D
Figure 3.14: Iso ost lines
36
y
Y
are
onsumed per day
DSC1520/SG001
A tivity
1. Consider the following linear inequality
onstraints:
5x + 4y ≤ 20
(1)
15x + 8y ≤ 48
(2)
7x + 3y ≤ 21
(3)
x, y ≥ 0
(a) Graph the inequality
(b) Find the
onstraints and shade the feasible area.
oordinates of the
orner points of the feasible area.
( ) If produ tion is measured in thousands of units of
P = 3x + 2y ,
2. A
ompany
x and y and the obje
tive fun tion is to maximise
nd the optimal solution.
an use two types of ma hine (A and B) in a manufa turing plant. The number of operators
required and the running
ost per day are given in the following table.
Cost per Available Floor area Prot per
ma hine
day
operators
(m2 )
Ma hine A
6
2
2
20
Ma hine B
3
4
2
30
360
280
160
Maximum available
Write down the prot fun tion and inequality
onstraints.
3. A gardener requires fertiliser with a minimum of 80 units of nitrogen, 15 units of potassium and 10
units of iron. Two fertiliser mixes are available, namely Brand X and Brand Y. Brand X
ontains 20
units of nitrogen, two units of potassium and one unit of iron per kilogram, while Brand Y
four units of nitrogen, one unit of potassium and two units of iron per kilogram. Brand X
per kilogram, while Brand Y
osts R180
osts R60 per kilogram.
(a) Dene the de ision variables for this problem.
(b) Write down the obje tive fun tion and inequality
( ) Solve this LP problem graphi ally.
37
ontains
onstraints for the gardener's problem.
PART II
NON-LINEAR FUNCTIONS
38
Chapter 4: Appli ations of non-linear fun tions
Learning obje tives
After you have
⊲
⊲
⊲
⊲
ompleted this
plot quadrati ,
set up quadrati
hapter, you should be able to
ubi , exponential, logarithmi
revenue,
optimise quadrati
and hyperboli
fun tions
ost and prot fun tions from given information
revenue,
ost and prot fun tions
determine market equilibrium for quadrati
determine break-even points for quadrati
demand and/or supply fun tions
(or
ubi )
explain, manipulate and interpret exponential and logisti
manipulate and interpret hyperboli
T R, T C
and/or
P
fun tions
growth fun tions
fun tions
The theory of non-linear fun tions dis ussed in this
hapter is
onsidered to be knowledge that you have
gained at s hool or elsewhere. If you feel un ertain about non-linear fun tions, please
onsult Appendix D.
4.1 Non-linear fun tions in e onomi s
Linear fun tions gave us a good introdu tion to modelling business
revenue,
on epts su h as demand and supply,
ost. However, these models are very limited in representing real-life situations.
For example, the linear total revenue fun tion
T R = 36q
rm whose demand fun tion takes only pri e into a
represents the revenue of a perfe tly
ount, so that demand is simply given by
For a monopolist, however, the demand fun tion is usually given by the fun tion
revenue is given by
T R = pq = (a − bq)q = aq −
bq 2 . This is a quadrati
p = a − bq ,
ompetitive
p = 36.
so that total
fun tion. If we take the demand
fun tion for a monopolist to be
p = 50 − 2q,
then the total revenue fun tion is
T R = pq = (50 − 2q)q = 50q − 2q 2 .
This is a quadrati
fun tion with its graph as shown in Figure 4.1.
Another example that illustrates the limitations of linear models is the linear
This
ost fun tion assumes that
ost fun tion
T C = 10 + 2q .
osts rise by the same amount (two units) for ea h extra unit of output
produ ed, even after 200 units have been produ ed, whi h is not realisti . What is more realisti , is that
after the initial setup, the
ost of produ ing an extra unit will start de reasing. Total
osts should rise at a
de reasing rate, until expansion might be required and further outlay might in rease the rate again.
Su h a situation is best des ribed by a
ubi
total
ost fun tion
T C = aq 3 − bq 2 + cq + d,
with
a, b, c
and
d
onstants, and
q
the number of units produ ed. The graph in Figure 4.2 illustrates su h
a situation:
39
DSC1520/SG001
TR
300
200
100
5
10
Figure 4.1:
15
20
25
q
T R = 50q − 2q 2
TC
In reasing rate
De reasing rate
q
T C = aq 3 − bq 2 + cq + d
Figure 4.2:
4.2 Quadrati fun tions
You need to know what the properties of quadrati
fun tions are and you must be able to plot su h fun tions.
The following is a summary of the information available in Appendix D.1:
⊲
The general form of a quadrati
fun tion is
f (x) = ax2 + bx + c,
where
a, b
and
c
are
onstants.
⊲
A quadrati
⊲
The roots of a quadrati
fun tion has a minimum (or a maximum) turning point if
equation
ax2 + bx + c = 0
nd the roots, we use the formula (often
alled the
x=
When the quadrati
⊲
equation
fun tion
are found where the graph
b
( )
a < 0).
rosses the
x
axis. To
minus formula)
√
b2 − 4ac
.
2a
2
− 4ac)
is
alled the dis riminant. The nature of the roots of
an be derived from the value of the dis riminant, namely:
b2 − 4ac > 0, the graph has two dierent real roots.
2
If b − 4ac = 0, the graph tou hes the x axis at a single
2
If b − 4ac < 0, the graph does not tou h the x axis.
(a) If
(b)
(or
an be fa torised, nd the roots by equating ea h fa tor to zero.
The expression under the square root (b
a quadrati
−b ±
a>0
40
point.
DSC1520/SG001
⊲
The turning point of the graph, that is either the maximum or minimum point, is at
x=
−b
.
2a
The graph is symmetri al about the verti al line through this point, whi h is
When the roots, the turning point and the
y
axis inter ept have been found, we
alled the vertex.
an plot a quadrati
fun tion.
4.2.1 Optimising quadrati revenue, prot and ost fun tions
In any business, the major obje tive is to maximise prot, where
profit P = T R − T C.
Equivalent to this, rms aim to maximise total revenue (T R), while minimising total
Let's work through a few examples to illustrate these
osts (T C ).
on epts with non-linear fun tions.
Examples
1. Consider the demand fun tion for a monopolist given by the fun tion
p = 100 − 2q,
where
p
is the pri e per unit of a good and
q
the quantity produ ed and sold.
From the demand fun tion, we nd the total revenue fun tion to be
T R = price per unit × number of units sold
=p×q
= (100 − 2q) × q
= 100q − 2q 2 .
To maximise total revenue, we need to nd the turning point of the parabola representing the
fun tion. From the graph of the
rea hed at
TR
fun tion in Figure 4.3, we
an estimate the maximum
TR
TR
to be
q = 25.
TR
1200
1000
800
600
400
200
10
Figure 4.3:
Comparing the
c = 0.
TR
20
30
40
TR
q
T R = 100q − 2q 2
fun tion with the standard quadrati
From this we nd that
50
fun tion, we nd that
is a maximum at its turning point at
q=
a = −2, b = 100
and
−100
−100
−b
=
=
= 25,
2a
2(−2)
−4
where
T R = 100(25) − 2(25)2 = 2 500 − 1 250 = 1 250.
This means that when 25 units are produ ed and sold, a maximum total revenue of R1 250 is made.
41
DSC1520/SG001
2. On the other hand,
onsider the total
ost fun tion for a
ompany given by
T C = 800 − 120q + 5q 2 ,
where
q
is the number of units produ ed. This fun tion is graphed in Figure 4.4.
TC
800
700
600
500
400
300
200
100
5
Figure 4.4:
Again, we estimate from the graph that
b = −120
and
c = 800,
10
15
20
25
q
T C = 800 − 120q + 5q 2
TC
is a minimum at
q = 12.
Algebrai ally, with
a = 5,
the minimum is found at the turning point
q=
−b
−(−120)
120
=
=
= 12.
2a
2(5)
10
At this point,
T C = 800 − 120(12) + 5(12)2 = 80,
whi h means that when 12 units are produ ed, the
3. A rm's total
q
ost of R80.
ost fun tion and demand fun tions are given by
T C = 200 + 3q
where
ompany has a minimum total
and
pd = 107 − 2q,
is the number of units of a produ t produ ed and sold.
From the demand fun tion, we nd the total revenue fun tion to be
T R = pq = (107 − 2q)q = 107q − 2q 2 .
The prot fun tion for the rm is therefore
P = TR − TC
= 107q − 2q 2 − (200 + 3q)
= −2q 2 + 107q − 3q − 200
= −2q 2 + 104q − 200.
The prot fun tion is shown in the graph from Maxima in Figure 4.5.
To maximise prot, we need to nd the turning point of
quadrati
fun tion, we nd
a = −2, b = 104
q=
and
c = −200.
P.
By
omparing
P
with the standard
The turning point is at
−b
−104
=
= 26,
2a
2(−2)
giving a maximum prot of
P = −2(26)2 + 104(26) − 200 = 1 152.
When the rm produ es and sells 26 units of the produ t, they will make a maximum prot of R1 152.
42
DSC1520/SG001
Figure 4.5:
P = −2q 2 + 104q − 200
4. You will often en ounter problems in assignments and/or examinations where the fun tions are not
readily provided as in the above examples. Consider, for example, the following problem statement:
The ACE
produ tion
ompany de ides to enter the market with a new mi rowave oven, the ACE2015.
osts entail xed
osts of R120 000 per month and a unit
The
ost of R400 per oven produ ed.
A market survey established that at a wholesale pri e of R600 per unit, demand will be 1 000 units per
month, but if the pri e is in reased to R1 000 per unit, there will be no demand.
Assuming that the demand fun tion is linear, they
on lude that demand is given by
q = 2 500 − 2,5p,
where
p
is pri e in rand and
q
is the number of units demanded.
We need to nd the wholesale pri e and the number of units to be produ ed for maximum prot.
To solve this problem, we need to nd the prot fun tion that needs to be maximised. We know that
prot
P = T R − T C,
so we rst need to nd the total revenue and
ost fun tions.
We know that total revenue is given by the number of units produ ed and sold times the pri e, that is
T R = pq = p(2 500 − 2,5p) = 2 500p − 2,5p2 .
Total
ost is given by xed
osts plus variable
ost, where xed
osts are given as R120 000 and variable
ost as R400 per unit. Therefore,
T C = 120 000 + 400q,
with
q
the given demand fun tion,
q = 2 500 − 2,5p.
Therefore,
T C = 120 000 + 400(2 500 − 2,5p) = 1 120 000 − 1 000p.
The prot fun tion is therefore
P = TR − TC
= 2 500p − 2,5p2 − (1 120 000 − 1 000p)
= −2,5p2 + 3 500p − 1 120 000.
This is a quadrati
fun tion with
a = −2,5, b = 3 500
and
c = 1 120 000.
The vertex of the prot fun tion is at
p=
−b
−3 500
=
= 700,
2a
2 × −2,5
43
DSC1520/SG001
P = −2,5p2 + 3 500p − 1 120 000
Figure 4.6: Prot fun tion
whi h means that prot is a maximum when
p=
R700. This is
(Maxima)
onrmed by the graph of the prot
fun tion in Figure 4.6.
The maximum prot at a pri e of R700 is
Pmax = −2,5(700)2 + 3 500(700) − 1 120 000 = R105 000.
4.2.2 Market equilibrium
In Se tion 3.1 linear demand and supply fun tions were
demand and supply fun tions and see how we
Consider, for example, the quadrati
onsidered. In this se tion, we look at non-linear
an nd market equilibrium.
demand and supply fun tions
pd = q 2 − 10q + 25
ps = q 2 + 6q + 9.
and
We need the properties as shown in the following table to plot these fun tions:
Turning point
pd = q 2 − 10q + 25
ps = q 2 + 6q + 9
a > 0 → minimum
a > 0 → minimum
q=
Discriminant
Vertical axis intercept
−b
2a
=
−(−10)
2
=5
q=
−b
2a
=
−6
2
= −3
b2 − 4ac = (−10)2 − 4(1)25 = 0 b2 − 4ac = 62 − 4(1)(9) = 0
touches axis
touches axis
p = 25
p=9
The demand and supply fun tions are plotted on the same axes in Figure 4.7, showing only the rst quadrant.
Market equilibrium is found where the graphs interse t.
supply fun tions equal to ea h other and solving for
q,
This point is found by setting the demand and
to nd
pd = ps
q 2 − 10q + 25 = q 2 + 6q + 9
−10q − 6q = 9 − 25
−16q = −16
q = 1.
At
q = 1,
both the demand fun tion and the supply fun tion give
44
p = 16.
DSC1520/SG001
p
60
50
40
ps = x2 + 6x + 9
30
20
10
pd = x2 − 10x + 25
1
2
Figure 4.7: Quadrati
3
4
5
q
demand and supply fun tions
4.2.3 Break-even analysis
Initially, it
osts the
ompany money to set up for the produ tion of a produ t the
ompany has to pay
for ma hines, spa e, employees, et . It is only after a number of units have been sold, that these
be
overed from the revenue. Thereafter, when more units of the produ t are sold, the
to make a prot. The point where total revenue and total
At break-even, total revenue (T R) is equal to total
ost are equal is
osts will
ompany will start
alled the break-even point.
osts (T C ), that is
T R = T C = F C + V C,
where
FC
is xed
osts and
VC
variable
osts.
Example
A
ompany produ es a produ t at a xed
for the produ t is given as
per unit.
q = 65 − 5p,
ost of R30 and variable
where
From the given information, we nd the total
q
ost of R2 per unit. The demand fun tion
is the number of units produ ed and sold, and
p
is the pri e
ost fun tion to be
T C = F C + V C = 30 + 2q.
We also need to nd the total revenue fun tion in terms of
terms of
p,
q.
The demand fun tion is, however, given in
so we transform it to nd
q = 65 − 5p
5p = 65 − q
p = 13 − 0,2q.
We now nd that
T R = price per unit × number of units produced and sold
= pq
= (13 − 0,2q)q
= 13q − 0,2q 2 .
Break-even o
urs where
T R = T C,
that is where
13q − 0,2q 2 = 30 + 2q
11q − 0,2q 2 − 30 = 0
−0,2q 2 + 11q − 30 = 0.
45
DSC1520/SG001
We solve this quadrati
Here,
a = −0,2, b = 11
fun tion by using the formula
c = −30,
and
q=
TR
and
TC
√
b2 − 4ac
2a
as given on page 40.
so we nd
−(11) ±
p
(11)2 − 4(−0,2)(−30)
2(−0,2)
√
−11 ± 97
=
−0,4
√
−11 + 97
= 2,9
=
−0,4
By plotting the
x=
−b ±
√
−11 − 97
q=
= 52,1.
−0,4
or
fun tions on the same axes, we nd the break-even points where these two
graphs interse t. This is demonstrated in Figure 4.8.
T R/T C
T R = −0,2q 2 + 13q
200
150
100
T C = 2q + 30
50
10
20
30
40
50
Figure 4.8: Break-even at interse tion of
TR
60
and
TC
q
fun tions
A tivity
1. Consider again the ACE
produ tion line.
to R300.
ompany's problem above.
ACE
If they do so, the saving on labour
However, the
onsiders in orporating two robots into the
ost
an redu e the unit
apital expenditure in urred will raise their xed
ost of the ACE2015
osts to R200 000.
How will
in orporating the robots ae t the optimal pri e and protability, and what would you advise ACE to
do?
2. Clan y's Chariots, a
to R150 per
ar-rental rm, has a eet of 100 identi al vehi les. The xed daily
ar, while ea h
ar used in urs an additional
the rent is set at R200 per day, all
osts amount
ost of R50 per day. Experien e shows that if
ars are rented out, whereas for ea h in rease of R20 the number of
ars rented out drops by one. Determine the optimal pri e, the number of
ars rented out per day at that
pri e and the maximum prot.
3. Find the e onomi ally meaningful equilibrium pri e and quantity if the demand and supply fun tions are
given as
p = q 2 − 0,5
p = −q 2 + 4.
and
4. The demand and supply fun tions for a produ t are given by
pd = −(q + 4)2 + 100
and
ps = (q + 2)2 .
(a) Plot both fun tions on the same axes and estimate the equilibrium pri e and quantity.
(b) Find the equilibrium pri e and quantity algebrai ally.
46
DSC1520/SG001
5. The demand fun tion for a
ertain produ t is given by
p
q = 600 − ,
8
where
p
is the pri e per unit and
q
is the number of units sold.
(a) Find the total revenue fun tion for the produ t.
(b) Determine the number of units that need to be sold to maximise total revenue.
( ) It is given that the total
ost fun tion for the produ t is given by
prot fun tion.
(d) Determine the number of units that the
T C = 800 − 120q + 5q 2 .
Find the
ompany should produ e and sell to break even.
4.3 Cubi fun tions
The general
ubi
fun tion is
f (x) = ax3 + bx2 + cx + d,
where
of
a, b, c
ubi
and
d
are
onstants. Consult Appendix D.2 for information about the form, graph and roots
fun tions.
The following is an example of
ubi
fun tions in a real-life situation.
Example
The demand and total
ost fun tions for a
ertain good are given by
p = 90 − q
T C = q3.
and
From this information, we nd the total revenue fun tion to be
T R = pq = (90 − q)q = 90q − q 2 ,
and the prot fun tion,
P = T R − T C = 90q − q 2 − q 3 .
We
an nd the break-even points by either setting
that is by setting
P = 0.
TR = TC
or by nding the roots of the prot fun tion,
Both these methods result in
90q − q 2 − q 3 = q(90 − q − q 2 ) = 0.
From this, we nd that either
q=0
or
90 − q − q 2 = 0.
Multiplying by
of the se ond equation, we nd the equivalent
−1
and rearranging the
omponents
q 2 + q − 90 = 0.
Fa torising this gives
(q + 10)(q − 9) = 0,
whi h gives
q = −10
or
q = 9.
Of
ourse, a negative number of units does not make sense, so we
on lude
that they break even when either nothing is produ ed, or when nine units are produ ed and sold.
This is
onrmed by the graph of
the graphs interse t at
q=0
and
T R and T C
q = 9.
The rm will make a prot as long as
TR
in Figure 4.9(a), where the break-even points are found where
is greater than
T C.
When
TC
will make a loss. From the graph, it is evident that they will only make a prot while
Figure 4.9(b), we see that prot be omes negative when
47
q > 9.
T R, however, they
0 ≤ q ≤ 9. Also, from
is greater than
DSC1520/SG001
T R/T C
2000
P
1500
300
1000
200
TR
500
2
4
100
TC
6
8
q
10 12
2
(a) T R = 90q − q 2 and T C = q 3
4
6
8
10
q
(b) P = 90q − q 2 − q 3
Figure 4.9: Break-even where
T R = T C,
that is where
P =0
4.4 Exponential and logarithmi fun tions
Complete notes on exponential and logarithmi
fun tions are provided in Appendi es D.3 and D.4.
Note that we expe t you to be a quainted with these notes. We also assume that you
solve equations
ontaining exponentials and logarithms by using the rules for working with su h fun tions.
Exponential fun tions to base
of growth we
an manipulate and
e des
ribe growth and de ay in a wide range of systems. The three main types
onsider are unlimited, limited and logisti
growth.
4.4.1 Unlimited growth
Unlimited growth is modelled by the fun tion
f (t) = aert ,
where
t
stands for time, while
a
and
r
are
onstants.
Examples of su h fun tions are investment and
population growth.
Example
The population (P op) of a village was 753 in 1980. It is given that the population grows a
fun tion
P op(t) = 753e0,03t ,
where
p
is the number of people in the population at time
P op
Figure 4.10 shows the graph of
The population in 1990 (when
t,
with
t=0
in 1980.
from 1980 to 2040.
t = 10)
is
al ulated as
P op(10) = 753e0,03×10 = 753e0,3 = 1 016
and in 2010 the population was
P op(30) = 753e0,03×30 = 753e0,9 = 1 852.
We round the numbers, sin e the population
onsists of people.
48
ording to the
DSC1520/SG001
P op
4000
3000
2000
1000
10
20
30
40
50
Figure 4.10: Unlimited growth fun tion
If we need to know when the population will be 3 500, we set
753e0,03t = 2 500
3 500
e0,03t =
753
ln e0,03t = ln 4,65
60
t
P op(t) = 753e0,03t
P op(t) = 3 500
and solve for
t,
that is
(Remember that ln(ex ) = x.)
0,03t = 1,54
t = 51,33.
This means that after 51 years (and 4 months) in 2031 the population is proje ted to be 3 500.
4.4.2 Limited growth
Limited growth is modelled by the fun tion
f (y) = a(1 − e−by ),
where
a
and
b
are
onstants.
Examples of limited growth are
onsumption fun tions, and ele tri al and
me hani al systems.
Example
The
onsumption of a protein against the in ome of households, is modelled by
where
y
Con(y) = 300 1 − e−0,3y ,
is the in ome measured in thousands of rand and
Con
is measured in kilograms per year.
By simply looking at the fun tion, we see that if a family has no in ome, they
onsume no protein, sin e
Con(0) = 300(1 − e0 ) = 300(1 − 1) = 0.
On the other hand, if someone has an in ome of R10 000 per month, they
onsume
Con(10) = 300(1 − e−0,3×10 ) = 300(1 − e−3 ) = 285 kg protein per year.
Figure 4.11 shows how the
It is
onsumption in reases as in ome in reases.
lear that as in ome in reases, the
onsumption of protein in reases at a de reasing rate towards an
upper limit of 300 kilogram per annum.
49
DSC1520/SG001
Con
300
250
200
150
100
50
5
10
15
Figure 4.11: Limited growth fun tion
20
y
C(y) = 300(1 − e−0,3y )
4.4.3 Logisti growth
Logisti
growth is modelled by
f (t) =
where
a, b
and
c
are
onstants and
t
is time.
a
,
1 + be−ct
Examples of logisti
growth are
onsumption fun tions,
onstrained populations, growth of epidemi s and sales.
Example
A virus spreads through a
hi ken farm a
ording to the fun tion
F (t) =
where
F
is the number of infe ted
hi kens and
infe ted, at whi h stage there were 800
Initially, when
t = 0,
F (0) =
After two weeks, when
t
is the number of days sin e the rst
hi ken has been
hi kens on the farm.
800
800
= 1 chicken is infected.
=
0
1 + 799e
800
t = 14,
F (14) =
and when
800
,
1 + 799e−0,1t
800
800
=
= 4 chickens are infected
−0,1(14)
294,93
1 + 799e
t = 100,
F (100) =
800
= 772 chickens are infected.
1 + 799e−0,1(100)
Figure 4.12 shows the growth fun tion.
From the graph it is
lear that in the long run, all the
hi kens will be infe ted.
A tivity
1. In a
ertain area, the number of people aged over 60 (in thousands) is expe ted to grow a
the fun tion
P (t) = 125,5e0,012t ,
where
t
is measured in years sin e 1995.
50
ording to
DSC1520/SG001
F
800
700
600
500
400
300
200
100
20
40
60
80
Figure 4.12: Logisti
100
120
140
t
growth fun tion
(a) Find the number of people older than 60 in 1995 and in 2020.
(b) Plot the graph of
P
from 1995 to 2095 and
omment on the general trend in the population.
( ) How long will it take for the population of people older than 60 to grow to one million?
2. Sales of a new sports magazine are expe ted to grow a
ording to the fun tion
S(t) = 200 000(1 − e−0,05t ),
where
t
is measured in weeks.
(a) Cal ulate the number of magazines sold after one week and after one year.
(b) Plot sales over the rst two years of the magazine's availability and
omment on the trend in
sales.
3. S ientists estimate that a lake
an sustain no more than 6 000 of a
ertain kind of sh. They formulated
the growth fun tion of the sh population to be
P (t) =
where
t
6 000
,
1 + 29e−0,4t
is measured in years.
(a) Find the initial sh population and the population after ten years.
(b) Plot the graph of
P
for
0 < t < 20.
( ) How long will it take for the population to rea h 4 000?
4.5 Hyperboli fun tions
The general hyperboli
fun tion is given by
f (x) =
where
a, b
and
c
are
a
,
bx + c
onstants.
Consult Appendix D.5 for information about the form and graph of a hyperboli
51
fun tion.
DSC1520/SG001
Example
The demand and supply fun tions for a good are given by
200
p
q+1=
where
q
is the number of units of the good and
and
p
ps = 5 + 0,5q.
is the pri e per unit.
To be able to plot these fun tions on the same axes, we need to have both with the same dependent variable.
Therefore, write the demand fun tion with
p
as dependent variable, that is
200
p
p(q + 1) = 200
200
pd =
.
q+1
q+1=
Figure 4.13 shows the graph of this fun tion.
p
180
160
140
120
100
80
pd =
60
200
q+1
40
ps = 5 + 0,5q
20
5
10
15
20
25
q
Figure 4.13: Demand and supply fun tions
To determine the pri e and quantity of the good at market equilibrium, equate the demand and supply
fun tions to nd
200
q+1
(5 + 0,5q)(q + 1) = 200
5 + 0,5q =
0,5q 2 + 5,5q + 5 − 200 = 0
q 2 + 11q − 390 = 0.
Solving this quadrati
equilibrium is when
fun tion gives
q = 15
q = 15
q = −26.
A negative number of units is not possible, so the
are produ ed and pri e
p=
This
or
(Multiply by 2,)
200
200
=
= R12,50.
q+1
15 + 1
orresponds with the point in Figure 4.13 where the demand and supply fun tions interse t.
52
DSC1520/SG001
A tivity
1. The value of a
ar (in thousands of rand)
t
years after it has been bought, is given by
V (t) = 1 +
840
.
1 + 2t
(a) Graph the value fun tion for the rst ten years. Des ribe how the
ar depre iates over time.
(b) How long will it take for the value to drop to R200 000?
2. A new model of laptop is gradually repla ing an earlier model. The proje ted sales (in thousands) for
the old and new models are given by
Sold =
where
t
9
t+3
and
Snew =
36
,
21 − t
is measured in months.
(a) Graph the sales for the old and new models on the same axes. Did the introdu tion of the new
model improve overall sales?
(b) Find the time and sales at the interse tion of the fun tions algebrai ally. What does this interse tion represent?
53
PART III
DIFFERENTIATION
54
Chapter 5: Dierentiation theory
Learning obje tives
After you have
⊲
⊲
⊲
⊲
⊲
⊲
⊲
⊲
⊲
ompleted this
explain the
hapter, you should be able to
on ept of dierentiation
apply the power rule to dierentiate given fun tions
dierentiate exponential and logarithmi
dierentiate
fun tions
omposite fun tions using the
hain rule
apply the produ t rule to dierentiate the produ t of two fun tions
apply the quotient rules to dierentiate the quotient of two fun tions
nd the rst, se ond and third derivative of a fun tion
use dierentiation to nd the optimal point(s) of a non-linear fun tion
nd the turning point(s) of a non-linear fun tion
use the se ond derivative to determine the nature of the turning point(s)
determine the intervals along whi h a non-linear fun tion is in reasing or de reasing
use dierentiation to sket h a non-linear fun tion
5.1 Slope of a urve and dierentiation
In Se tion C.1 the slope of a line is dened as the verti al
hange (∆y ) divided by the horizontal
hange
(∆x), that is
Slope =
The slope of a straight line is
∆y
y2 − y1
=
.
∆x
x2 − x1
onstant at any point on the line. For a
urve, however, the slope at dierent
points varies.
1 are drawn between points
Consider, for example, the
urve shown in Figure 5.1, where
D.
is quite steep and negative, while the slope of
The slope of
the slope of
CD
hord
AB
B with oordinates (x; y) and point C a short distan
hord BC is given by
Slope =
If
hord
BC
slope of the
BC
A, B, C
and
is atter and positive, while
is steep and positive.
Consider point
The slope of
hords
B.
When point
C
oordinates
(x + ∆x; y + ∆y).
∆y
vertical change
=
.
horizontal change
∆x
is progressively shortened, so that
urve at point
e from it with
∆x
and
∆y get smaller and smaller, we ome
B , the slopes of hord BC and the
rea hes point
loser to the
urve are the
same.
At this point, the horizontal distan e,
∆x,
is almost zero, but not equal to zero (to avoid dividing by zero,
whi h is not allowed). We say ∆x tends to zero and write
1
The line between two points on a urve is alled a hord.
55
∆x → 0.
DSC1520/SG001
y = f (x)
D
A
C(x + ∆x; y + ∆y)
C∗
B(x; y)
C ∗∗
x
Figure 5.1: Chords with varying slopes and the tangent at
The pro ess of moving
loser and
i ally as the limit when
∆x
loser to the point where
∆x
is very
B
lose to zero, is des ribed mathemat-
tends to zero. This is written as
∆y
.
∆x→0 ∆x
lim
The straight line that tou hes the
tangent is also the slope of the
Of
ourse, the slope of the
y = f (x)
urve at point
B
is
tangent
alled the
to the
urve at
B.
The slope of the
urve at that point.
urve varies as one moves along the
is known as the derivative of the
as
urve with respe t to
urve. The formula for the slope of the
x.
urve
In mathemati al notation, this is written
∆y
dy
= f ′ (x) = lim
.
∆x→0 ∆x
dx
To illustrate, we work through the following example to nd the derivative of a fun tion from rst prin iples.
Example
Consider the fun tion
y = f (x) = x2
with its graph shown in Figure 5.2.
To nd the derivative of
f (x)
from rst prin iples, we need to nd an expression for
lim
∆x→0
∆y
.
∆x
We start with the given fun tion
y = x2
and repla e
x
and
y
with the
oordinates of point
C,
that is
y + ∆y = (x + ∆x)2 .
Squaring the RHS gives
y + ∆y = x2 + 2x∆x + (∆x)2 .
When we subtra t
y
(whi h is equal to
x2 )
from both sides of the equation, we nd
y + ∆y − y = x2 + 2x∆x + (∆x)2 − x2 ,
56
∆y
∆x and then determine
DSC1520/SG001
y
C (x + ∆x; y + ∆y)
4
3
∆y
2
∆x
B (x; y)
1
1
Figure 5.2:
y = x2
with points
x
2
B
and
C
on the
urve
or
∆y = 2x∆x + (∆x)2 .
Dividing by
∆x
on both sides of the equation, gives
∆y
= 2x + ∆x.
∆x
When
∆x
now be omes smaller and smaller (tends to zero), we nd the derivative to be
dy
∆y
= lim
= lim (2x + ∆x) = 2x.
dx ∆x→0 ∆x ∆x→0
The slope of the
urve
y = x2
is
alled its derivative and is given by
dy
= 2x
dx
By using this formula, we
an nd the slope of the
⊲
At
x = −1
the slope is
f ′ (−1) = −2.
⊲
At
x = 0,5
the slope is
f ′ (0,5) = 2 × 0,5 = 1.
⊲
At
x=2
the slope is
or
f ′ (x) = 2x.
urve at any point on the
urve. Examples:
f ′ (2) = 2 × 2 = 4.
The derivative of a fun tion gives the rate at whi h the fun tion is
hanging at ea h point on the fun tion.
5.2 Dierentiation rules
The pro ess of nding the derivative of
y
with respe t to
x,
denoted by
dy
dx , is
alled
dierentiation.
In the
following se tions we dis uss the dierent rules for dierentiation.
5.2.1 The power rule for dierentiation
The power rule is the basi
rule to determine the derivative of
power rule states the following:
57
xn ,
where
n
may be any real number. The
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y = f (x) = xn
If
then
f ′ (x) =
dy
= nxn−1 .
dx
Consider, for example, the fun tion
y = x5
In this
ase
n = 5,
so a
or
f (x) = x5 .
or
f ′ (x) = 5x5−1 = 5x4 .
ording to the power rule,
dy
= 5x5−1 = 5x4
dx
The following additional rules are appli able when working with the power rule:
⊲
When a mathemati al term is multiplied (or divided) by a
multiplied (or divided) by the
For any term of the form
kxn ,
the derivative is
k×
The
onstant
k
onstant, the term is dierentiated and then
onstant.
d n
x = k × nxn−1 .
dx
an be positive, negative or a fra tion.
Examples:
d
(10x3 ) = 10
dx
⊲
d
(−3x2 ) = −3
dx
d 2
x
dx
= 10 × 3x2
= −3 × 2x
= 30x2 .
= −6x.
The derivative of a
A
d 3
x
dx
onstant like 5
d
dx
1 5
x
2
1 d 5
x
=
2 dx
1
= × 5x4
2
5 4
= x .
2
onstant term is zero.
an be written as
zero is equal to one.
5 × 1,
whi h is also equal to
5 × x0 ,
be ause anything to the power
The derivative of 5 is therefore
d
5=5
dx
d 0
x
dx
= 5 × 0x0−1 = 0,
sin e anything multiplied by zero, is zero.
⊲
The derivative of a polynomial
y = f (x) = axn + bxn−1 + · · · + cx2 + dx1 + e,
with
a, b, c, d and e
onstants, is the sum of the derivatives of the terms.
Example:
d 4
d
d
d
d
d 4
(x + 2x3 − 5x2 + 3x + 100) =
(x ) +
(2x3 ) −
(5x2 ) +
(3x) +
100
dx
dx
dx
dx
dx
dx
=
d
d
d
d 4
(x ) + 2 (x3 ) − 5 (x2 ) + 3 (x) + 0
dx
dx
dx
dx
= 4x3 + 2 × 3x2 − 5 × 2x + 3 × 1
= 4x3 + 6x2 − 10x + 3.
58
DSC1520/SG001
Examples
1. Dierentiating
f (x) = 10x8 ,
gives
f ′ (x) = 10
2. To dierentiate
f (x) =
d 8
x = 10 × 8x7 = 80x7 .
dx
1
x2 , we rst simplify to get
f (x) = x−2 .
The derivative is then
f ′ (x) = −2x−2−1 = −2x−3 = −
3. The derivative of
f (x) = x2 + 7x + 5
is
f ′ (x) =
d
d
d 2
x +7 x+
5 = 2x + 7.
dx
dx
dx
f (x) = x − 4 x1 + 8 x22 ,
4. To dierentiate the fun tion
2
.
x3
we simplify to get
f (x) = x − 4x−1 + 16x−2 .
d
d
d
x − 4 x−1 + 16 x−2
dx
dx
dx
= 1 − 4(−1x−1−1 ) + 16(−2x−2−1 )
f ′ (x) =
= 1 + 4x−2 − 32x−3
4
32
= 1 + 2 − 3.
x
x
5. To dierentiate the fun tion
f (x) =
5x+2
x , we rst simplify to get
f (x) =
5x 2
+ = 5 + 2x−1 .
x
x
Now, by using the power rule we nd
f ′ (x) =
6. Simplify the fun tion
d
d
2
5 + 2 x−1 = 0 + 2(−1x−2 ) = − 2 .
dx
dx
x
y = 10 + 5x +
1
to nd
x2
y = 10 + 5x + x−2 .
Dierentiation gives
2
dy
= 0 + 5 − 2x−3 = 5 − 3 .
dx
x
7. Dierentiating the fun tion
P (q) =
q3
3
+ 700q − 15q 2
gives
d
d
1 d 3
q + 700 q − 15 q 2
3 dq
dq
dq
1
= (3q 2 ) + 700(1) − 15(2q 1 )
3
= q 2 + 700 − 30q.
P ′ (q) =
59
Then,
DSC1520/SG001
8. To apply the power rule to the fun tion
y=
√1 , we need to simplify it rst to get
x
1
y = x− 2 .
Dierentiating this fun tion gives
1
1
dy
= − x(− 2 −1)
dx
2
1 3
= − x− 2
2
1
=− 3
2x 2
1
=− √ .
2 x3
9. Simplifying the fun tion
√
q
f (q) = 4 q2
gives
1
1
3
f (q) = 4q 2 q −2 = 4q 2 −2 = 4q − 2 .
Dierentiation gives
3
f (q) = 4 −
2
′
5
3
q (− 2 −1)
= −6q − 2
6
=− 5
q2
6
= −p .
q5
A tivity
1. Find the slope of the
urve
y = x3 + 4x − 7
at
x = 3.
2. Dierentiate ea h of the following fun tions by using the power rule:
(a)
√
y =4+x+2 x
(b)
f (x) =
( )
f (x) = 7 +
(d)
p(q) =
1
x
−
5
x2
+ 10
√7
x
q 2 −q
q3
5.2.2 Exponential and natural logarithmi fun tions
The derivative of the basi
exponential fun tion
y = ex
is the fun tion itself, that is
d x
dy
=
e = ex .
dx
dx
60
DSC1520/SG001
When the exponential fun tion
ontains a
onstant, the derivative is multiplied by the
onstant, that is
d ax
e = aeax .
dx
For example, if
f (x) = 10ex + 5e3x ,
then
f ′ (x) = 10ex + 5(3)e3x = 10ex + 15e3x .
The derivative of the natural logarithmi
fun tion (the log to base
e),
y = ln(x)
is 1 over
x,
that is
1
d
ln(x) = .
dx
x
5.2.3 The hain rule
The
hain rule for dierentiation is used to dierentiate
The following are examples of su h
1.
y = (2x + 3)5
2.
y = e4x+5
3.
y = ln(5x − 1)
4.
y=
The
the inner fun tion is
u = 2x + 3
u = 4x + 5
the inner fun tion is
1
x+7 the inner fun tion is
and the outer fun tion is
u = 5x − 1
u=x+7
dus
and the outer fun tion is
y=
x
of a
ln(u).
1
u.
omposite fun tion
y = f (u)
is given by
in the rst equation
or f ′ (x) = f ′ (u)g′ (x).
an el out.
In words we say that the derivative of a
omposite fun tion is the derivative of the outer fun tion,
multiplied by the derivative of the inner fun tion.
Now let's dierentiate ea h of the
2
y = eu .
and the outer fun tion is
dy du
dy
=
·
dx
du dx
Note that the
y = u5 .
and the outer fun tion is
hain rule of dierentiation states that the derivative in terms of
u = g(x)
fun tion of a fun tion).
omposite fun tions:
the inner fun tion is
with inner fun tion
2
omposite fun tions (the
omposite fun tions given above.
Refer to Se tion D.6 on page 184.
61
DSC1520/SG001
Examples
1. For the
omposite fun tion
f (x) = y = (2x + 3)5 ,
we denote the inner fun tion by
This outer fun tion
The inner fun tion
By
y(u)
u
u = 2x + 3,
so that the fun tion be omes
u
is dierentiated in terms of
is dierentiated
y = f (u) = u5 .
by using the power rule, namely
d 5
dy
=
u = 5u4 .
du
du
in terms of x to nd
d
du
=
(2x + 3) = 2.
dx
dx
ombining these results, the derivative is found to be
d 5 d
d
(2x + 3)5 =
u ·
(2x + 3)
dx
du
dx
= 5u4 · 2
= 10(2x + 3)4 .
2. For the
(since u = 2x + 3)
omposite fun tion
f (x) = e4x+5 ,
the inner fun tion is
u = 4x + 5,
so that the outer fun tion be omes
f (u) = eu
with
u = 4x + 5.
The derivatives of the outer and inner fun tions are
f ′ (u) =
d u
e = eu
du
u′ (x) =
and
d
(4x + 5) = 4.
dx
ombining these gives
f ′ (x) = f ′ (u)u′ (x) = eu · 4 = 4e4x+5 .
f (x) = y = ln(5x − 1)
3. Transform the outer fun tion
that is
f (u) = ln u
to be in terms of the inner fun tion
with
u = 5x − 1,
u = 5x − 1.
The derivatives of the outer and inner fun tions are
f ′ (u) =
d
1
ln u =
du
u
u′ (x) =
and
ombining these results in
f ′ (x) = f ′ (u)g′ (x) =
4. Transform the outer fun tion
f (x) = y =
f (u) =
d
(5x − 1) = 5.
dx
5
1
·5=
.
u
5x − 1
1
x+7 to be in terms of the inner fun tion
1
= u−1
u
with
u = x + 7,
u = x + 7.
The derivatives of the outer and inner fun tions are
f ′ (u) =
1
d −1
u = −u−2 = − 2
du
u
and
u′ (x) =
d
(x + 7) = 1,
dx
ombining these results in
f ′ (x) = f ′ (u)g ′ (x) = −
62
1
1
·1=−
.
2
u
(x + 7)2
that is
DSC1520/SG001
A tivity
Dierentiate ea h of the following fun tions by using the
2.
y = (4 − 5x)3
√
f (x) = x2 + 12
3.
y=
4.
q = 500 ln(p3 + 8p)
5.
g(x) = 5e1,5 +
1.
hain rule:
15
1+ex
2
x+2
− ln(2x − 1)
5.2.4 The produ t rule
Fun tions may
onsist of the produ t of two or more fun tions. To dierentiate su h a fun tion, we apply
the produ t rule. This rule states that for the fun tion
f (x) = y = u(x)v(x),
the derivative
f ′ (x) = u′ (x)v(x) + u(x)v ′ (x) or
du
dv
dy
=
v(x) + u(x) .
dx
dx
dx
In words: To dierentiate the produ t of two fun tions, multiply the derivative of
the rst fun tion with the se ond fun tion and add the rst fun tion, multiplied by
the derivative of the se ond fun tion.
Examples
1. The fun tion
f (x) = ex (x2 + 5)
onsists of the following distin t, independent fun tions of
u(x) = ex
x
that are multiplied together:
and v(x) = x2 + 5
The derivatives of these fun tions are
u′ (x) = ex
and
v ′ (x) = 2x.
Applying the produ t rule gives
f ′ (x) = u′ (x)v(x) + u(x)v ′ (x)
= ex (x2 + 5) + ex 2x
= ex (x2 + 5 + 2x).
2. The fun tion
f (x) = (x2 + 2x + 1)(3x + 2)
onsists of the produ t of the fun tions
u(x) = x2 + 2x + 1 with
du
= 2x + 2
dx
63
and
v(x) = 3x + 2 with
dv
= 3.
dx
DSC1520/SG001
The derivative of
f,
a
ording to the produ t rule, is
dv
du
v+u
dx
dx
= (2x + 2)(3x + 2) + (x2 + 2x + 1) × 3
f ′ (x) =
= 6x2 + 10x + 4 + 3x2 + 6x + 3
= 9x2 + 16x + 7.
3. The fun tion
f (x) = x2 e2x+1
onsists of the produ t of the fun tions
u(x) = x2
The derivative of
with
f
du
= 2x
dx
and
v(x) = e2x+1
with
dv
= e2x+1 · 2 (chain rule).
dx
is
dv
du
v+u
dx
dx
= 2x · e2x+1 + x2 · 2e2x+1
f ′ (x) =
= 2xe2x+1 + 2x2 e2x+1
= 2xe2x+1 (1 + x).
4. The fun tion
(factorisation)
√
P (q) = 2 q(q + 5)
onsists of
1
√
u(q) = 2 q = 2q 2
with u′ (q) =
2 −1
1
q 2 =√
2
q
and
v(q) = q + 5 with v ′ (q) = 1.
The derivative of
P
is
P ′ (q) = u′ (q)v(q) + u(q)v ′ (q)
√
1
= √ (q + 5) + 2 q · 1
q
√
5
√
= q+ √ +2 q
q
5
√
=3 q+√
q
3q + 5
= √ .
q
5. The fun tion
C(y) = (y + 4) ln(y)
onsists of the produ t of
u(y) = y + 4
and
v(y) = ln(y)
with derivatives
u′ (y) = 1
and
64
v ′ (y) =
1
.
y
DSC1520/SG001
The derivative of
C
is
C ′ (y) = u′ (y)v(y) + u(y)v ′ (y)
1
= 1 · ln y + (y + 4)
y
y 4
= ln y + +
y y
4
= ln y + 1 + .
y
A tivity
Dierentiate the following fun tions by applying the produ t rule:
1.
√
Q(p) = p p + 5
2.
AC(q) =
3.
f (x) =
4.
f (x) = x3 ln(x3 )
1
q
− q ln q
3
x−5
20 xe
5.2.5 The quotient rule
When we have a fun tion
onsisting of the quotient of two fun tions, that is
f (x) =
u(x)
v(x)
we apply the quotient rule given by
u′ (x)v(x) − u(x)v ′ (x)
.
v(x)2
f ′ (x) =
Examples
1. Consider the fun tion
f (x) =
x2 + 2x + 1
.
3x + 2
Here,
u(x) = x2 + 2x + 1 with u′ (x) = 2x + 2
Now, a
and
v(x) = 3x + 2 with v ′ (x) = 3.
ording to the quotient rule,
dv
− u dx
v2
(2x + 2)(3x + 2) − (x2 + 2x + 1)(3)
=
(3x + 2)2
6x2 + 10x + 4 − (3x2 + 6x + 3)
=
(3x + 2)2
3x2 + 4x + 1
.
=
(3x + 2)2
f ′ (x) =
du
dx v
65
DSC1520/SG001
2. The fun tion
y = f (x) =
x2
1 + ex
onsists of the quotient of
u(x) = x2
The derivative of
f
with u′ (x) = 2x
v(x) = 1 + ex
and
with v ′ (x) = ex .
is
u′ (x)v(x) − u(x)v ′ (x)
v(x)2
2x(1 + ex ) − x2 ex
=
(1 + ex )2
2x + 2xe− x2 ex
=
(1 + ex )2
x(2x + 2ex − xex )
.
=
(1 + ex )2
f ′ (x) =
3. The fun tion
P (q) =
q
3q + 5
onsists of the quotient of
u(q) = q
with u′ (q) = 1
and
v(q) = 3q + 5 with v ′ (q) = 3.
A
ording to the quotient rule,
u′ (q)v(q) − u(q)v ′ (q)
v(q)2
1 · (3q + 5) − q · 3
=
(3q + 5)2
3q + 5 − 3q
=
(3q + 5)2
5
=
.
(3q + 5)2
P ′ (q) =
Note:
The quotient of two fun tions
an always be transformed to be the produ t of two fun tions. For
example, the fun tion
f (x) =
2x + 3
= (2x + 2)(3x + 2)−1 .
3x + 2
We now have the produ t of
u(x) = 2x + 2 with
u′ (x) = 2
and
v(x) = (3x + 2)−1
with v ′ (x) = −1(3x + 2)−2 · 3 = −
66
3
.
(3x + 2)2
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Applying the produ t rule results in
f ′ (x) = u′ (x)v(x) + u(x)v ′ (x)
= 2(3x + 2)−1 + (2x + 2) · (−3(3x + 2)−2 )
2
3(2x + 2)
=
−
3x + 2 (3x + 2)2
2(3x + 2) − 3(2x + 2)
=
(3x + 2)2
−2
=
.
(3x + 2)2
A tivity
Use the quotient rule to dierentiate the following fun tions:
ln x
3x
1.
f (x) =
2.
C(y) = 1 −
3.
F (x) =
2x ln x
3x2
4.
P (q) =
50−q 2
50+q 2
e−0,8y
y2
5.3 Higher derivatives
Up to now, we have found only the rst derivative of fun tions. When we now dierentiate this result again,
we get the se ond derivative and we
an derive this again to nd the third derivative.
The rules in the previous se tion are always appli able, regardless of whether the rst, se ond or third
derivative is found.
The higher-order derivatives are used to determine and/or
onrm maximum, minimum and ine tion points,
as will be seen in following se tions.
The notation used for higher-order derivatives are as follows:
First derivative
Second derivative
d df
d2 f
=
dx dx
dx2
df
dx
f ′ (x)
f ′′ (x)
Third derivative
d d2 f
d3 f
=
dx dx2
dx3
Consider, for example, the fun tion
f (x) = 25x4 − 10x2 + 200.
The rst, se ond and third derivatives of
f
are
f ′ (x) = 25(4)x3 − 10(2)x + 0 = 100x3 − 20x,
f ′′ (x) = 100(3)x2 − 20 = 300x2 − 20
and
f ′′′ (x) = 300(2)x − 0 = 600x.
67
f ′′′ (x)
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A tivity
1. Find the rst and se ond derivatives of ea h of the following fun tions:
(a)
(b)
( )
f (x) = (150 − 2x)x
√
f (x) = 10x + x
f (x) = x4 −
1
x4
2. Find the rst, se ond and third derivatives of the following fun tion:
C(x) =
x3
5x
− 8x2 +
+ 180.
5
2
5.4 Optimisation of fun tions in one variable
Optimisation is the pro ess of nding maximum and minimum points. Stationary points are those points
on a
urve where the slope
hanges from positive to negative (at a maximum point), or from negative to
positive (at a minimum point). Obviously, at the point where the slope
Consider the
hanges, it rst has to be ome zero.
urve in Figure 5.3.
f (x)
A
ive
Pos
it
Pos
it
ive
e
tiv
ga
Ne
x
B
Figure 5.3: Turning points
At point A (the maximum), the slope
hanges from positive to negative and at B (the minimum), the slope
hanges from negative to positive. At ea h of the stationary points A and B, the slope of
We know that the slope of a fun tion
f
is given by the derivative of
Slope = f ′ (x) =
Also, as shown above, the slope of
′
of a fun tion by setting f (x)
f
f,
f
is zero.
that is
df
.
dx
is zero at the turning points. We
an therefore nd the turning points
= 0.
The turning points of a fun tion
f
is found by setting the derivative of
zero, that is
f ′ (x) =
df
= 0.
dx
We use the following method to nd the turning point(s) of a fun tion
68
f.
f
equal to
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Find the rst derivative of
⊲
Set
⊲
Determine the fun tion value at the resulting points
f ′ (x) = 0
f,
f ′ (x).
⊲
and solve for
∗
the turning point are (x ;
that is
x;
x∗ ,
that is, nd
f (x∗ ),
so that the
oordinates of
f (x∗ )).
Example
Let's nd the turning point(s) of the fun tion
f (x) = 3x2 − 18x + 34.
We start by dierentiating the fun tion and then set the result equal to zero, that is
f ′ (x) = 6x − 18 = 0.
When we solve this equation, we nd
x=
18
= 3, (with) f (3) = 3(3)2 − 18(3) + 34 = 7.
6
The turning point is therefore at (3; 7).
From the fun tion we see that the
oe ient of
x2
is positive, so we
3
to be a minimum point. This is onrmed by the graph of
f
on lude that the stationary point has
in Figure 5.4.
f (x)
30
20
10
1
2
Figure 5.4:
3
4
5
6
x
f (x) = 3x2 − 18x + 34
5.4.1 The nature of stationary points
On e we have found the stationary points of a fun tion, we need to determine whether they are minimum
or maximum points. We
an, of
ourse, graph the fun tion to see this, but it may be quite
not really pra ti al. We need an algebrai
When we
way to determine this.
onsider a maximum point, we know that the slope
stationary point.
umbersome and
In Figure 5.5, we see that the slope of
maximum from the left, until it eventually rea hes zero.
f
hanges from positive to negative at the
de reases gradually as we move
loser to the
After the maximum point, the slope be omes
in reasingly negative, meaning that it keeps on de reasing.
The slope of
f
hanges from positive, through zero to negative as we move through the maximum the value
′
of the slope (f ) is therefore de reasing through the maximum. This means that the derivative of the slope
′′
(f ) at the maximum is negative.
3
See property (a) on page 169.
69
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f (x)
4
3
2
1
0
−3
−2
−1
−1
1
x
−2
f (x) = −x2 − 2x + 3
Figure 5.5:
When we
onsider the minimum turning point in Figure 5.6, we see that the slope of
less negative) until it rea hes zero and then starts to in rease. The value of the slope
f ′′
through the minimum point, meaning that
f in reases (be omes
f ′ therefore in reases
is positive.
f (x)
8
7
6
5
4
3
2
1
−1
1
derivative test:
If
if
f ′′ (x∗ ) =
f ′′ (x∗ ) =
d2 f
dx2
d2 f
dx2
< 0,
x∗
3
x
f (x) = x2 − 2x + 3
Figure 5.6:
To determine whether a stationary point at
2
is a maximum or minimum point, we use the following
then
f
has a maximum turning point at
x∗ ;
x∗
> 0,
then
f
has a minimum turning point at
x∗
70
x∗ .
and
se ond
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However, when
f ′′ (x) = 0,
we
annot
ome to any
on lusion.
If the slope
negative at this point, or from negative to positive, it is a turning point.
negative) throughtout the point, it probably is an ine tion point
f′
hanges from positive to
If the slope stays positive (or
4 and will need more investigation.
5.4.2 Intervals along whi h a fun tion is in reasing or de reasing
When we know whether the stationary points of
where
f
are maximum or minimum points, we
an determine
is in reasing and where it is de reasing.
f (x) = −x2 − 2x + 3, we nd that f has a turning point where f ′ (x) = −2x − 2 = 0, that is
x = −1. From f ′′ (x) = −2 < 0 we on lude that the turning point is a maximum. This is onrmed by
If we
at
f
onsider
Figure 5.5 on the previous page.
For a maximum at
x = −1,
we
an say that to the left of
mathemati al terms,
f
is in reasing for
Furthermore, to the right of
f
x < −1
x = −1, f
is de reasing for
f
or
x = −1
the fun tion is in reasing. Therefore, in
in reases over the interval
(−∞; −1).
is de reasing. Therefore,
x > −1
or
f
de reases over the interval
(−1; ∞).
Example
Consider the fun tion
f (x) = −x3 + 9x2 − 24x + 26.
To nd the stationary points of
f,
we nd the derivative of
f,
set it equal to zero and solve for
x.
The derivative is
f ′ (x) = −3x2 + 18x − 24.
Now, set it equal to zero and solve
−3x2 + 18x − 24 = 0
x2 − 6x + 8 = 0
(divide by − 3)
(x − 4)(x − 2) = 0
(factorise)
x − 4 = 0 or
Therefore,
f
has stationary points at
x=4
x − 2 = 0.
and
x = 2.
We now need to determine whether these points are maximum or minimum points, by using the se ond
derivative, namely
f ′′ (x) =
d
(−3x2 + 18x − 24) = −6x + 18.
dx
⊲
Sin e
f ′′ (4) = −6(4) + 18 = −6 < 0,
⊲
Sin e
f ′′ (2) = −6(2) + 18 = 6 > 0,
With this information at hand, we
(−∞; 2)
and
(4; ∞),
we have a lo al
we have a lo al
6 minimum point at
an easily draw a rough graph of
while it in reases over
Figure 5.7 shows the graph of
5 maximum point at
f
x = 4.
x = 2.
and spe ify that
f
is de reasing over
(2; 4).
f.
4
Ine tion points fall outside the s ope of this module. For more detail google ine tion points al ulus.
Sin e f has larger values to the left, the turning point is a lo al maximum.
6
Sin e f has smaller values to the right, the turning point is a lo al minimum.
5
71
DSC1520/SG001
f (x)
24
20
16
12
8
4
1
Figure 5.7:
2
3
4
5
x
6
f (x) = f (x) = −x3 + 9x2 − 24x + 26
A tivity
1. Find the turning point(s) of ea h of the following fun tions. Use the se ond derivative to determine
the nature of the turning point(s).
(a)
f (x) = x2 − 6x + 6
(b)
f (t) = 13 t3 − 2t2 − 5t + 8
( )
Q(p) = 64p +
(d)
f (x) = 200 − 2(x − 4)2
1
p
2. Determine the intervals along whi h ea h of the following fun tions in reases or de reases:
(a)
f (x) = x2 − 18x + 11
(b)
f (x) = 3x2 − 0,1x3
5.4.3 Sket hing fun tions using dierentiation
If we
an nd the turning points and the inter epts on the
x and y axes, we should be able to sket
h non-linear
fun tions.
Examples
1. If we qui kly need to draw the graph of the fun tion
y = f (x) = x2 − 6x + 6,
we start by dierentiation the fun tion and setting the result equal to zero to nd the turning point(s)
of
f.
Here,
f ′ (x) = 2x − 6 = 0,
The fun tion therefore has a turning point at
The
oordinates of the turning point are
x = 3,
(3; −3).
72
giving
where
x = 3.
f (3) = (3)2 − 6(3) + 6 = −3.
DSC1520/SG001
Sin e
f
is a quadrati
fun tion with
a > 0,
we know that it has a minimum turning point.
This is
onrmed by the se ond derivative,
f ′′ (x) = 2 > 0,
Furthermore, from
f
which indicates a minimum turning point.
we see that when
x = 0,
then
y = 6.
This is the
y
axis inter ept.
√
x axis inter epts, set y = 0 and solve for x. Maxima gives the roots as x =√2 + 3 = 4,73
√
b2 −4ac
x = 3 − 3 = 1,27, whi h is the same result as found by using the formula x = −b± 2a
.
To nd the
or
f.
We now have enough information to draw a rough graph of
Figure 5.8 shows the graph of
f
with the points we have found above, indi ated with dots.
y
9
8
7
6
5
4
3
2
1
0
−1
−2
−3
1
2
Figure 5.8:
2. For the
ubi
3
4
5
x
6
f (x) = x2 − 6x + 6
fun tion
y = f (x) = 2x3 − 3x2 ,
the rst derivative is
f ′ (x) = 6x2 − 6x.
We set
f ′ (x) = 6x2 − 6x = 6x(x − 1) = 0,
to nd the turning points of
f
to be at
x=0
and
x = 1.
The turning points are at
(0; 0)
and
(1; −1).
Using the se ond derivative, namely
f ′′ (x) = 12x − 6,
we nd
The
y
f ′′ (0) = −6 < 0
We
x = 0 and
x = 0 or x = 1,5.
inter ept is at
whi h gives
(a maximum turning point) and
the
x
f ′′ (1) = 6 > 0
inter epts are found by setting
( a minimum turning point).
2x3 − 3x2 = 0,
an now draw a rough graph through these points, as shown in Figure 5.9.
73
or
x2 (2x − 3) = 0,
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y
3
2
1
0
1
−1
−2
−3
Figure 5.9:
f (x) = 2x3 − 3x2
74
x
Chapter 6: Appli ations of dierentiation
Learning obje tives
After you have
⊲
explain the
hapter, you should be able to
on epts of average and marginal revenue and
use dierentiation to nd marginal revenue and
ost
ost fun tions from total revenue and
ost fun -
tions
⊲
⊲
ompleted this
nd average revenue and
nd total revenue and
ost fun tions from total revenue and
ost fun tions
ost fun tions from given average revenue and
ost fun tions
nd the marginal produ t of labour, given a labour fun tion
use dierentiation to optimise e onomi
fun tions like revenue,
ost and prot fun tions
use dierentiation to determine and interpret point elasti ity of non-linear demand
6.1 Marginal and average fun tions
In Chapter 5, the derivative of a fun tion was des ribed as the equation for the rate of
In e onomi s, the rate at whi h total revenue (T R)
at whi h total
Therefore,
ost (T C )
hanges, is
alled
hanges, is
alled
marginal ost (M C ).
hange of the fun tion.
marginal revenue (M R)
marginal revenue (M R) is the rate of hange in total revenue
T R fun tion in terms of q , that is
and the rate
per unit in rease in output,
q,
and
is found by dierentiating the
MR =
Likewise,
marginal ost (M C ) is the rate of
T C fun tion in terms
by dierentiating the
hange in total
of
q,
TR
TC
FC
VC
ost per unit in rease in output,
q,
and is found
that is
MC =
Throughout this
dT R
.
dq
dT C
.
dq
hapter, the variables in the following table are used:
Total revenue
Total
ost
Fixed
ost
Variable
ost
MR
MC
MV C
Marginal revenue
Marginal
ost
Marginal variable
ost
AR
AC
AF C
AV C
Average revenue
Average
ost
Average xed
ost
Average variable
ost
6.1.1 Marginal revenue
Marginal revenue is dened as the additional in ome that is generated from the sale
of one more unit of a good or servi e.
75
DSC1520/SG001
If the demand fun tion for a
ertain produ t is given by
p = a − bq ,
then we
an nd the total revenue
fun tion by multiplying the demand fun tion by the number of units demanded, that is
T R = pq = (a − bq)q = aq − bq 2 .
TR
The marginal revenue fun tion is then found by dierentiating the
MR =
fun tion, that is
d
(aq − bq 2 ) = a − 2bq.
dq
Example
Suppose we have the demand fun tion
p = 6 − 0,5q.
From this, we nd the total revenue fun tion to be
T R(q) = pq = (6 − 0,5q)q = 6q − 0,5q 2 .
By dierentiation, we nd the marginal revenue fun tion to be
MR =
From the
MR
d
dT R
=
(6q − 0,5q 2 ) = 6 − q.
dq
dq
fun tion, we nd that when ve units have been sold, the revenue from the next unit sold,
will be
M R(5) = 6 − 5 = R1.
This means that the sixth unit to be sold, will generate additional in ome of R1.
6.1.2 Marginal ost
Marginal
ost refers to the in rease or de rease in the
unit or serving one more
The total
ost fun tion
ost of produ ing one more
ustomer.
onsists of xed
osts and variable
osts, and is given by
T C = F C + V C.
Marginal
ost is found by dierentiating the
MC =
Note that, sin e xed
Marginal
ost is
TC
fun tion, that is
d
d
d
d
dT C
=
(F C + V C) =
FC + V C =
V C.
dq
dq
dq
dq
dq
onstant and is not inuen ed by the number of units produ ed,
ost is therefore given by the derivative of the variable
marginal variable
ost, that is
M C = M V C.
76
ost fun tion, so that
MC
dF C
dq
= 0.
is the same as
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Example
1. Given the total
ost fun tion
T C = 100 + 40q,
we nd that marginal
ost is
MC =
This means that
2. For the total
MC
is
onstant and does not vary with output ea h additional unit will
ost R40.
ost fun tion
TC =
we nd marginal
q = 15,
1 3
q − 8q 2 + 120q,
3
ost to be
MC =
When
d
(100 + 40q) = 40.
dq
dT C
= q 2 − 16q + 120.
dq
then
M C = (15)2 − 16(15) + 120 = 105.
This means that the
ost to produ e the 16th unit, is R105.
6.1.3 Average revenue and ost fun tions
Average fun tions give expressions for the average value of e onomi
Average revenue (AR)
variables over an interval.
is dened as the average revenue per unit for the rst
q
su
essive units sold, and is
found by dividing total revenue by the number of units sold, that is
AR =
TR
.
q
From this formula, we nd that
T R = AR × q.
But, sin e total revenue is dened as
T R = p × q,
it follows that
AR × q = p × q
This means that
AR
Average ost (AC )
is equal to pri e, with
is given by total
p
or
given by the demand fun tion.
ost, divided by the number of units produ ed, that is
AC =
From this follows that if average
AR = p.
TC
.
q
ost is known, then total
number of units produ ed, that is
T C = AC × q.
77
ost is given by average
ost multiplied by the
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Examples
1. A perfe tly
1 demand fun tion is given by
ompetitive rm's
p = 20,
and we nd that
T R = pq = 20q.
From this follows that marginal revenue and average revenue are
MR =
This means that
d
20q = 20
dq
and
AR =
TR
20q
=
= 20.
q
q
M R = AR = p.
2 is given by
2. The demand fun tion for a monopolist
p = 50 − 2q.
From this demand fun tion we nd that
T R = pq = (50 − 2q)q = 50q − 2q 2 .
Marginal revenue and average revenue are
MR =
d
(50q − 2q 2 ) = 50 − 4q
dq
and
From the graph in Figure 6.1, we see that the slope of
AR =
MR
50q − 2q 2
= 50 − 2q.
q
is double the slope of
AR.
AR
MR
50
40
30
20
AR = 50 − 2q = p
10
M R = 50 − 4q
5
10
15
Figure 6.1: A monopolist's
3. Given that the average
ost fun tion for a
AR
20
and
1
MR
fun tions
ertain produ t is
AC = 2q + 5 +
we nd the total
q
25
30
,
q
ost fun tion to be
T C = AC × q = 2q 2 + 5q + 30.
A perfe tly ompetitive rm hooses output levels to maximise prots, that is where M C = p.
In a monopolisti market only one ompany may oer produ ts and servi es to the publi . This is the opposite of a perfe tly
ompetitive market, in whi h an innite number of rms operate.
2
78
DSC1520/SG001
By simply looking at this total
ost fun tion, we nd that xed and variable
F C = 30
Marginal
ost fun tion, that is
d
T C = 4q + 5.
dq
MC =
The
V C = 2q 2 + 5q.
and
ost is the derivative of the total
osts are
ost to produ e one additional unit of the produ t, after 50 units have been produ ed, is
M C(50) = 4(50) + 5 = R205.
A tivity
1. The demand fun tion for a monopolist is
q = 120 − 3p.
(a) Find expressions for
(b) Find the value of
q
T R, M R
for whi h
and
AR.
AR = 0.
Does it make sense for the monopolist to sell this number
of units? Explain.
2. A rm's xed
osts amount to R1 000 and their variable
(a) Write down the total
(b) Find the marginal
of marginal
ost fun tion and
ost fun tion and
ost in this
al ulate total
(b) Find the
TC
de reases as
MC
q = 20.
when
q = 20.
q
10
.
q
in reases. Use dierentiation.
osts.
ost fun tion. Comment on the relationship between
4. The average revenue for a
Des ribe the meaning
ost fun tion:
fun tion and dedu e xed and variable
( ) Find the marginal
ost when
al ulate the value of
AC = 50 +
AC
V C = 3q .
ase.
3. A rm has the following average
(a) Show that
ost is
TC
and
M C.
ertain rm is given by
AR = 180 − 12q.
(a) Write down the total revenue fun tion and nd the marginal revenue fun tion.
(b) Comment on the relationship between
AR
5. Consider the following total revenue and total
T R = 200q − 4q 2
and
M R.
ost fun tions of a rm:
and
TC =
q3
− 12q 2 + 164q + 100.
3
(a) Derive the formulae for marginal and average revenue.
(b) Derive the formulae for marginal and average
ost.
( ) Find the rm's prot fun tion and determine the prot if 20 units are produ ed and sold.
79
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6.1.4 Produ tion fun tions
During any produ tion pro ess, inputs are transformed into units of output.
There may be a variety of
inputs, su h as following:
l
k
r
te
labour number of workers employed, either per day or per hour
physi al
apital - buildings, ma hinery, et .
raw materials
te hnology, in luding information te hnology
A produ tion fun tion gives the relationship between input and output. The general form of su h a fun tion
states the level of output,
q,
that depends on the amounts of inputs used in the produ tion pro ess, that is
q = f (l, k, r, te , . . .).
For this module we assume that the inputs
fun tion of labour, l , that is
As before, we nd the
3
k, r, te , . . .
are xed , therefore the level of output is only a
q = f (l).
marginal produ t of labour
by dierentiating the labour fun tion, that is
MPL =
dq
.
dl
In e onomi s, the marginal produ t of labour (M P L) is the
hange in output that
results from employing an additional worker.
We also nd the
average produ t of labour, AP L,
labour units, l , that is
to be the produ tion fun tion divided by the number of
q
AP L = .
l
The average produ t of labour (AP L) is a measure of the average amount of output
ea h worker
an produ e.
Example
Suppose we have the daily produ tion fun tion
q = 15l2 − 0,5l3 ,
where
l
is the number of workers employed per day.
We nd the marginal produ t of labour by dierentiation, that is
MPL =
d
d
q = (15l2 − 0,5l3 ) = 30l − 1,5l2 .
dl
dl
When ten workers are employed per day, that is
l = 10,
then
M P L = 30(10) − 1,5(10)2 = 150.
This means that, when ten workers are employed per day, employing an additional worker will in rease daily
produ tion at the rate of 150 units.
3
This is normally the ase in the short run.
80
DSC1520/SG001
The average produ t of labour is
AP L =
q
15l2 − 0,5l3
=
= 15l − 0,5l2 .
l
l
When ten workers are employed per day, then
AP L = 15(10) − 0,5(10)2 = 100.
This means that the average produ tivity per worker is 100 units of output per day.
The total daily output of the rst ten workers is
q = AP L × l = 100 × 10 = 1 000 units.
A tivity
A daily produ tion fun tion is given as
where
l
1
q = 225l − l3 ,
3
is the number of workers employed per day.
1. Find the marginal produ t of labour when
l = 5.
What does this value mean?
2. Find the average produ t of labour fun tion when
l = 5.
What is the meaning of this value?
6.2 Optimisation of e onomi fun tions
In Chapter 5.3, we found the turning point(s) of a non-linear fun tion to be where the rst derivative is equal
to zero. Furthermore, the nature of su h turning point(s) is determined by nding the se ond derivative at
the point(s). If
f ′′ (x) > 0,
f ′′ (x) < 0,
it is a minimum point, and when
We now apply this to e onomi
fun tions like
T R, T C
and
it is a maximum point.
P.
Examples
1. The demand fun tion for a good is given as
p = 100 − 2q,
where
p
is the pri e per unit and
q
is the number of units produ ed and sold.
From the demand fun tion, we nd the total revenue fun tion to be
T R = price per unit × number of units
=p·q
= (100 − 2q)q
= 100q − 2q 2 .
The marginal revenue is given by
MR =
The
TR
d
d
TR =
(100q − 2q 2 ) = 100 − 4q.
dq
dq
fun tion has a turning point where
M R = 0,
100 − 4q = 0
81
that is where
or
q = 25.
DSC1520/SG001
The se ond derivative is
T R′′ (q) = −4.
Sin e this is less than zero, the turning point is indeed a maximum point.
This means that when 25 units are sold, total revenue is a maximum of
T R(25) = 100(25) − 2(25)2 = R1 250.
2. The demand fun tion for a
ertain produ t is given by
p = 24 − 6 ln q.
From the demand fun tion, we nd the total revenue fun tion to be
T R(q) = pq = (24 − 6 ln q)q = 24q − 6q ln q.
To nd the number of units that should be produ ed and sold to maximise total revenue, we set
M R = T R′ (q) = 0,
with
d
q ln q
dq
d
d
= 24 − 6
q · ln q + q ln q
dq
dq
1
= 24 − 6 ln q + q
q
= 24 − 6 ln q − 6
T R′ (q) = 24 − 6
(product rule)
= 18 − 6 ln q.
Setting this equal to zero, gives
18 − 6 ln q = 0
3 − ln q = 0
ln q = 3
q = e3 = 20,09 ≈ 20 units.
Sin e the se ond derivative of
T R,
T R′′ (q) =
for all
q > 0, T R
d
6
(18 − 6 ln q) = − < 0
dq
q
is an in reasing fun tion.
3. The output over time for a rm is given by the fun tion
Q(t) =
where
t is in
years and
Q is the
t3
− 5t2 + 9t + 90,
3
output in hundreds of units produ ed. We need to determine the years
when output is a maximum and when it is a minimum, that is the turning points of
The rst derivative of
Setting
Q′ = 0,
Q
is
Q′ (t) = t2 − 10t + 9.
we nd by fa torisation
t2 − 10t + 9 = (t − 9)(t − 1) = 0.
82
Q.
DSC1520/SG001
Therefore,
Q
has turning points at
t=9
and
t = 1,
with
Q(9) = 9
and
Q(1) = 94,33.
The se ond derivative is
d 2
(t − 10t + 9) = 2t − 10.
dt
′′
This gives at t = 1, Q (1) = 2 − 10 = −8 < 0, indi ating a maximum
′′
Q (9) = 18 − 10 = 8 > 0, indi ating a minimum turning point.
Q′′ (t) =
We
turning point, and at
t = 9,
an therefore say that output will be a maximum in year one and it will be a minimum in year
nine.
This is
onrmed by the graph in Figure 6.2.
Figure 6.2:
Q(t) =
t3
3
− 5t2 + 9t + 90
4. A monopolist has a demand fun tion
p = 152,5 − 3q,
and
osts given by
T V C = 0,5q 3 − 15q 2 + 175q
and
T F C = 300.
The monopolist's total revenue fun tion is
T R = pq = 152,5q − 3q 2 ,
while his total
ost fun tion is
T C = T V C + T F C = 0,5q 3 − 15q 2 + 175q + 300.
From these, we nd the prot fun tion as
P = T R − T C = (152,5q − 3q 2 ) − (0,5q 3 − 15q 2 + 175q + 300) = −0,5q 3 + 12q 2 − 22,5q − 300.
To determine the output for whi h prot is a maximum or minimum, we use the derivatives
P ′ (q) = −1,5q 2 + 24q − 22,5
and
P ′′ (q) = −3q + 24.
Set the rst derivative of the prot fun tion equal to zero and solve for
q,
P ′ (q) = −1,5q 2 + 24q − 22,5 = 0,
−2
By using the minus
b
whi h is simplied by multiplying by
formula with
a = 3, b = −48 and c = 45, we
√
−b ± b2 − 4ac
q=
p2a
48 ± (−48)2 − 4(3)(45)
=
2(3)
48 ± 42
=
6
= 15
or
1.
83
nd
that is
to get
3q 2 − 48q − 45 = 0.
DSC1520/SG001
q=1
The prot fun tion therefore has turning points when
Sin e
P ′′ (1) = −3(1) + 24 = 21 > 0,
Sin e
P ′′ (15) = −3(15) + 24 = −21 < 0,
This is
onrmed by the graph of
P
Figure 6.3:
Note:
We
and when
p = 1,
prot is a minimum at
prot is a maximum at
q = 15.
with
q = 15,
P (1) = −311.
with
P (15) = 375.
in Figure 6.3.
P (q) = −0,5q 3 + 12q 2 − 22,5q − 300
ould also have used the fa t that prot has turning points where
M R = M C.
By
dierentiation, we nd
MR =
d
T R = 152,5 − 6q
dq
and
MC =
d
T C = 1,5q 2 − 30q + 175.
dq
Equating these gives
152,5 − 6q = 1,5q 2 − 30q + 175
This is the same equation as
5. The demand and total
P ′ (q)
− 1,5q 2 + 24q + 22,5 = 0.
or
above, giving the same results.
ost fun tions for a rm are given by
p = 50 − 2q
and
T C = 160 + 2Q.
From this, we nd the total revenue fun tion to be
T R = pq = (50 − 2q)q = 50q − 2q 2 .
From Se tion 3.2, we know that break-even o
urs when total revenue is equal to total
therefore nd the break-even points for the rm by setting
T R = T C,
ost. We
an
that is
50q − 2q 2 = 160 + 2Q
−2q 2 + 48q − 160 = 0
(6.1)
2
q − 24q + 80 = 0
(q − 20)(q − 4) = 0,
q = 20
(multiply by − 2)
or
q = 4.
The rm therefore breaks even when either four, or 20 units of the good are produ ed and sold.
Figure 6.4 shows the
TR
and
TC
fun tions, the points where break-even o
urs and the areas where
prot or loss is made.
T R > T C , that is between the break-even
where T C > T R, that is when less than four
We see that prot is made everywhere in the areas where
points. On the other hand, loss is made over the area
units or more than 20 units are sold.
84
DSC1520/SG001
TR
TC
TR
300
Prot
200
Loss
100
Loss
Break-even points
q = 4 and q = 20
5
Figure 6.4:
From the
TR
TC
and
TC
10
15
T R = 50q − 2q 2
20
and
q
25
T C = 160 + 2q
fun tions, we nd the prot fun tion to be
P = 50q − 2q 2 − (160 + 2q) = −2q 2 + 48q − 160.
We nd the turning point of
P
where
P ′ (q) = −4q + 48 = 0
Sin e
P ′′ (q) = −4 < 0,
that is at
q = 12.
prot is indeed a maximum at this point, with
P (12) = 128.
If we need to sket h the prot fun tion, we already know that the turning point is at (12; 128) and the
y
inter ept is at
(0; −160).
We also need the
x
axis inter epts, whi h we nd by setting
P (q) = −2q 2 + 48q − 160 = 0.
This equation is the same as setting
q = 20.
TR = TC
as in Equation 6.1 above, whi h resulted in
These are the break-even points at (4; 0) and (20; 0) on the graph of
P
(12; 128)
100
50
0
−50
−100
5
10
15
20
q
Break-even points
q = 4 and q = 20
−150
Figure 6.5:
P (q) = −2q 2 + 48q − 160
85
P
q=4
in Figure 6.5.
and
DSC1520/SG001
A tivity
1. Mary is a vendor at a ea market. She sells hand-dyed T-shirts over weekends. Her demand and total
ost fun tions are
p = 240 − 10q
and
T C = 120 + 8q.
(a) Write down the total revenue and prot fun tions.
(b) Find the number of T-shirts that should be sold to maximise prot.
( ) Find the
MC
(d) Plot the
TR
and then
and
and
MR
TC
fun tions, and show that
MR = MC
when prot is a maximum.
fun tions on the same graph. From the graph, estimate the break-even points
onrm your answer algebrai ally. Also show the area on the graph where prot is made.
Explain what these values mean to Mary.
(e) Plot the
M R and M C fun
tions on the same graph. What is signi ant about the point of interse tion
of these fun tions?
2. The average
ost and revenue fun tions for a brand of water bottle are given as
AC = 15 +
8 000
q
and
AR = 25.
(a) Write down the fun tions for total revenue (T R), total
marginal
ost (T C ), marginal revenue (M R) and
ost (M C ).
(b) How many water bottles should be produ ed and sold to break even?
( ) Write down the prot fun tion. Is it possible to nd the maximum prot? Explain this by using
dierentiation and the
3. A
MR
ompany supplies kit hen
and
MC
fun tions.
abinets. Their demand and total
p = 5 504 − 0,8q
(a) Determine the pri e and quantity of
well as
MR
and
M C.
and
ost fun tions are given as
T C = 608 580 + 120q.
abinets for whi h prot is maximised, using dierentiation as
Give the number of
abinets to be supplied, the pri e per
prot to be made when prot is a maximum.
(b) Find the break-even points.
4. A rm's average revenue fun tion is given by
AR =
(a) Find the
TR
and
MR
50
.
e0,5q
fun tions.
(b) Determine the number of units to be produ ed and sold to maximise revenue.
5. The total
ost of produ ing a
ertain good is given by
T C = 120 ln(q + 10).
(a) Find the total xed
(b) Find the marginal
ost fun tion,
T F C.
ost fun tion.
( ) How many units should be produ ed to minimise total
86
ost?
abinet and the
DSC1520/SG001
6.3 Elasti ity of demand non-linear demand fun tions
In Se tion 2.3 the
on ept of elasti ity was introdu ed for linear demand fun tions.
se tion to refresh your understanding of the
Please
onsult this
on ept of elasti ity.
Point elasti ity of demand for linear fun tions is dened as
εd =
1 p
∆q p
· =− · .
∆p q
b q
For non-linear demand fun tions, we use derivatives, so elasti ity is given by
εd =
dq p
· .
dp q
dp
dq
dq or dp , depending on the format of the fun tion.
dp
dq
in terms of q , nd the derivative
dq and invert it to nd dp , whi h is needed
The derivative of a demand fun tion is given by either
If the demand fun tion gives
p
for the denition of elasti ity. We may to do this, be ause
1
dq
= dp .
dp
dq
Depending on the
absolute value
of the elasti ity
oe ient at a
ertain pri e, denoted by
ases may be identied:
⊲
If
|εd | > 1
at a
|εd | < 1
at a
|εd |4 , the
following
elasti
at that pri e, meaning that if the pri e in reases
inelasti
at that pri e, meaning that if the pri e in reases
ertain pri e, then demand is
by 1%, demand de reases by more than 1%.
⊲
If
ertain pri e, then demand is
by 1%, demand de reases by less than 1%.
⊲
If
|εd | = 1,
⊲
If
|εd |
is
we have unitary elasti ity. When pri e in reases by 1%, demand de reases by 1%.
onstant for all posible pri es, and
onstant elasti
Note:
Dierentiation
or inelasti .
|εd | > 1
|εd | < 1,
or
an also be used for a linear demand fun tion,
then we say demand is respe tively
q = a − bp,
sin e the derivative,
∆q
∆p , whi h is needed to determine elasti ity.
gives the slope of the demand fun tion, whi h is the same as
Examples
1. Consider the demand fun tion
dp
= −2q.
dq
p = 60 − q 2 , with
For the elasti ity formula, we need
dq
dp . We therefore invert the derivative to nd
1
dq
=
.
dp
−2q
Elasti ity of demand is therefore given by
εd =
Transforming the demand fun tion
4
1 p
p
dq p
· = − · = − 2.
dp q
2q q
2q
p = 60 − q 2 ,
q 2 = 60 − p
we nd
or
q=
p
60 − p.
Absolute value is the magnitude of a quantity, irrespe tive of sign. For example, | − 2| = 2.
87
dq
dp ,
DSC1520/SG001
Then,
p
p
.
=−
2
−2q
2(60 − p)
εd =
When the pri e is R45 per unit, that is at
p = 45,
εd = −
Sin e
|ε| = 1,5 > 1,
elasti
demand is
45
= −1,5.
2(60 − 45)
at the pri e of R45. This means that if the pri e in reases by
1%, demand de reases by 1,5%.
2. For the exponential demand fun tion
q = 45e−0,04p ,
we nd
dq
= 45(−0,04)e−0,04p = −1,8e−0,04p .
dp
Elasti ity of demand is found to be
εd =
dq p
·
dp q
= −1,8e−0,04p ·
p
45e−0,04p
= −0,04p.
At
p = 10, |εd | = 0,4 < 1.
This indi ates that demand is
inelasti
when the pri e is R10 if the pri e
in reases by 1%, demand de reases by 0,4%.
3. The derivative of the demand fun tion
q=
200
= 200p−2
p2
is
d
200p−2 = (−2)200p−3 = −400p−3 .
dp
Therefore,
εd =
Sin e
|εd | = 2 > 1,
p
−400p−2
dq p
· = −400p−3 ·
=
= −2.
dp q
200p−2
200p−2
demand is elasti
everywhere this is
4. The demand for one-litre bu kets of i e
onstant elasti ity.
ream is given by
p = 50 − 0,5q
or
q = 100 − 2p.
From the demand fun tion, we nd the total revenue fun tion in terms of
q
to be
T R = pq = q(50 − 0,5q) = 50q − 0,5q 2 .
From this, marginal revenue is found to be
MR =
and average revenue is
AR =
d
T R = 50 − q
dq
50q − 0,5q 2
TR
=
= 50 − 0,5q.
q
q
We nd the pri e and quantity at whi h revenue is a maximum by setting
M R = 50 − q = 0,
that is at
q = 50,
88
which gives
p = 50 − 0,5(50) = 25.
DSC1520/SG001
dq
dp
Sin e
= −2,
we nd the
oe ient of pri e elasti ity of demand in terms of
εd =
and in terms of
εd =
p = 25,
to be
dq p
p
p
p
· = −2 ·
=−
=
,
dp q
100 − 2p
50 − p
p − 50
q,
At maximum revenue,
p
50 − 0,5q
100
dq p
· = −2 ·
=1−
.
dp q
q
q
and
25
= | − 1| = 1,
25 − 50
|εd | =
whi h indi ates unitary elasti ity, meaning that if the pri e in reases by 1%, demand de reases by 1%.
A tivity
1. Consider the demand fun tion
q = 25 − 3p.
(a) Find the
oe ient of elasti ity for the demand fun tion.
(b) Cal ulate elasti ity of demand when the pri e is R6,00 and
omment on the meaning of the value
you nd.
2. Find the
oe ient of elasti ity for the demand fun tion
q = 80 − 10 ln p
and
omment on the elasti ity at pri e R60.
3. The demand for ti kets to the So
er World Cup is given by
q = 192 − p2 ,
where
q
is the number of ti kets bought and
p
is the pri e per ti ket in hundreds of rand.
(a) Find the expression for pri e elasti ity of demand in terms of
(b) Cal ulate pri e elasti ity of demand if ti kets
( ) If ten seats be ome available when
the new pri e.
4. The demand fun tion for a
εd = −1,
p.
ost R800 ea h. What does this mean?
determine the per entage
hange in pri e. Cal ulate
ertain wine is given by
p = 1 500e−0,025q ,
where
p
is the pri e per bottle of the wine and
q
is the number of bottles demanded.
(a) Find the marginal revenue fun tion for the wine.
(b) Determine pri e and quantity at whi h
TR
is a maximum.
( ) Find pri e elasti ity at maximum revenue.
89
PART IV
INTEGRATION
90
Chapter 7: Integration theory
Learning obje tives
After you have
⊲
⊲
⊲
⊲
⊲
ompleted this
explain the
hapter, you should be able to
on ept of integration as the reverse of dierentiation
apply the power rule of integration to integrate given
omposite fun tions
apply integration by substitution to integrate given fun tions
understand the
on ept of denite integration
apply integration to determine the area between a
urve and the
x
axis over a given interval
7.1 Integration as the reverse of dierentiation
In mathemati s, we often en ounter operations that reverse ea h other dire tly, like addition and subtra tion;
and multipli ation and division. In a similar way, integration is the reverse of dierentiation. This means
that when a fun tion
f (x) = x2
is differentiated to find f ′ (x) = 2x,
then, when we integrate the result, it will again give the original fun tion, that is
Z
A s hemati
′
f (x) dx =
Z
2x dx = x2 = f (x).
representation of this relation is shown in Figure 7.1.
Dierentiate
d
f (x)
dx
f ′ (x)
f (x)
Z
f ′ (x) dx
Integrate
Figure 7.1: Integration reverses dierentiation
7.2 Integration rules
7.2.1 The power rule for integration
Dierentiating a fun tion su h as
f (x) = x3
by using the power rule of dierentiation, gives
f ′ (x) =
d 3
x = 3x3−1 = 3x2 .
dx
91
DSC1520/SG001
Integration reverses this operation. Therefore, to get
x3
3x2 ,
from
we need to add one to the exponent of
x
and divide by three, that is
Z
f ′ (x) dx =
Z
Now, suppose we need to integrate the fun tion
3x2+1
= x3 .
3
3x2 dx =
f (x) = x5 .
This means that we should nd the fun tion
5
whose derivative is x .
We know that
d 6
x = 6x5 ,
dx
whi h is six times the answer we need. We therefore need to divide by six to get
Z
Sin e the derivative of any
ould
ontain a
x5 dx =
x5 .
Thus,
x6
.
6
1
onstant is zero, the fun tion that we have to dierentiate to nd the integral ,
onstant and the integration would not be
orre t.
Example:
d 3
(x + 2) = 3x2 ,
dx
We
d 3
(x + 200) = 3x2
dx
and
d 3
(x − 2 000) = 3x2
dx
an therefore not determine through integration whether the fun tion should
ontain a
onstant, or what
the value should be. It may be zero, but it may also be something else.
The result of integration should therefore always in lude a
Z
2
3
3x dx = x + c
onstant. Our two examples above should be
Z
and
x5 dx =
x6
+ c.
6
In general, the power rule for integration is given by
Z
xn dx =
xn+1
+ c.
n+1
Note the following rules for working with the power rule for integration:
1. The integral of any
onstant term,
Z
2. The integral of a
k,
k dx =
is
Z
kx + c
(when integrating with regard to
k × x0 dx = k ×
x).
Sin e
x0 = 1,
x0+1
+ c = kx + c.
0+1
onstant multiplied by a variable term, is the
onstant multiplied by the integral of
the variable term, that is
Z
kf (x) dx = k
Z
f (x) dx.
3. The integral of the sum of a number of terms is the sum of the integrals of the terms, that is
1
Z
(f (x) + g(x)) dx =
Z
f (x) dx +
Z
g(x) dx.
The result of integrating a fun tion is alled the integral, while the fun tion that is being integrated is alled the integrand.
92
DSC1520/SG001
4. We know that
Therefore, the integral
Applying the power rule to
1
x
whi h is not dened.
1
d
ln x = .
dx
x
Z
Z
1
−1
dx = ln |x| + c.
x dx =
x
R
R
= x−1 results in division by zero, x1 dx = x−1 dx =
5. The integral of the natural exponential fun tion
Z
f (x) = ex ,
x−1+1
−1+1
+c =
x0
0
=?,
is the fun tion itself, that is
ex dx = ex + c.
Examples
1. The integral of
2. The integral of
f (x) = x3 + 3x2 + x + 1 is
Z
x3+1
x2+1
x1+1
x0+1
f (x) dx =
+3
+
+
+c
3+1
2+1 1+1 0+1
x2
x4
+ x3 +
+ x + c.
=
4
2
√
f (x) = 2x3 + x − x12 is
Z
Z
Z
Z
1
1
−2
3
3
2
2
(2x + x − x ) dx = 2 x dx + x dx − x−2 dx
1
x−2+1
x3+1
x 2 +1
−
=2×
+ 1
+c
3+1
−2 + 1
2 +1
3
=2
x−1
x4 x 2
+ 3 −
+c
4
−1
2
3
1
x4 2x 2
+
+ + c.
=
2
3
x
3. The integral of
1
3
f (x) = √ + is
x x
Z 1
1 + 3
x
x2
1
dx =
Z
1
(x− 2 + 3x−1 ) dx
1
x− 2 +1
+ 3 ln x + c
= 1
−2 + 1
1
=
x2
1
2
+ 3 ln x + c
√
= 2 x + 3 ln x + c.
4. It is always best to simplify a fun tion before attempting to integrate it.
fun tion
f (x) =
Simplifying
f
gives
f (x) =
1
x
+5
x + 5x2
.
x2
and
Z 1
+ 5 dx = ln x + 5x + c.
x
93
Consider for instan e the
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5. The integral of
f (x) = 3ex + 5
Z
6. Simplifying
f (x) =
is
x
(3e + 5) dx = 3
et + e2t
et
Z
x
e dx +
Z
5 dx = 3ex + 5x + c.
et e2t
+ t = 1 + et . Now,
et
e
Z t
Z
e + e2t
dt = (1 + et ) dt = t + et + c.
et
gives
f (x) =
A tivity
Integrate the following fun tions:
1.
f (x) = x + x3 + x3,5
2.
f (x) =
3.
f (x) =
4.
f (x) = 2x +
5.
f (q) = q(q −
√1
x5
√
x+ x
x
1
2x
√
q)
7.2.2 Integration by substitution
To integrate
Consider the
omposite fun tions, we need to reverse the
hain rule for dierentiation.
2
omposite fun tion
f (x) = (4x + 3)3 .
If
4x + 3
is seen as an entity, say we set
be integrated with regard to
u
u = 4x + 3,
to
x,
u = 4x + 3
u3 du =
By substituting
4x + 3
Z
by
u
whi h
an
u4
+ c.
4
also needs to be taken into a
ount. If we dierentiate
we nd
du
= 4,
dx
2
f (u) = u3 ,
to nd
Z
However, the inner fun tion
then we have the simple fun tion
which results in
and
3
dx
by
du = 4 dx
dx =
du
.
4
du
4 , we get
du
u3
4
Z
1
=
u3 du
4
1 u4
+c
=
4 4
(4x + 3)4
+ c.
=
16
(4x + 3) dx =
or
Z
(substitute)
(substitute back)
Refer to Se tion 5.2.3 on page 61.
94
u with
regard
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Examples
1. Consider the fun tion
f (x) =
To integrate
f,
we set
u = 2x − 1
du
=2
dx
10
.
2x − 1
and dierentiate to get
or
du = 2 dx,
which gives
dx =
du
.
2
dx =
du
.
5
dx =
du
.
3
Now,
Z
10
dx
2x − 1
Z
10 du
=
u 2
Z
1
du
=5
u
= 5 ln u + c
f (x) dx =
Z
= 5 ln(2x − 1) + c.
2. To integrate the fun tion
f (x) = e5x−2 ,
we set
u = 5x − 2
and dierentiate to nd
du
=5
dx
or
du = 5 dx,
which gives
Now,
Z
5x−2
e
du
eu
5
Z
1
=
eu du
5
1
= eu + c
5
1 5x−2
= e
+ c.
5
dx =
Z
3. To integrate
f (x) =
we set the inner fun tion to be
du
=3
dx
u = 3x + 4
or
√
3x + 4,
and dierentiate to nd
du = 3 dx,
95
which gives
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Integrating
f
gives
√
Z
Z
3x + 4 dx =
1
(3x + 4) 2 dx
1 du
u2
3
Z
1
1
=
u 2 du
3
=
Z
=
1 u2
+c
3 32
3
3
12
(3x + 4) 2 + c
33
2p
(3x + 4)3 + c.
=
9
=
4. For the integration
Z
we set
u = 8q + 1
1
dq,
8q + 1
and dierentiate to nd
du
=8
dq
or
du = 8 dq,
which gives
dq =
du
.
8
Now,
Z
1
dq =
8q + 1
1 du
u 8
Z
1
1
du
=
8
u
Z
=
1
ln u + c
8
=
1
ln(8q + 1) + c.
8
5. To integrate the fun tion
f (x) =
√
1
x+1+ √ ,
x
we rst simplify to nd
1
1
f (x) = (x + 1) 2 + x− 2 .
The rst term is a
omposite fun tion, with inner fun tion
du
= 1, which gives
dx
96
u = x + 1.
du = dx.
This is dierentiated as
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Now,
Z
f (x) dx =
=
=
Z
Z
1
(x + 1) 2 dx +
1
1
u 2 du +
x2
1
2
Z
1
x− 2 dx
+c
3
u2
1
+ 2x 2 + c
3
2
3
1
2
= (x + 1) 2 + 2x 2 + c
3
√
2p
(x + 1)3 + 2 x + c.
=
3
6. For the integration
we set
u = 5 − 8y
and dierentiate to nd
du
= −8 or
dy
2
dy,
5 − 8y
Z
du = −8 dy, which gives
dy = −
du
.
8
Now,
Z
2
dy =
5 − 8y
2
du
×−
u
8
Z
2
1
=−
du
8
u
Z
1
= − ln |u| + c
4
1
= − ln |5 − 8y| + c.
4
A tivity
Integrate ea h of the following fun tions:
1.
f (x) =
√
3
2.
p(q) =
1
2−7q
3.
f (x) = 10e2−5q
4.
f (t) =
1
e2t
9x − 5
+ 2et+2 + 3
7.3 The denite integral and the area under a urve
When it is ne essary to nd the area under a
urve, one
an nd an approximation by dividing the area into
intervals, determining the areas of the re tangles under the
In Figure 7.2(a) the area under the
∆x
and height
y.
urve between
x=a
urve and adding them.
and
x=b
is divided into four intervals of width
The total area is then approximated by the sum of the areas of the four re tangles, that is
Area ≈
b
X
x=a
97
y∆x.
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We see that there are large areas between the
urve and the re tangles not being taken into a
ount in
al ulating the total area.
When
∆x
is halved, as in Figure 7.2(b), the areas omitted in the summation be ome smaller.
y = f (x)
y = f (x)
a
∆x
∆x
∆x
∆x
∆x ∆x ∆x ∆x ∆x ∆x ∆x ∆x
x
b
(a) Area divided into four intervals
Figure 7.2: Area under
a
urve approximated by sum of re tangles
∆x
When we halve the intervals again and again, making
sum of the areas of the re tangles (
∆x
Pb
a y∆x) gets
smaller and smaller, the area represented by the
loser and
loser to the true area under the
be omes innitely small (tends to zero), we nd the exa t area under the
urve, between
This innite summation is represented mathemati ally by the integration symbol
urve between
a
and
b
x
b
(b) Area divided into eight intervals
Z
a
urve. When
and
b.
b
and the area under the
a
is given by
Area =
Z
b
f (x) dx.
a
The area shown in Figure 7.3 is the exa t area between the
urve
f
and the
x
axis, from point
a
to
b.
f (x)
Z
b
f (x) dx
a
a
Figure 7.3: Area under
b
urve determined exa tly by integration
This area is found mathemati ally by integrating the fun tion
integral
F
at the upper bound,
x = b,
x
f
as usual, and then nding the value of the
minus the value at the lower bound,
98
x = a.
This is
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Area =
Z
b
b
f (x) dx = F (x)
a
a
= F (b) − F (a).
Consider for example the fun tion
f (x) = x2 − 2x + 3.
We nd the area under this parabola between
Z
The area between the
Z
1
5
(x2 − 2x + 3) dx =
urve
f
and the
(x2 − 2x + 3) dx =
x
x=1
x=5
and
by rst integrating
f
normally to nd
x3
x2
x3
− 2 + 3x + c =
− x2 + 3x + c.
3
2
3
axis from
x=1
to
x3
2
− x + 3x + c
3
5
is determined as
5
1
(5)3
3
(1)
− (5)2 + 3(5) + c −
− (1)2 + 3(1) + c
3
3
= (41,67 − 25 + 15 + c) − (0,33 − 1 + 3 + c)
=
= 31,67 − 2,33
= 29,34.
onstant disappears (c
Note that the
The graph of
f
− c = 0).
and the area determined above are shown in Figure 7.4.
f (x)
15
10
5
1
2
Figure 7.4:
3
4
x
5
f (x) = x2 − 2x + 3
When the area between the graph of a fun tion and the
x
axis is above the
x
axis, the area is positive, but
when it falls below the axis, it is negative.
The graph in Figure 7.5 shows the fun tion
f (x) = x2 − 1.
The following
al ulations show that it does not matter whether we
al ulate the areas separately and add
them together, or integrate over the entire area, the result is the same.
* The area between the
urve and the
Z
0
1
x
0 ≤ x ≤ 1 is under the x axis
1 1
x3
2
=
−x
−1 −0=− .
3
3
3
0
axis over
(x2 − 1) dx =
99
with
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f (x)
4
3
2
1
**
0
−2
* 1
−1
−1
f (x) = x2 − 1
Figure 7.5:
1≤x≤2
** The area over
Z
2
1
2
(x − 1) dx =
is
x3
−x
3
2
=
1
1
8 6
23
2
1 3
2
4
−2 −
−1 =
−
−
−
= − −
= .
3
3
3 3
3 3
3
3
3
Adding these results gives the net area between
Sin e
f
is symmetri
x
2
about the
y
Therefore, the total net area over
f
and the
x
axis, the total net area to the left of the
−2 ≤ x ≤ 2
is
2×
2
3
=
y axis as − 23 + 34 = 23 .
2
over −2 ≤ x ≤ 0 is also .
3
axis to the right of the
y
axis
4
3.
The same result is found by simply integrating over the interval, that is
Z
2
−2
2
(x − 1) dx =
x3
−x
3
2
=
−2
−8
8
2
2
2 −2
4
−2 −
− (−2) = 2 − 2 − −2 + 2 = −
= .
3
3
3
3
3
3
3
Examples
1. To nd the area under the
urve
f (x) = 40e2x−3
x = 0 and x = 2, we rst integrate f normally.
du
u = 2x − 3 and dierentiate to nd du
dx = 2 so that dx = 2 .
between
Z
40e2x−3 dx =
Z
40eu
Sin e
f
2
2
40e2x−3 dx = 20e2x−3
0
0
= 20(e2(2)−3 − e2(0)−3 )
= 20(e1 − e−3 )
= 20(2,668)
= 53,37.
The graph of
f
omposite fun tion, we set
du
= 20eu + c = 20e2x−3 + c.
2
The area is therefore
Z
is a
Now,
is shown in Figure 7.6(a).
100
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2. To nd the area under the
normally to nd
Z The area is therefore
Z
1
0,2
1
x +
x
2
1
x between
f (x) = x2 −
urve
1
x +
x
2
dx =
dx =
x3
+ ln x
3
(1)3
x = 0,2
and
x = 1,
we rst integrate
f
x3
+ ln x + c.
3
1
0,2
(0,2)3
=
+ ln(1) −
+ ln(0,2)
3
3
= (0,3333 + 0) − (0,0027 − 1,6094)
= 0,3333 + 1,6067 = 1,94.
The graph of
f
is shown in Figure 7.6(b).
(a) f (x) = 40e2x−3
(b) f (x) = x2 + x1
Figure 7.6: Areas between
f
and
x
axis
A tivity
Evaluate ea h of the following denite integrals. (Work to two de imal pla es.)
1.
Z
3
(x + 5) dx
1
10
x+5
dx
2
10
x2 + 8x
dx
x3
1,3
4.
Z
10
dx
x
5.
Z
8p
2.
Z
3.
Z
1
3
0,3
1
6.
Z
1
q + 8 dq
500e0,4x dx
0
7. Consider the fun tion
f (q) = 16 − q 2 .
(a) Plot the fun tion over the interval
(b) Cal ulate the net area between
( ) Find the net area between
above and below the
q
f
f
0 ≤ q ≤ 10.
and the
and the
q
q
axis over the interval
axis over the interval
axis separately.
101
0 ≤ q ≤ 10
0 ≤ q ≤ 10
by using integration.
by
al ulating the areas
Chapter 8: Appli ations of integration
Learning obje tives
After you have
⊲
⊲
⊲
ompleted this
hapter, you should be able to
apply integration to nd total revenue and
ost fun tions from marginal revenue and
apply integration to determine total quantity, given the rate of
apply denite integration to determine
ost fun tions
hange in the quantity
onsumer surplus, produ er surplus and total surplus
8.1 From marginal to total fun tions
In Chapter 6 marginal fun tions were dened as the derivative of total fun tions.
integration is the reverse of dierentiation. Total
ost and total revenue fun tions
We also know that
an therefore be found by
integrating marginal fun tions.
8.1.1 Total ost from marginal ost
The marginal
ost for a
ertain produ t is given by the derivative of the total
M C = T C ′ (q) =
where
We
q
10
,
q
is the number of units produ ed. It is also known that the total
an therefore nd the total
ost fun tion by integrating the marginal
TC =
Z
ost fun tion, that is
ost of produ ing 100 units is R500.
ost fun tion, that is
M C(q) dq
10
dq
q
= 10 ln q + c.
=
It is given that
T C = 500
when
q = 100.
Z
We therefore nd the
onstant,
c,
by setting
10 ln 100 + c = 500
to get
c = 500 − 10 ln 100 = 453,95.
The total
ost fun tion is therefore
T C = 10 ln q + 453,95.
8.1.2 Total revenue from marginal revenue
As for total
ost, marginal revenue is found by dierentiating the total revenue fun tion, that is
MR =
d
T R.
dq
102
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Suppose we know that the marginal revenue fun tion for a good is
M R = 100 − 4q,
where
q
Now, we
is the number of units sold, and that total revenue from selling 25 units is R1 250.
an nd the total revenue fun tion by integrating the marginal revenue fun tion, that is
TR =
=
Z
Z
M R dq
(100 − 4q) dq
4q 2
+c
2
= 100q − 2q 2 + c.
= 100q −
From the information given,
T R = 1 250
when
q = 25.
Therefore, we nd the value of
c
as follows:
100(25) − 2(25)2 + c = 1 250
2 500 − 1 250 + c = 1 250
c = 1 250 − 2 500 + 1 250 = 0.
The total revenue fun tion is therefore
T R = 100q − 2q 2 .
8.1.3 Rate of hange to total quantity
Pra ti al situations are often des ribed in terms of the rate at whi h a quantity
hanges over time, for
example
⊲
⊲
⊲
⊲
the rate at whi h the demand for a good
the rate at whi h the volume of sales
hanges over a period of time
hanges
the rate at whi h natural resour es are depleted
the rate at whi h pollutants invade the environment
We know that the rate at whi h a fun tion
hanges, is given by the derivative (the slope) of the fun tion,
that is
Rate of change in T R =
d
T R.
dq
Examples
1. At a small tool-hire
ompany, the estimated rate of in rease in the maintenan e
given by
C(t) = 2 + t1,5 ,
where
C
is in rand and
Total maintenan e
t
is time in weeks.
osts during the rst ve weeks (0
Z
≤ t ≤ 5)
5
(2 + t1,5 ) dt
0
t2,5 5
= 2t +
2,5 0
TC =
52,5
−0
2,5
= R32,36.
= 2(5) +
103
are given by
ost of power drills is
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During the following ve weeks (5
< t ≤ 10), total maintenan e osts amount
t2,5 10
T C = 2t +
2,5 5
(5)2,5
(10)2,5
− 2(5) +
= 2(10) +
2,5
2,5
= 146,49 − 32,36
to
= R114,13.
2. The annual rate at whi h parran is
onsumed at an informal settlement is given by
d
K = 1 560e0,012t ,
dt
where
when
K is the amount
t = 0, no parran
To nd the
of paran
is
onsumed per year (in litres) and
onsumed (K
t
is in years. We assume that
= 0).
onsumption fun tion, we use integration to nd
K=
Z
(1 560e0,012t ) dt
e0,012t
+c
0,012
= 130 000e0,012t + c.
= 1 560
The value of
c
is found by using the assumed values, that is
0 = 130 000e0 + c or
The
c = −130 000.
onsumption fun tion is therefore
K = 130 000e0,012t − 130 000.
The amount of paran
onsumed during the rst ten years is given by
K=
Z
0
10
d
K dt.
dt
10
= 130 000e0,012t
0
0,012(10)
= 130 000 e
− e0,012(0)
= 130 000(0,1275)
= 16 574,6 litres.
To nd the amount
onsumed during the next ten years we integrate over the interval
that is
K=
Z
20
10
3. The marginal
10 < t ≤ 20,
d
K dt = 130 00 e0,012(20) − e0,012(10) = 18 694 litres.
dt
ost of a good is given by
M C = −q 2 + 80q,
where
q
is the number of units produ ed. It is also given that the xed
104
ost of produ tion is R500.
DSC1520/SG001
From the marginal fun tion we nd the total
TC =
Z
ost fun tion to be
(−q 2 + 80q)d q = −
From the given information we dedu e that
q2
q3
+ 80 + c = −0,33q 3 + 40q 2 + c.
3
2
c = 500.
Therefore,
T C = −0,33q 3 + 40q 2 + 500.
The total
ost of produ ing su
essive units from
q =3
to
q = 12,
is found by integrating over this
interval, that is
TC =
Z
12
(−q 2 + 80q) dq
3
12
= −0,33q 3 + 40q 2
3
3
= −0,33(12) + 40(12)2 − (−0,33(3)3 + 40(3)2 )
(fixed costs cancel out)
= 5 189,76 − 351,09
= 4 838,67.
A tivity
1. The marginal revenue fun tion for a rm is given by
M R = 120 − 2q,
where
q
is the number of units sold.
(a) Find the total revenue fun tion if it is given that
TR = 0
when
q = 0.
(b) Cal ulate the rm's total revenue when 50 units are sold.
2. The rate at whi h Johnnie
an memorise a sequen e of items is given by
N ′ (t) =
where
t
10
,
1+t
is in minutes.
(a) Sket h the rate of memory retention roughly and
omment on it.
(b) Find the fun tion of the total number of items memorised over an interval in time, given that at
time
t=0
no items have been memorised.
( ) Cal ulate the number of items memorised over the rst and se ond ten minutes.
3. The marginal
ost and revenue fun tions for a
M C = 80
where
q
ertain produ t are
and
M R = 200 − 20q,
is the number of units produ ed and sold.
(a) Find the prot fun tion, given that xed
(b) Determine the value of
q
osts amount to R500 and total revenue is zero when
q = 0.
for whi h prot is a maximum.
8.2 Consumer and produ er surplus
In Se tion 3.3, the
on epts of
onsumer surplus and produ er surplus for linear demand and supply fun tions
were dis ussed. In this se tion we
onsider
onsumer surplus and produ er surplus for non-linear fun tions.
105
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8.2.1 Consumer surplus
As stated before,
onsumer surplus,
CS ,
is the dieren e between the total amount that
willing to spend on a good or servi e (indi ated by the demand
onsumers are
urve) and the total amount that they
a tually spend (at market pri e, that is at equilibrium).
p0 and q0 , onsumer surplus is represented graphi ally by the area en losed
q = 0 to q = q0 , minus the area of the re tangle representing revenue at
market equilibrium), p0 × q0 .
If market pri e and quantity are
by the demand fun tion, from
market pri e and quantity (at
Consumer surplus is therefore given by
CS = Area under demand curve − area of rectangle
Z q0
(demand function) − p0 q0 .
=
0
Examples
1. Consider the demand fun tion
p = 60 − 2q.
It is known that market equilibrium,
E0 ,
is at pri e
p0 = 12
and quantity
q0 = 24.
Figure 8.1 shows
onsumer surplus graphi ally:
p
60
40
CS
20
E0
p0
5
10
15
Figure 8.1: Consumer surplus for
We determine
CS
20
q0
25
p = 60 − 2q
30
at
q
p0 = 12
by subtra ting the area of the re tangle at equilibrium from the area of the triangle
under the demand line, as was done in Se tion 3.3. We
CS =
=
Z
Z
an also use integration as follows:
q0
(demand function)dq − p0 q0
0
24
0
(60 − 2q)dq − 12 × 24
= 60q − q
24
2
0
− 288
= (60(24) − (24)2 ) − (60(0) − (0)2 ) − 288
= 864 − 288 = 576.
2. Consider the quadrati
demand fun tion
p = 100 − q 2 .
106
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p
100
CS
50
E0 (8; 36)
p0
2
4
Figure 8.2: Consumer surplus for
Market equilibrium is given to be at
q0 = 8.
In Figure 8.2 the demand fun tion and
The area is
q0
6
8
q
10
p = 100 − q 2
at
q0 = 8
Substituting this into the demand fun tion, gives
p0 = 36.
onsumer surplus are shown graphi ally.
al ulated algebrai ally as follows:
Z
8
(100 − q 2 )dq − 36 × 8
0
q3 8
= 100q −
− 288
3
0
CS =
= 629,33 − 288 = 341,33.
3. The area representing
onsumer surplus for the hyperboli
p=
with market equilibrium at
q0 = 3
and
p0 =
100
3+2
demand fun tion
100
,
q+2
= 20,
is shown as the shaded area in Figure 8.3.
p
50
CS
E0 (3; 20)
p0
1
q0
2
Figure 8.3: Consumer surplus for
107
q
3
p=
100
q+2 at
q=3
DSC1520/SG001
We nd
CS
algebrai ally as follows:
CS =
Z
3
0
100
dq − p0 q0
q+2
3
0
= 100 ln(q + 2) − 20 × 3
= 100[ln(3 + 2) − ln(0 + 2)] − 60
d
(q + 2) = 1
dx
= 100(0,9163) − 60
= 31,63
8.2.2 Produ er surplus
Produ er surplus is the dieren e between the revenue the produ er re eives for
pri e per unit,
p0 ,
and the revenue they are willing to a
ept for su
q0
units of a good at market
essive units over the interval (0; q0 ).
P S = (revenue at market price) − (acceptable revenue at a lower price)
= p0 q0 − (area under supply function over (0; q0 ))
Z q0
(supply function)dq
= p0 q 0 −
0
Examples
1. For the linear supply fun tion
p = 20 + 4q,
produ er surplus is given by the shaded area in Figure 8.4.
p
40
p0 = 36
30
20
10
E0
PS
(4; 36)
Revenue a eptable
to produ er
1
2
3
Figure 8.4: Produ er surplus for
108
q0
4
5
p = 20 + 4q
at
q0 = 4
DSC1520/SG001
Cal ulating
PS
by using integration gives
P S = Area of rectangle − area under curve
Z 4
(20 + 4q) dq
= 36 × 4 −
0
4q 2 4
= 144 − 20q +
2
0
= 144 − (20(4) + 2(4)2 − 0)
= 144 − 122
= 32.
2. Produ er surplus for the supply fun tion
p = q 2 + 6q
at market equilibrium
E0 = (q0 ; p0 ) = (4; 40)
is
al ulated as
Z
4
(q 2 + 6q) dq
0
3
6q 2 4
q
+
= 40 × 4 −
3
2
0
3
(4)
2
= 160 −
+ 3(4) − 0
3
= 160 − (21,33 + 48)
P S = p0 q 0 −
= 90,67.
A graphi al representation of this situation is shown in Figure 8.5.
p
40
p0
E0
(4; 40)
30
PS
20
Revenue
a eptable
to produ er
10
1
2
3
Figure 8.5: Produ er surplus for
4
q0
q
5
p = q 2 + 6q
at
q0 = 4
8.2.3 Total surplus
To measure total e onomi
welfare, we add the
onsumer surplus to the produ er surplus to arrive at the
total surplus.
For example,
onsider the demand and supply fun tions for a good given respe tively by
pd = 300e−0,2q
and
109
ps = 2e0,8q .
DSC1520/SG001
We nd the equilibrium pri e and quantity by setting
pd = ps ,
that is
300e−0,2q = 2e0,8q
150e−0,2q = e0,8q
e0,8q
= 150
e−0,2q
e0,8q e0,2q = 150
eq = 150
q0 = ln 150 = 5,01.
Substituting this value into the demand (or the supply) fun tion, results in
p0 = 300e−0,2(5,01) = 300e−1,002 = 110,14.
Consumer surplus at market equilibrium is
CS =
Z
q0
(pd ) dq − p0 q0
0
=
Z
5,01
0
(300e−0,2q ) dq − 110,14 × 5,01
300e−0,2q 5,01
− 551,80
−0,2 0
300 −0,2(5,01)
(e
− e0 ) − 551,80
=
−0,2
= −1 500(0,367 − 1) − 551,80
=
= 949,28 − 551,80
= 397,48.
Produ er surplus at market equilibrium is found to be
Z
q0
ps dq
Z
= 110,14 × 5,01 −
P S = p0 q 0 −
0
2e0,8q
= 551,8 −
0,8
5,01
(2e0,8q ) dq
0
5,01
0
= 551,8 − 2,5(e4,008 − e0 )
= 551,8 − 2,5(55,04 − 1)
= 416,7.
Total surplus in the market is the sum of
CS
and
P S,
that is
397,48 + 416,7 = 814,18.
The total shaded
area in Figure 8.6 represents the total surplus in the market.
A tivity
1. Sket h ea h of the following demand fun tions, shade the area representing
al ulate
(a)
CS
to two de imal pla es:
pd = 100 − q 2
at
q0 = 8
110
onsumer surplus and
DSC1520/SG001
p
300
200
pd = 300e−0,2q
CS
0,8q
E0 ps = 2e
p0
100
PS
q0
1
Figure 8.6: Produ er and
(b)
pd =
100
q+1 at
2
3
4
5
q
onsumer surplus at market equilibrium.
q0 = 9
2. Sket h ea h of the following supply fun tions, shade the area representing produ er surplus and
late
PS
to two de imal pla es:
1
q+1 at
(a)
ps = 10 −
(b)
ps = q 2 + 4
at
q0 = 4
q0 = 5
3. The demand and supply fun tions for a good are given as
pd = 100 − 0,5q
and
ps = 10 + 0,5q .
(a) Determine pri e and quantity at equilibrium.
(b) Cal ulate
onsumer surplus, produ er surplus and total surplus at equilibrium.
111
al u-
Chapter 9: Solutions to a tivities
9.1 Se tion 2.1 (Revenue, ost and prot fun tions)
1a
The total
ost fun tion:
T C = F C + V C = 250 + 25q .
1b T C(20) = 250 + 25 × 20 = R750.
1
Substituting into the formula
q = 46.
TC = FC + V C
we nd
1 400 = 250 + 25q ,
whi h results in
25q = 1 150
or
They provided 46 lessons.
2a T C = 1 000 + 15q
and
T R = 35q .
2b T C(400) = 1 000 + 15(400) = R7 000.
2
When
2d
Prot fun tion:
When
T R = 1 750,
q = 100,
then
35q = 1 750
or
q=
= 50.
P = T R − T C = 35q − (1 000 + 15q) = 20q − 1 000.
then prot is
P (100) = 20(100) − 1 000 = R1 000.
3a T C = F C + V C = 900 + 30q , T R = 60q
3b T R = 60q = 4 200
4a
1 750
35
so that
and
P = T R − T C = 60q − (900 + 30q) = 30q − 900.
q = 70.
From the given information we nd that
F C = 1 000
and
V C = 15.
The total weekly
ost fun tion is
therefore given by
T C(q) = F C + V C = 1 000 + 15q,
where
4b
q
is the number of
From the total
al ulators produ ed per week.
T C axis is 1 000. So, when
T C = 1 000+15×100 = 2 500.
ost fun tion, we see that the slope is 15 and the inter ept on the
nothing is produ ed, the
ost is R1 000; and when 100
al ulators are produ ed,
The graph is shown in Figure 9.1.
TC
(100; 2 500)
2500
2000
1500
1000
500
10 20 30 40 50 60 70 80 90 100
Figure 9.1:
4
The total
ost of produ ing 25
al ulators is
q
T C = 1 000 + 15q
T C(25) = 1 000 + 15(25) = 1 375.
112
DSC1520/SG001
4d
To determine the number of
7 000 = 1 000 + 15q
4e
15q = 6 000,
Total revenue is given by the number of
T R fun
T R = 0, that
To graph the
are sold,
whi h gives
we solve for
q
in the equation
al ulators sold/demanded (q ) multiplied by the pri e per
al u-
T R = 35q.
lator. Therefore,
4f
T C = 7 000,
q = 400.
al ulators produ ed when
and nd that
tion for
q = 0 to 100,
is the
oordinate (0; 0).
we need to nd two points on the line. When no
al ulators
T R = 3 500,
giving the
When 100
al ulators are sold,
oordinate (100; 3 500).
The graph of
TR
is shown in Figure 9.2.
TR
(100; 3 500)
3500
3000
2500
2000
1500
1000
500
10 20 30 40 50 60 70 80 90 100
Figure 9.2:
4g
To nd the number of
q = 50.
4h
Therefore, 50
When 80
al ulators sold when
T R = 35q
T R = 1 750,
4i
we set
1 750 = 35q ,
from whi h it follows that
al ulators are sold.
al ulators are sold,
T C(80) = 1 000 + 15(80) = 2 200
Sin e
q
TR > TC
we
The prot when
on lude that revenue ex eeds
q = 80
is
and
T R(80) = 35(80) = 2 800.
osts, so a prot is made.
P (80) = T R(80) − T C(80) = 2 800 − 2 200 = R600.
9.2 Se tion 2.2 (Appli ations: demand, supply, ost, revenue)
1a
When
1b
The demand fun tion is graphed in Figure 9.3.
p = 0,
then
q = 64,
and when
q = 0,
then
p=
64
4
= 16,
giving the
oordinates
(0; 64)
and
(16; 0).
q
64
16
Figure 9.3: Demand fun tion
1
When the pri e
p
in reases by R1, demand
q
p
q = 64 − 4p
de reases by four rides per hour. (The slope is
113
−4.)
DSC1520/SG001
1d
Writing the demand fun tion
p = 16 − 14 q = 16 − 0,25q
2a
Writing the supply fun tion
results in
2b
.
q = 64 − 4p
with
p = 0,25q + 22,5
p
on the left-hand side gives
q
with
as the subje t, we nd
q = −90 + 4p.
To graph the demand and supply fun tions, we nd the
fun tion
(60; 0).
q = 210 − 3,5p,
For the supply fun tion
oordinates
(0; −90)
and
when
p = 0, q = 210
and when
4p = 64 − q ,
resulting in
0,25q = −22,5 + p,
whi h
oordinates on the axes for ea h. For the demand
q = 0, p = 60,
giving the
oordinates
(0; 210)
p = 0,25q + 22,5, when p = 0, q = −90 and when q = 0, p = 22,5,
(22,5; 0). The fun tions are graphed on the same axes as follows:
and
giving the
q
210
180
150
120
90
(40; 70)
60
30
0
−30
10 20 30 40 50 60 70
p
−60
−90
Note:
Sin e negative pri es and quantities do not make sense, only the quadrant where both
positive are shown in demand and supply graphs. The
p
and
q
are
oordinate on the negative axis only assists in drawing
the line.
2
The point of interse tion
simultaneously.
with
3a
q = 70.
an either be read o the graph or it
1 This point is at
(40; 70),
whi h says that when pri e
p = 40,
We have two distin t points on the supply line, namely
We rst nd the slope
d,
and
(q2 ; p2 ) = (90; 110).
The
110 − 60
50
p2 − p1
=
=
= 1,25.
q2 − q1
90 − 50
40
p axis (c), we
60 = c + 1,25(50),
Now, to nd the inter ept on the
p = c + 1,25q ,
(q1 ; p1 ) = (50; 60)
p = c + dq .
that is
d=
that is
substitute one of the given points, say (50; 60), into the
from whi h it follows that
is therefore
3b
demand and supply are equal
(When demand and supply are equal, we say the market is in equilibrium.)
equation of the supply fun tion is in the form
equation
an be found by solving the fun tions
c = −2,5.
The supply fun tion
p = −2,5 + 1,25q.
The number of T-shirts that will be supplied additionally for ea h R1 in rease in pri e is given by the slope
of the supply fun tion with
to have
q
q
as subje t. We therefore need to transform the equation from
at the left, that is
1,25q = p + 2,5 or
p = −2,5 + 1,25q
q = 0,8p + 2.
Therefore, for ea h R1 in rease in pri e, 0,8 additional T-shirts are supplied.
3
When the pri e is R85, then
3d
When 120 T-shirts are supplied, the pri e is
1
See Se tion C.5 on page 163.
q(85) = 0,8(85) + 2 = 70
T-shirts are supplied.
p(120) = −2,5 + 1,25(120) = R147,50
114
per T-shirt.
DSC1520/SG001
9.3 Se tion 2.3 (Elasti ity of linear demand and supply fun tions)
1a
From the given demand fun tion,
by
q = 250 − 5p,
εd = −b ·
1b
|εd | =
2
3
< 1,
b=5
and point elasti ity of demand is given
p
p
p
p
= −5 ·
=−
=
.
q
250 − 5p
50 − p
p − 50
Using the formula derived in (a) with
Sin e
we see that
p = 20,
we nd
we say that demand is inelasti .
εd =
20
20−50
= − 32 .
This means that a
hange in pri e will not have a
signi ant ee t on demand. In fa t, if the pri e is in reased by 1%, demand will de rease by 0,667%.
When
Sin e
30
30−50
= − 23 .
p = 30,
then
εd =
3
2
> 1,
we say that demand is elasti . This means that a
|εd | =
hange in pri e will have a signi ant
ee t on demand. In fa t, if the pri e is in reased by 1%, demand will de rease by 1,333%.
1
We need the values of demand at these pri es, that is
Ar
εd = −b ·
Sin e
2a
If
|εd | > 1,
p = 90 + 0,05q ,
p is
then
2
Using the formula derived in (a) with
if
Ar
with
d = 20
25 ≤ p ≤ 35.
( he k for yourself ). Point elasti ity of supply in
p
p
p
= 20 ·
=
.
q
−1 800 + 20p
p − 90
p = 70,
we nd that
εs =
70
70−90
= −3,5.
Sin e
at pri e R70. When the pri e in reases by 1%, supply will in rease by 3,5%.
+p2
εs = d · pq11 +q
. From the supply fun
2
when p = 60, then q = −600. Therefore,
elasti ity of supply is given by
p = 40,
q = −1 000,
then
and
εs = 20 ·
Sin e
over the interval
q = −1 800 + 20p
εs = d ·
elasti
p1 + p2
25 + 35
= −5 ·
= −1,5.
q1 + q2
125 + 75
demand is on average elasti
terms of
2b
q(25) = 250−125 = 125 and q(35) = 250−5×35 = 75.
elasti ity of demand is therefore
|εs | > 1,
supply is elasti
tion
|εs | > 1,
q = 20p − 1 800,
supply is
we nd that
100
40 + 60
= 20 ·
= −1,25.
−1 000 − 600
−1 600
over the given pri e interval.
9.4 Se tion 3.1 (Equilibrium and break-even)
1a
The demand fun tion
ps = −40 + 8qs
1b
pd = 800 − 2qd is transformed
qs = 5 + 0,125ps .
to
transformed is
To nd the equilibrium, we set
pd = ps
qd = 400 − 0,5pd
and the supply fun tion
to nd
800 − 2q = −40 + 8q
−2q − 8q = −800 − 40
−10q = −840
q = 84.
At equilibrium
q = 84
rings are supplied/demanded. When we substitute this into
the equilibrium pri e to be
p = 800 − 2(84) = R632.
115
pd = 800 − 2q ,
we nd
DSC1520/SG001
1
When the pri e
p = 720,
whi h is higher than the equilibrium pri e, then Lindiwe will supply
qs − qd = (5 + 0,125(720)) − (400 − 0,5(720)) = 95 − 40 = 55
more rings than the demand at this pri e.
1d
The level of ex ess demand when
P = 560
is:
Qd − Qs = (400 − 0,5(560)) − (5 + 0,125(560))
= 120 − 75
= 45.
If the pri e per ring is lowered to R560,
2a
ustomers will demand 45 more rings.
The demand and supply fun tions with
l
as subje t, are
ld = 17,5 − 0,25wd
2b
At equilibrium,
wd = ws ,
and
ls = −5 + 0,5ws .
that is
70 − 4ld = 10 + 2ls
−4l − 2l = 10 − 70
−6l = −60
l = 10.
Therefore, at equilibrium, ten labourers are working. When we substitute
the equilibrium wage to be
2
w = 70 − 4(10) = R30
The ex ess demand for labour when
w = 20
l = 10
into
per hour.
wd = 70 − 4l,
we nd
is
ld − ls = (17,5 − 0,25(20)) − (−5 + 0,5(20))
= 12,5 − 5
= 7,5.
If the wages are lowered to R20, fewer workers will be willing to work and
needed to work.
2d
The ex ess supply for labour when
w = 40
7,5 ≈ 8
more workers will be
is
ls − ld = (−5 + 0,5(40)) − (17,5 − 0,25(40))
= 15 − 7,5
= 7,5.
If wages are raised to R40, then eight more workers will be willing to work.
9.5 Se tion 3.1.4 (Complementary and substitute goods)
1a
To nd equilibrium, equate the demand and supply fun tions of ea h produ t and solve for
pX
and
pY ,
that is
qdX = qsX
190 − 2pX − 2pY = −10 + 2pX
−4pX − 2pY = −200
2pX + pY = 100,
116
(9.1)
DSC1520/SG001
and
qdY = qsY
240 − 2pX − 4pY = −40 + pY
−2pX − 5pY = −280
2pX + 5pY = 280.
From Equation 9.1 we nd that
pY = 100 − 2pX .
(9.2)
Substituting this into Equation 9.2 gives
2pX + 5(100 − 2pX ) = 280
(9.3)
2pX + 500 − 10pX = 280
(9.4)
−8pX = −220
(9.5)
pX = 27,5,
(9.6)
and
pY = 100 − 2(27,5) = 45.
The pri es at equilibrium are therefore R2 750 for pit hing wedges and R4 500 for putters.
1b
At equilibrium,
qX = −10 + 2pX = −10 + 2(27,5) = 45
pit hing wedges and
qY = −40 + pY = −40 + 45 = 5
putters should be sto ked.
9.6 Se tion 3.1.5 (Taxes and subsidies)
1
To nd the equilibrium, we set
pd = s d
to nd
p = 140.
Graphi ally the equilibrium is determined as
E0
200 − 5q = 92 + 4q
or
−9q = −108.
This gives
q = 12
and
in Figure 9.4.
p
ps = 101 + 4qs
200
ps = 92 + 4qs
ET
150
b
b
E0
100
pd = 200 − 5qd
50
10
20
30
40
q
Figure 9.4: Equilibrium before and after tax
2a
The supplier pays the tax, so the supply fun tion be omes
2b
Equilibrium is found when
200 − 5q = 101 + 4q
The equilibrium after tax is shown as
2
After tax, the
−9q = 99,
whi h gives
or
q = 11
ps = 101 + 4qs .
and
p = 145.
in Figure 9.4.
ustomer pays R145, whi h is R5 more than before tax. The supplier re eives the pri e after
tax minus the tax, that is
5:4 between
ET
or
ps − 9 = 92 + 4qs ,
145 − 9 = R136,
whi h is R4 less than before tax. The tax is therefore distributed
ustomer and supplier.
117
DSC1520/SG001
9.7 Se tion 3.2 (Break-even analysis)
1a
It is given that the pri e per unit
Fun tions for total revenue and total
p = 30,
T R = p × q = 30q
At the break-even point,
T R = T C,
xed
osts
F C = 200
and variable
ost
VC = 5
per unit.
ost therefore are
so
and
T C = F C + V C = 200 + 5q.
30q = 200 + 5q ,
whi h gives
25q = 200
or
q = 8.
The rm should therefore produ e and sell eight units to break even.
1b
At break-even,
2a
Total revenue
2b
Break-even o
T R = 30(8) = 240
and
T C = 200 + 5(8) = 240,
whi h are equal, as expe ted.
T R = number of watches sold × price per watch = q × 6,60 = 6,6q .
urs when
T C = T R,
that is when
800 + 0,2q = 6,6q, which gives 6,4q = 800 or q = 125.
2
Suppose the pri e at break-even is given by
TR = TC
gives
P =
832
160
P,
then
= R5,20.
T R = P × 160
and
T C = 800 + 0,2 × 160.
Setting
9.8 Se tion 3.3 (Consumer and produ er surplus)
1a
Demand and supply fun tions are given as
pd = 58 − 0,2q
and
ps = 4 + 0,1q .
58 − 0,2q = 4 + 0,1qQ, whi h results in 0,3q = 54 or q0 = 180. Substituting
p0 = 22. This means that at equilibrium 180 seats on the bus will be sold for
At equilibrium,
this into either
fun tion gives
R22 ea h.
Figure 9.5 shows equilibrium,
onsumer surplus and produ er surplus for
pd = 58 − 0,2q
and
p
A = 58
CS
p0 = 22
E0
PS
B=4
q0 = 180
Figure 9.5: Market equilibrium;
1(b)i
At equilibrium
onsumers pay
p0 × q0 = 22 × 180 = R3 960.
1(b)iii CS = 7 200 − 3 960 = R3 240.
At equilibrium, the bus
q
onsumer surplus and produ er surplus
1(b)ii p0 × q0 + 0,5 × 22 × (58 − 22) = 3,960 + 3 240 = R7 200.
1( )i
290
ompany re eives R3 960.
1( )ii 180 × 4 + 0,5 × 180 × (22 − 4) = 720 + 1 620 = 2 340.
1( )iii P S = 3 960 − 2 340 = 1 620.
118
ps = 4 + 0,1q .
DSC1520/SG001
9.9 Se tion 3.4 (Linear programming)
1a
Constraint (1) goes through the points (0; 5) and (4; 0);
onstraint (2) goes through (0; 6) and (3,2; 0);
onstraint (3) goes through (0; 7) and (3; 0). These are plotted in Figure 9.6.
y
7
6
5
4
(3)
A
B
3
2
(1)
C
1
D
1
2
(2)
3
x
4
Figure 9.6: Feasible area
1b
Coordinates at A are (0, 5) and D is at (3; 0).
At B, inequalities (1) and (2)
ross, whi h is where
5x + 4y = 20
and
15x + 8y = 48
interse t. From (1) we
5
5
nd that y = 5 − x. Substituting this into (2) gives 15x + 8(5 − x) = 48, resulting in
4
4
5×1,6
8
or x =
= 3. The oordinates of B are (1,6; 3).
5 = 1,6, so that y = 5 −
4
15x + 40 − 10x = 48
ross, whi h is where 15x + 8y = 48 and 7x + 3y = 21 interse t. From (3)
y = 7 − 37 x. Substituting into (2) gives 15x + 8(7 − 37 x) = 48, resulting in 15x − 56
3 x = −8 or
that y = 1,91. The oordinates of C are (2,18; 1,91).
At C, inequalities (2) and (3)
we nd that
x = 2,18,
1
so
The obje tive fun tion's value at the
orner points are at A
P = 10,00;
P = 9,00. The maximum prot is found at point B, giving
x and 3 000 units of y for a maximum prot of R10 800.
and at D
units of
2
If the de ision variables are
20a + 30b.
The inequality
a
and
b
the optimal solution as produ ing 1 600
(3)
be the number of kilograms of Brand X and
The obje tive fun tion is to minimise
The inequality
P =
(2)
y
the number of kilograms of Brand Y the gardener
should pur hase.
3b
P = 10,36;
(1)
2a + 2b ≤ 160
x
at C
for ma hines A and B, respe tively, then the prot fun tion is
2a + 4b ≤ 280
Let
P = 10,80;
onstraints are the following:
6a + 3b ≤ 360
3a
at B
ost, that is Minimise Cost
= 180x + 60y .
onstraints are the following:
20x + 4y ≥ 80
(nitrogen)
2x + y ≥ 15
(potassium)
x + 2y ≥ 10
(iron)
119
DSC1520/SG001
y
A
20
15
(1)
B
10
5
(2)
(3)
C
2
4
D
6
8
x
10
Figure 9.7: Corner points, feasible area and iso ost lines
3
The inequalities, the feasible area and the iso ost lines are shown in Figure 9.7. The
orner points are
A(0; 20), B(1,67; 11,65), C(6,67; 1,66) and D(10; 0). The solution is found at point B, where 1,67 kilogram of
Brand X and 11,65 kilogram of Brand Y should be pur hased at minimum
ost of
180(1,67) + 60(11,65) = R999,60.
9.10 Se tion 4.1 (Quadrati fun tions in e onomi s)
1
If the robots are pur hased, the total
ost fun tion is
T C = 200 000 + 300p.
The demand and total revenue fun tions remain the same as
q = 2 500 − 2,5p
and
T R = 2 500 − 2,5p2 .
The prot fun tion now be omes
P = TR − TC
= 2 500p − 2,5p2 − (200 000 + 300(2 500 − 2,5p))
= 2 500p − 2,5p2 − 200 000 − 750 000 + 750p
= −2,5p2 + 3 250p − 950 000.
Here,
a = −2,5, b = 3 250
and
c = −950 000.
pm =
The pri e that maximises prot is
−b
−3 250
=
= R650,
2a
2(−2,5)
whi h is R50 per unit less than before. The maximum prot at this pri e is
P = −2,5(650)2 + 3 250(650) − 950 000 = R106 250.
We would advise ACE to pur hase the robots, sin e their prot will be
120
106 250 − 105 000 = R1 250
higher.
DSC1520/SG001
2
Total
ost is
al ulated as xed
osts plus variable
ost, therefore
T C = 100 × 150 + 50q = 15 000 + 50q.
To nd the demand fun tion, we need two points on the line. It is given that when rent is R200, all the
ars are rented out, that is the point
rented out drops by ten. Therefore,
The slope of the demand fun tion
(p1 ; q1 ) = (200; 100). Also, when the rent goes up
(p2 ; q2 ) = (220; 99) is also on the demand line.
qd = a − bp
b=
by R20, the number
is
99 − 100
−1
=
= −0,05,
220 − 200
20
giving the demand fun tion to be
q = a − 0,05p.
To nd the value of
a,
we use the point (200; 100) to nd
demand fun tion is therefore
100 = a − 0,05(200) = a − 10, giving a = 110.
The
q = 110 − 0,05p.
The total revenue fun tion is given by
T R = pq = p(110 − 0,05p) = 110p − 0,05p2
and the prot fun tion is
P = TR − TC
= 110p − 0,05p2 − (15 000 + 50q)
= 110p − 0,05p2 − (15 000 + 50 (110 − 0,05p))
= 110p − 0,05p2 − 20 500 + 2,5p
= −0,05p2 + 112,5p − 20 500.
This is a quadrati
fun tion with
a < 0,
p=
If the rent is R1 125, then
to rent out half a
−112
112,5
−b
=
=
= 1 125.
2a
2(−0,5)
0,1
q = 110 − 0,5(1 125) = 53,75
ar, we round the number of
The prot when the rent is R1 125 per
This means that if they rent out 54
3
therefore it has a maximum turning point. The vertex is at
ar is
ars will be rented out per day. Sin e it is impossible
ars to 54.
P (1 125) = −0,05(1 125)2 + 112,5(1 125) − 20 500 = 42 781,25.
ars per day for R1 125 ea h, they will make a daily prot of R42 781,25.
The demand and supply fun tions are shown graphi ally in the following gure:
121
DSC1520/SG001
To nd the equilibrium, equate the demand and supply fun tions to nd
q 2 − 0,5 = −q 2 + 4
2q 2 = 4,5
p
q = ± 2,25 = 1,5
A negative quantity does not make sense, so
4a
or
− 1,5.
q = 1,5, giving the equilibrium pri
e to be
p = (1,5)2 −0,5 = 1,75.
To plot the fun tions, we rst need to simplify them and then nd
pd = −(q + 4)2 + 100 = −(q 2 + 8q + 16) + 100 = −q 2 − 8q + 84 and ps = (q + 2)2 = q 2 + 4q + 4.
These fun tions are plotted in Figure 9.8. Equilibrium seems to be at the point (4; 35).
Figure 9.8:
4b
To determine the
pd = −q 2 − 8q + 84
and
ps = q 2 + 4q + 4
ompany's equilibrium pri e and quantity, we equate the demand and supply fun tions
to nd
−q 2 − 8q + 84 = q 2 + 4q + 4
−2q 2 − 12q + 80 = 0
q 2 + 6q − 40 = 0
(q + 10)(q − 4) = 0
(divide by − 2)
(factorise)
q = −10
or
q = 4.
A negative number of items does not make sense, therefore we
and sold for
5a
p=
(4)2
+ 4(4) + 4 = R36
on lude that four items should be produ ed
to break even.
Total revenue is given by the pri e per unit, times the number of units sold, that is
We
an also nd
TR
in terms of
q
p2
p
= 600p − .
T R = pq = p 600 −
8
8
by transforming the demand fun tion:
p
q = 600 −
8
p
= 600 − q
8
p = 4 800 − 8q.
Now,
T R = pq = (4 800 − 8q)q = 4,800q − 8q 2 .
122
DSC1520/SG001
5b
Here,
T R = −8q 2 + 4 800q
is a quadrati
fun tion with a negative
oe ient of
q2,
whi h indi ates that
the graph has a maximum turning point. This point is at
q=
−b
−(4 800)
4 800
=
=
= 300,
2a
2(−8)
16
where
T R = −8(3002 ) + 4 800(300) = 720 000.
The
ompany therefore needs to produ e and sell 300 units of the produ t to get a maximum revenue of
R720 000.
Figure 9.9 shows that the turning point of the total revenue fun tion is indeed at
Figure 9.9:
5
Prot is
q = 300
units.
T R = 8q 2 + 4 800q
al ulated as the dieren e between total revenue and total
ost, that is
P = TR − TC
= −8q 2 + 4 800q − (800 − 120q + 5q 2 )
= (−8 − 5)q 2 + (4 800 + 120)q − 800
= −13q 2 + 4 920q − 800.
5d
To break even, total revenue should be equal to total
ost. This is also the point where prot is zero. We
therefore nd the break-even point by setting the prot fun tion equal to zero, that is
P = −13q 2 + 4 920q − 800 = 0.
This is a quadrati
fun tion with
q=
=
=
=
a = −13, b = 4 920
and
c = −800.
By using the formula as before, we nd
√
b2 − 4ac
2a p
−4 920 ± 4 9202 − 4(−13)(−800)
2(−13)
−4 920 ± 4 915,77
−26
−4 920 − 4 915,77
−4 920 + 4 915,77
= 0,16
or
q=
= 378,3,
−26
−26
−b ±
whi h means that prot is zero (or break-even is rea hed) when either no units are produ ed, or when 378
units are produ ed and sold.
This is
onrmed by the graphs of the
T R, T C
and prot (P ) fun tions in Figure 9.10.
123
DSC1520/SG001
(a) T R and T C
(b) Prot
Figure 9.10: Break-even where
TR = TC
and/or
P =0
9.11 Se tion 4.4 (Exponential and logarithmi fun tions)
1a
In 1995,
t=0
and
P (0) = 125,5e0,012(0) = 125,5,
whi h means that there were 125 500 people older than
60.
In 2020,
t = 25
and
P (25) = 125,5e0,012(25) = 206,9,
whi h means that there are 206 900 people older than
60.
1b
Figure 9.11 shows the growth fun tion from 1995 to 2095.
Figure 9.11: Unlimited growth fun tion
This is an unlimited growth fun tion, meaning the number of people older than 60 will in rease at a faster
rate as time goes by.
1
We need to nd
t
when
P (t) = 1 000,
that is
125,5e0,012t = 1000
e0,012t = 7,968
0,012t = ln 7,968
2,075
= 172,95.
t=
0,012
Therefore, after 173 years this population will be one million.
2a
After one week,
After 52 weeks,
2b
t = 1,
and
S(1) = 200 000(1 − e−0,05 ) = 9 754
S(52) = 200 000(1 − e−0,05×52 ) = 185 145
magazines.
magazines.
Graph 9.12 shows the sales of magazines for 104 weeks.
The sales will grow at a de reasing rate with an upper limit of 200 000 magazines.
124
DSC1520/SG001
Figure 9.12: Sales of magazines
3a
Initially,
t = 0,
with
After ten years, when
3b
P (1) =
6 000
1+29e−0,4
t = 10, P (10) =
≈ 294
sh.
6 000
1+29e−0,4(10)
= 3 918
sh.
Graph 9.13 shows the growth of the sh population.
Figure 9.13: Fish population
3
The population will be 4 000 when
6 000
1 + 29e−0,4t
−0,4t
4 000(1 + 29e
) = 6 000
4 000 =
1 + 29e−0,4t = 1,5
1,5 − 1
= 0,017
e−0,4t =
29
−0,4t = ln 0,017
−4,06
t=
= 10,15,
−0,4
that is after ten years.
9.12 Se tion 4.5 (Hyperboli fun tions)
1a
Figure 9.14 shows the graph of this fun tion. The
ar depre iates very fast over the rst three years and
after that the rate of depre iation de reases.
125
DSC1520/SG001
Figure 9.14: Value fun tion
1b
The value will be R200 000 when
200 = 1 +
840
1 + 2t
199(1 + 2t) = 840
199 + 398t = 840
840 − 199
= 1,6 years.
t=
398
2a
Figure 9.15 shows the graph of these fun tions.
Figure 9.15: Sales of old and new models of laptops
The sales of the new model grows at an in reasing rate, whi h shows a rapid improvement in rate of sales.
2b
The sales of the models are equal at the interse tion, where
36
9
=
t+3
21 − t
9(21 − t) = 36(t + 3)
(−9 − 36)t = 36 × 3 − 21 × 9
−81
t=
= 1,8.
−45
When
t = 1,8,
then
Sold = Snew = 1,875.
This means that after 1,8 months (about 54 days) 1 875 of both
the old and the new laptops are sold.
9.13 Se tion 5.2.1 (The power rule for dierentiation)
1
The slope is given by the derivative of the fun tion, that is
dy
d 3
=
(x + 4x − 7) = 3x2 + 4.
dx
dx
126
DSC1520/SG001
At
2a
x = 3,
dy
= 3(3)2 + 4 = 31.
dx
Simplify the fun tion to get
1
y = 4 + x + 2x 2 .
Dierentiation gives
1
dy
1
1 1
= 0 + 1 + 2 × x 2 −1 = 1 + x− 2 = 1 + √ .
dx
2
x
2b
Simpli ation gives
f (x) = x−1 − 5x−2 + 10.
Dierentiating
f
gives
f ′ (x) = (−1)x−1−1 − 5(−2)x−2−1 + 0 = −x−2 + 10x−3 = −
2
Simpli ation results in
f (x) = 7 +
√7
x
=7+
7
1
x2
1
= 7 + 7x− 2 .
10
1
+ 3.
2
x
x
Dierentiation gives
1
1
7
7 3
f ′ (x) = 0 + 7 −
x− 2 −1 = − x− 2 = − √ .
2
2
2 x3
2d
We rst simplify to get
p(q) =
q 2 −q
q3
=
q2
q3
−
q
q3
= q −1 − q −2 .
Dierentiation gives
p′ (q) = −q −2 − (−2)q −3 = −
1
2
+ 3.
2
q
q
9.14 Se tion 5.2.3 (The hain rule)
1
For
y = (4 − 5x)3 ,
set
u = 4 − 5x
so that
y = u3 .
dy
= 3u2
du
Combining these gives
2
For
f (x) =
√
x2 + 12,
set
f ′ (u) =
Combining these gives
For
y=
15
1+ex , set
dy
1 1
1
= u− 2 = √
du
2
2 u
u = 1 + ex
Combining these gives
For
q = 500 ln(p3 + 8p),
so that
y=
15
u
= 15u−1 .
du
= 2x.
dx
The derivatives are
dy
15
= −15u−2 = − 2 .
du
u
and
dy du
15
15ex
dy
=
·
= − 2 · ex = −
.
dx
du dx
u
(1 + ex )2
set
u = p3 + 8p
so that
du
= 3p2 + 8
dp
Combining these gives
and u′ (x) =
x
1
.
f ′ (x) = f ′ (u) · u′ (x) = √ · 2x = √
2
2 u
x + 12
du
= 0 + ex
dx
4
du
= −5.
dx
and
dy
dy du
=
·
= 3u2 · (−5) = −15(4 − 5x)2 .
dx
du dx
√
1
u = x2 + 12 so that y = u = u 2 .
The derivatives are
3
The derivatives are
q = 500 ln u.
and
The derivatives are
dq
500
=
.
du
u
dq
dq du
500
500(3p2 + 8)
=
·
=
· (3p2 + 8) =
.
dp
du dp
u
p3 + 8p
127
DSC1520/SG001
5
We know that the derivative of the sum of a number of terms is simply the derivative of the terms added
together. We therefore dierentiate ea h term and add.
⊲
The derivative of the rst term is
⊲
For the se ond term,
f (x) =
d
1,5
dx 5e
2
x+2 , we set
= 0 be
ause it does not
u=x+2
so that
ontain
f (u) =
u′ (x) = 1 and f ′ (u) = −2u−2 =
resulting in
f ′ (x) = f ′ (u) · u′ (x) =
⊲
For the third term,
f (x) = ln(2x − 1),
we set
f ′ (u) =
= 2u−1 .
onstant.
The derivatives are
−2
,
u2
−2
−2
·1=
.
2
u
(x + 2)2
u = 2x − 1
1
u
2
u
x and is therefore a
so that
f (u) = ln(u).
The derivatives are
and u′ (x) = 2,
resulting in
f ′ (x) = f ′ (u) · u′ (x) =
2
1
·2=
.
u
2x − 1
Adding these derivatives, gives
−2
2
g (x) =
+
=2
2
(x + 2)
2x − 1
′
1
1
−
2x − 1 (x + 2)2
.
9.15 Se tion 5.2.4 (The produ t rule)
√
1 Q(p) = p p + 5
v ′ (p)
=
1
2 (p
− 12
+ 5)
onsists of
u(p) = p
with derivative
u′ (p) = 1
and
1
v(p) = (p + 5) 2
with derivative
.
The produ t rule gives
Q′ (p) = u′ (p)v(p) + u(p)v ′ (p)
1
1
1
= 1 · (p + 5) 2 + p · (p + 5)− 2
2
p
p
= p+5+ √
2 p+5
2(p + 5) + p
√
=
2 p+5
3p + 10
= √
.
2 p+5
2 AC(q) =
1
q , is dierentiated using the power rule to nd
−2
−q . The se ond term, q ln q , is the produ t of u(q) = q and v(q) = ln q , with u′ (q) = 1 and
1
q
− q ln q
onsists of two terms. The rst term,
d −1
=
dq q
′
v (q) = 1q . The derivative of
AC
is
AC ′ (q) = −q −2 − [u′ (q)v(q) + u(q)v ′ (q)]
1
−1
= 2 − 1 · ln q + q ·
q
q
−1
= 2 − ln q − 1.
q
128
DSC1520/SG001
3
3
3
f (x) = 20
xe2x−5 onsists of the produ t of u(x) = 20
x with derivative
2x−5
′
2x−5
e
with derivative v (x) = e
· 2. The derivative of f is
The fun tion
v(x) =
u′ (x) =
3
20 and
f ′ (x) = u′ (x)v(x) + v ′ (x)u(x)
3
3
=
· e2x−5 + 2e2x−5 · x
20
20
3 2x−5
= e
(1 + 2x).
20
4 f (x) = x3 ln(x3 )
u(x) = x3
of f is
onsists of
( hain rule). The derivative
with derivative
u′ (x) = 3x2
and
v(x) = ln x3
with
v ′ (x) =
1
x3
· 3x2
f ′ (x) = u′ (x)v(x) + v ′ (x)u(x)
3x2 3
·x
x3
= 3x2 (ln x3 + 1).
= 3x2 · ln x3 +
9.16 Se tion 5.2.5 (The quotient rule)
1 f (x) =
ln x
3x
onsists of the quotient of
u(x) = ln x with u′ (x) =
1
x
and
v(x) = 3x
with v ′ (x) = 3.
Applying the quotient rule results in
u′ (x)v(x) − u(x)v ′ (x)
=
f (x) =
v(x)2
′
2
The se ond term of
C(y) = 1 −
u(y) = e−0,8y
e−0,8y
y2
1
x (3x)
− ln x(3)
3 − 3 ln x
1 − ln x
=
=
.
(3x)2
9x2
3x2
onsists of the quotient of
with u′ (y) = e−0,8y (−0,8)
and
v(y) = y 2
with v ′ (y) = 2y.
Applying the quotient rule for the se ond term, results in
u′ (y)v(y) − u(y)v ′ (y)
(v(y))2
−0,8e−0,8y y 2 − 2ye−0,8y
=−
(y 2 )2
−0,8y
e
(0,8 + 2y)
.
=
y4
C ′ (y) = 0 −
3 F (x) =
2x ln x
3x2
onsists of the quotient of
u(x) = 2x ln x
with u′ (x) = 2 ln x +
2x
(product rule)
x
and
v(x) = 3x2
Applying the quotient rule results in
u′ (x)v(x) − u(x)v ′ (x)
v(x)2
(2 ln x + 2)(3x2 ) − (2x ln x)(6x)
=
(3x2 )2
6x2 (ln x + 1 − 2 ln x)
=
9x4
2(1 − ln x)
.
=
3x2
F ′ (x) =
129
with v ′ (x) = 6x.
DSC1520/SG001
4 P (q) =
50−q 2
50+q 2
onsists of the quotient of
u(q) = 50 − q 2
with u′ (q) = −2q
v(q) = 50 + q 2
and
v ′ (q) = 2q.
with
Applying the quotient rule results in
u′ (q)v(q) − u(q)v ′ (q)
v(q)2
(−2q)(50 + q 2 ) − (50 − q 2 )(2q)
=
(50 + q 2 )2
−100q − 2q 3 − 100q + 2q 3
=
(50 + q 2 )2
−200q
.
=
(50 + q 2 )2
P ′ (q) =
9.17 Se tion 5.3 (Higher derivatives)
1a
f (x) = (150 − 2x)x = 150x − 2x2 . Then,
= 150 − 4x and f ′′ (x) = 0 − 4 = −4.
√
1
1b Simpli ation gives f (x) = 10x + x = 10x + x 2 . Then,
1
1 3
1
1 1
1 1 3
f ′ (x) = 10 + x− 2 = √
and
f ′′ (x) = 0 − · x− 2 = − x− 2 = − √ .
2
2 x
2 2
4
4 x3
Simplify to nd
f ′ (x)
1
= x4 − x−4 . Then,
x4
4
f ′ (x) = 4x3 − (−4)x−5 = 4x3 + 4x−5 = 4x3 + 5 and
x
20
′′
2
−6
2
−6
f (x) = 4(3)x + 4(−5)x = 12x − 20x = 12x2 − 6 .
x
1
Simplify to nd
f (x) = x4 −
2 C ′ (x) = 35 x2 − 16x + 25 , C ′′ (x) = 65 x − 16 and C ′′′ (x) =
6
5
= 1,2.
9.18 Se tion 5.3 (Optimisation of fun tions in one variable)
1a
From
1b
Dierentiate
1
Dierentiate
f ′ (x) = 2x − 6 = 0 we nd that f has a stationary
= 2 > 0, the turning point is a minimum.
point at
x = 3.
′′
Sin e f (x)
f (t) = 31 t3 − 2t2 − 5t + 8 to nd f ′ (t) = t2 − 4t − 5 and f ′′ (t) = 2t − 4.
′
2
Setting f (x) = t − 4t − 5 = 0 and solving for t, gives (t − 5)(t + 1) = 0, resulting in turning points
′′
at t = 5 and t = −1. Sin e f (5) = 2(5) − 4 = 6 > 0, f has a lo al minimum at x = 5 and sin e
′′
f (−1) = 2(−1) − 4 = −6 < 0, f has a lo al maximum at t = −1.
Setting
Q(p) = 64p + p−1
Q′ (p) = 64 −
1
p2
=0
gives
1
′′
2 and Q (p)
pq
1
1
1
2
or p =
64 so that p = ±
64 , that is
p2
0, Q has a lo al minimum at p = 81 .
to nd
64 =
Q′ (p) = 64 − p−2 = 64 −
2
′′ 1
Sin e Q ( ) = 1 3 = 2(512) = 1 024 >
8
(8)
1
2
′′
Sin e Q (− ) =
= −2(512) = −1 024
8
(− 1 )3
< 0, Q
has a lo al maximum at
8
1d
Dierentiate
2a
Dierentiate
′
Setting f (x)
p=
2
.
p3
− 81 .
= −(−2)p−3 =
p=
1
8 or
p=
1
8.
f (x) = 200 − 2(x2 − 8x + 16) = −2x2 + 16x + 168: f ′ (x) = −4x + 16 and f ′′ (x) = −4.
= −4x + 16 = 0 gives x = 4. Sin e f ′′ (4) = −4 < 0, f has a maximum at x = 4.
f (x) = x2 − 18x + 11: f ′ (x) = 2x − 18 and f ′′ (x) = 2.
′
′′
Setting f (x) = 2x − 18 = 0 gives turning point at x = 9. Sin e f (9) = 2 > 0, f
Therefore, f de reases over (−∞; 9) and in reases over (9; ∞).
130
has a minimum at
x = 9.
DSC1520/SG001
2b
Dierentiate
f (x) = 3x2 − 0,1x3 : f ′ (x) = 6x − 0,3x2
and
f ′′ (x) = 6 − 0,6x.
6
′
2
Setting f (x) = 6x − 0,3x = 0 gives x(6 − 0,3x) = 0, so the turning points are at x = 0 and x =
0,3 = 20.
′′
′′
Sin e f (0) = 6 > 0, f has a minimum at x = 0 and sin e f (20) = 6 − 0,6(2) = −6 < 0 it has a maximum
at
x = 20.
Therefore,
f
de reases over
(−∞; 0)
and
(20; ∞),
(0; 20).
while it in reases over
9.19 Se tion 6.1 (Marginal and average fun tions)
1a
We need the demand fun tion to be in terms of
q,
so we transform it to nd
Now,
3p = 120 − q
or
p = 40 −
q
3.
q2
q
q = 40q − ,
T R = pq = 40 −
3
3
d
q2
2
MR =
40q −
= 40 − q
dq
3
3
and
AR =
1b
q
3
q
TR
= 40 − = p.
q
3
M R = 40 − 23 (120) = 40 − 80 = −40 < 0.
monopolist should stop selling the produ t when M R = 0, be ause he will start losing money after
2q
point. They should therefore stop selling when M R = 40 −
3 = 0, that is when q = 60.
AR = 40 −
When
= 0, q = 40 × 3 = 120.
2a T C = F C + V C = 1 000 + 3q
2b M C =
d
dq T C
=3
and
3a
AC
The derivative of
q>0
and average
is
The
this
T C(20) = 1 000 + 3(20) = 1 060.
M C(20) = 3.
TC
and
ase. It gives the amount by whi h
Here,
Marginal
ost is given by the slope of
T C,
whi h is
onstant in this
in reases for ea h extra unit produ ed.
d
−1
dq (50 + 10q )
= 0 − 10q −2 = − 10
q2 .
The slope of
AC
is therefore negative for all
ost de reases as the number of units produ ed in reases. (e onomies of s ale).
3b T C = AC × q = 50q + 10, with F C = 10 and V C = 50q .
3
Marginal
TC
ost,
MC =
d
dq T C
=
The slope of
5a M R =
d
dq T R
5b M C =
d q3
dq ( 3
5
= 50, is the slope of the T C
fun tion. For ea h unit produ ed,
in reases by R50.
4a T R = AR × q = 180q − 12q 2
4b
d
dq (50q + 10)
Prot
When
MR
=
and
MR =
is double the slope of
d
dq (200q
d
dq T R
AR
− 4q 2 ) = 200 − 8q
= 180 − 24q .
a monopolist.
and
AR =
TR
q
= 200 − 4q .
2
− 12q 2 + 164q + 100) = q − 24q + 164 and AC = TqC = q3 − 12q + 164 + 100
q .
3
3
P = T R − T C = 200q − 4q 2 − q3 − 12q 2 + 164q + 100 = − q3 + 8q 2 + 36q − 100.
3
2
q = 20, P = − (20)
3 + 8(20) + 36(20) − 100 = R1 153,33.
9.20 Se tion 6.1.4 (Produ tion fun tions)
1 MPL =
d
dl q
=
d
dl (225l
− 31 l3 ) = 225 − l2 .
M P L(5) = 225 − 52 = 200.
This means that when ve workers are employed per day, an additional worker
will in rease produ tion at a rate of 200 units per day.
131
DSC1520/SG001
2 AP L =
When
225l− 13 l3
l
= 225 −
l2
3.
l = 5, AP L(5) = 225 −
52
3
q
l
=
= 216,67.
This means that when ve workers are employed per day, the
average produ tivity is 217 output units per worker per day.
9.21 Se tion 6.2 (E onomi appli ations of optimisation)
1a
Total revenue is the pri e per unit times the number of units sold, that is
T R = pq = (240 − 10q)q = 240q − 10q 2 .
Prot is total revenue minus total
ost, that is
P = 240q − 10q 2 − (120 + 8q) = −10q 2 + 232q − 120.
1b
dP
dq
Prot is a maximum when
= −20q + 232 = 0,
that is when
T-shirts, she will make a prot of R1 224.
1 MR =
q = 11,6,
1d
d
dq T R
=
q = 11,6 ≈ 12
d
dq (240q
we nd that
d
− 10q 2 ) = 240 − 20q and M C = dq
(120 + 8q) = 8.
M R = 240 − 20(11,6) = 8, whi h is the same as M C .
TR
Figure 9.16 shows the graphs of
and
T C,
units. If Mary sells 12
At maximum prot, when
the break-even points and the area where prot is made.
TR
TC
1500
1000
TR
Prot
500
TC
Figure 9.16:
5
10
15
T R = 240q − 10q 2
From the graph, we estimate the break-even points to be at
by settinge
T R = T C,
240q − 10q 2 = 120 + 8q ,
q = 0,53 and q = 22,67.
that is
Solving this equation gives
20
and
q≈1
whi h
q
25
T C = 120 + 8q
and
q ≈ 23,
whi h are found algebrai ally
an be simplied to
5q 2 − 116q + 60 = 0.
Mary should produ e and sell either one or 23 T-shirts to break even. When she sells more than one, but
fewer than 23 T-shirts, she will make a prot.
1e
Figure 9.17 shows
MR
and
MC
MR
interse t at
and
MC
q = 11,6,
fun tions and their interse tion.
that is the point where prot is a maximum.
2a T R = AR × q = 25q ,
T C = AC × q = (15 +
8 000
q )q
MR =
d
dq T R
=
d
dq (25q)
MC =
d
dq T C
=
d
dq (15q
2b
Break-even o
= 25
= 15q + 8 000,
and
+ 8 000) = 15.
urs where
T R = T C,
that is where
25q = 15q + 8 000,
132
giving
q = 800.
DSC1520/SG001
MR
MC
250
200
150
MR
100
50
MC
2
4
6
8
M R = 240 − 20q
Figure 9.17:
2 P = T R − T C = 25q − (15q + 8 000) = 10q − 8 000.
The derivative of
P
is
MR
onstant, while
MC
q
and
therefore say that prot will in rease indenitely as
3a
10
and
This is a linear fun tion without a turning point.
are both
onstants and will never interse t. We
P ′′ (q)
= −1,6 < 0
P ′ (q) = −1,6q + 5 384 = 0,
that is when
5 504 − 1,6q = 120
For maximum prot of
The
−0,8q 2
that is when
M R = M C , that
5 504−120
q=
= 3 365.
1,6
or
P = 8 450 000, q = 3 365
ompany breaks even when
+ 5 384q − 608 580 = 0.
Total revenue is given by
T R = T C,
and pri e is
d
dq (5 504q
5 384
1,6
= 3 365.
− 0,8q 2 ) =
Sin e
5 504q − 0,8q 2 = 608 580 + 120q ,
q = 115 and q = 6 615.
50q
.
e0,05q
M R = T R′
d 50q
=
dq e0,05q
d
qe−0,05q
= 50
dq
= 50 1 · e−0,05q + q · (−0,05)e−0,05q
−0,05q
= 50e
Revenue is a maximum when
(product rule)
(1 − 0,05q).
T R′ = M R = 0,
that is where
50e−0,05q (1 − 0,05q) = 0
1 − 0,05q = 0
(divide by 50e−0,05q )
q = 20.
The rm needs to produ e and sell 20 units to maximise revenue.
133
d
dq (608 580
+
p = 5 504 − 0,8(3 365) = 2 812.
that is when
Solving this equation gives
T R = AR × q =
is when
Marginal revenue
4b
q=
and the prot
prot is indeed a maximum at this point.
Marginal analysis: Prot is a maximum when
120q),
an
in reases.
T R = pq = (5 504 − 0,8q)q = 5 504q − 0,8q 2
P = (5 504q − 0,8q 2 ) − (608 580 + 120q) = −0,8q 2 + 5 384q − 608 580.
Dierentiation: Prot is a maximum when
4a
MC = 8
From the demand fun tion, we nd that
fun tion is
3b
q
12
whi h gives
DSC1520/SG001
5a
Simplifying the
Fixed
TC
ost is therefore
5b
Marginal
5
To minimise
ost
fun tion, gives
T C = 120(ln q + ln 10) = 120 ln q + 120 ln 10.
T F C = 120 ln 10 = 276,3.
MC =
d
dq 120 ln(q
ost, we set
+ 10) =
MC = 0
to get
120
q+10 .
120
q+10
=0
120 = 0(q + 1) = 0,
or
whi h does not have a solution.
9.22 Se tion 6.3 (Elasti ity of demand non-linear demand fun tions)
1a
This is a linear fun tion with
dq
dp
=
εd =
1b
At
d
dp (25
− 3p) = −3,
giving elasti ity as
dq p
p
3p
· = −3 ·
=−
.
dp q
25 − 3p
25 − 3p
3(6)
p = 6, |εd | = − 25−3(6)
= | − 2,57| = 2,57 > 1.
Demand is elasti . For every 1% in rease in pri e,
demand will de rease by 2,57%.
2
The derivative of the logarithmi
demand fun tion is
dq
dp
=
−10
p , giving the
oe ient of elasti ity as
dq p
·
dp q
−10 p
·
=
p
q
10
=−
q
10
.
=−
80 − 10 ln p
εd =
When pri e is R60,
10
εd = − 80−10
ln 60 = −0,256.
Sin e
|εd | = 0,256 < 1,
demand is
inelasti
at
p = 60.
This
means that at pri e R60, demand de reases by 0,256% for every 1% in rease in pri e if pri e in reases by
5%, demand de reases by 1,28%.
3a
The derivative of the demand fun tion is
εd =
3b
p = 8,
If pri e is R800,
unitary elasti . If
p
d
dp (192
in reases by one unit,
When
4a
From the demand fun tion, we nd
giving pri e elasti ity of demand as
dq p
p
−2p2
· = −2p
=
.
dp q
192 − p2
192 − p2
giving elasti ity as
3
− p2 ) = −2p,
q
εd =
−2(8)2
192−(8)2
= −1.
This means that at this pri e, demand is
de reases by one unit.
p = 8, q = 192 − 82 = 128 seats are available, meaning that the ten additional seats represent an
10
in rease of
128 = 0,0781 or a 7,81%. Sin e we have unitary elasti ity, the pri e will de rease by 7,81%, that
is by R800 × 0,0781 = 62,48, to R737,52.
T R = pq = 1 500qe−0,025q .
Dierentiating this gives
d
1 500qe−0,025q
dq
= 1 500 1 · e−0,025q + q · (−0,025e−0,025q )
MR =
(product rule)
= 1 500e−0,025q (1 − 0,025q).
4b
M R = 0, that is where 1 500e−0,025q (1−0,025q) = 0. This is true when
1 − 0,025q = 0, whi h gives q = 40 and p = 1500e−0,025(40) = R551,82.
Total revenue is a maximum where
−0,025q
either e
=0
(impossible) or
134
DSC1520/SG001
4
Dierentiating the demand fun tion gives
dp
dq
= 1 500(−0,025)e−0,025q .
The
oe ient of elasti ity of
demand is
dq p
·
dp q
|εd | =
1 500e−0,025q
1
·
1 500(−0,025)e−0,025q
q
1
=−
0,025q
40
=− .
q
=
At maximum revenue,
q = 40,
|εd | = | − 1| = 1.
giving
by 1%, demand de reases by 1%.
This indi ates unitary elasti ity if pri e in reases
9.23 Se tion 7.2.1 (The power rule for integration)
1
2
3
R
R
(x + x3 + x3,5 ) dx =
√1
x5
Simplify
Now,
4
dx =
R
f
R
5
x2
2
x− 2 dx =
to nd
R
f (x) =
+
x4
4
+
x4,5
4,5
x−2,5 dx =
√
x+ x
x
(1 + x−0,5 ) dx = x +
x0,5
0,5
=
x
x
+c
x−1,5
−1,5
+
√
x
x
+c=x+
3
+c=
x− 2
− 23
=1+
√1
x
√
x
1
2
=−
√2
3 x3
+ c.
= 1 + x−0,5 .
√
= x + 2 x + c.
2
(2x + 0,5x−1 ) dx = 2x2 + 0,5 ln x + c = x2 + 0,5 ln x + c.
3
√
5 Simplify f to nd f (q) = q(q − q) = q 2 − q 2 .
p
Z 5
3
q3 q 2
q3 2 q5
2
Now,
q − q 2 dq =
− 5 +c=
−
+ c.
3
3
5
2
R
2x +
1
2x
dx =
R
9.24 Se tion 7.2.2 (Integration by substitution)
1
Simplify to nd
Set
u = 9x − 5
R √
3
9x − 5 dx =
R
1
(9x − 5) 3 dx .
and dierentiate to nd
Z
du
dx
=9
or
dx =
1
3
(9x − 5) dx =
du
9 . Then,
Z
1
u3
du
9
4
1 u3
= 4 +c
9 3
=
13 4
u3 + c
94
4
(9x − 5) 3
+ c.
=
12
2
Set the inner fun tion as
Z
u = 2 − 7q
1
dq =
2 − 7q
Z
and dierentiate to nd
1 du
1
·
=
u −7
−7
Z
du
dq
= −7,
or
dq =
du
−7 . Then,
1
1
1
du = − ln |u| + c = − ln |2 − 5q| + c.
u
7
7
135
DSC1520/SG001
3
Set
u = 2 − 5q
du
dq
so that
= −5,
Z
4
Simplify to nd
1
e2t
f (t) =
For the rst term, set
dv
dierentiate to nd
dt
giving
2−5q
10e
dq =
dq =
Z
du
−5 .
10eu
du
= −2eu + c = −2e2−5q + c.
−5
+ 2et+2 + 3 = e−2t + 2et+2 + 3.
u = −2t so that du
dt = −2
= 1 and dt = du. Then,
Z and
dt =
du
−2 ; for the se ond term set
v = t+2
Z
1
t+2
+ 2e
+ 3 dt = (e−2t + 2et+2 + 3) dt
e2t
Z
Z
u du
= e
+ 2 ev dv + 3t + c
−2
1
= − eu + 2ev + 3t + c
2
= −0,5e−2t + 2et+2 + 3t + c.
9.25 Se tion 7.3 (The denite integral and the area under a urve)
1
3
2
R3
(x + 5) dx = x2 + 5x 1 = 29 + 5(3) − ( 21 + 5) = 19,5 − 5,5 = 14.
R 10 x
R
10
100
x2
2 110 x+5
2 dx = 1 2 + 2,5 dx = 4 + 2,5x 1 = ( 4 + 2,5(10)) − (0,25 + 2,5) = 50 − 2,75 = 47,25.
R
R 10
2
10
8
3 310 x x+8x
) − (ln 3 − 83 ) = 1,503 − (−1,568) = 3,07.
dx = 3 x1 + 8x−2 dx = ln x − x8 2 = (ln 10 − 10
3
R 1,3 10
1,3
4 0,3
x dx = 10 ln x|0,3 = 10(ln 1,3 − ln 0,3) = 10(0,262 − (−1,204)) = 10(1,466) = 14,66.
5
1
We set
u=q+8
and nd that
Z
du
dq
=1
1
(q + 8) 2 dq =
and
du = dq .
6
Set
u = 0,4x
so that
Z
du
dx
1
1
u 2 du =
u2
3
2
2(q + 8) 2
q + 8 dq =
3
0,4x
and
dx =
dx =
Z
1
8
1
= 42,67 − 18 = 24,67.
du
0,4 . Then,
500
Z
eu
500 u
du
+c=
e + c = 1 250e0,4x + c,
0,4
0,4
and
Z
3
2u 2
2(q + 8) 2
+c=
+ c,
3
3
+c =
3
8p
= 0,4
500e
3
3
Z
and
Z
Therefore,
1
500e0,4x dx = 1 250e0,4x
0
0
= 1 250(1,4918 − 1) = 614,78.
7a
The graph as plotted by Maxima is shown in Figure 9.18.
7b
The net area is
Z
0
10
2
(16 − q ) dq =
q3
16q −
3
136
10
0
= −173,33.
and
DSC1520/SG001
Figure 9.18:
7
The
urve
uts the
q
axis at
Z
0
The area below the
Z
4
10
q
2
x = 4.
4
2
f (q) = 16 − q 2
The area above the
(16 − q ) dq =
q3
16q −
3
q
axis is
4
0
= 16(4) −
43
= 42,67.
3
axis is
(16 − q ) dq =
q3
16q −
3
The net area is therefore
10
4
43
103
− 16(4) −
= −173,33 − 42,67 = −216.
= 16(10) −
3
3
42,67 − 216 = −173,33.
This is equal to the net area
al ulated in (b).
9.26 Se tion 8.1 (From marginal to total fun tions)
2
1a T R = M Rdq = (120 − 2q) dq = 120q − 2 q2 + c = 120q − q 2 (c = 0).
R
R
1b T R(50) = 120(50) − 502 = 3 500.
2a
The sket h is shown in Figure 9.19:
Figure 9.19:
N (t) =
10
1+t
The rate of memory retention de lines rapidly during the rst ve minutes, after whi h it de lines gradually
to an assymptote at zero.
2b
The fun tion of the a tual items memorised is determined by integrating the given rate fun tion, that is
R 10
N (t) = 1+t
dt = 10 ln(1 + t) + c. Sin e N = 0 when t = 0, N (0) = 10 ln 1 = 0.
N (t) = 10 ln(1 + t).
R
= 10 ln(11) − 10 ln 1 = 24 items.
2 010 N ′ (t) dt = 10 ln(1 + t) 10
0
R 20 ′
20
10 N (t) dt = 10 ln(1 + t) 10 = 10 ln(21) − 10 ln(11) = 6,5 items.
137
Therefore,
c=0
and
DSC1520/SG001
3a
To nd the prot fun tion, we need to nd the total
ost and total revenue fun tions.
R
R
T R = M R dq = (200 − 20q) dq = 200q − 10q 2 + c. Sin e T R = 0 when q = 0, c = 0, giving
T R = 200q − 10q 2 .
R
R
The total ost fun tion is T C =
M C dq = 80 dq = 80q + c. Sin e xed ost is R500, T C = 80q + 500.
Here,
The prot fun tion is given by
3b
P ′ (q) = 0 or M R = M C . Dierentiating P = −10q 2 + 120q − 500 gives
results in q = 6. Setting M R = M C gives 200 − 20q = 80, whi h also results
Prot is a maximum when either
P ′ (q)
in
P = T R − T C = 200q − 10q 2 − (80q + 500) = −10q 2 + 120q − 500.
= −20q + 120 = 0,
q = 6.
whi h
9.27 Se tion 8.2 (Consumer and produ er surplus)
1a
At
1b
At
q0 = 8, p0 = 100 − 82 = 36. The graph of pd is shown in Figure 9.20(a). CS
demand urve and the line at p0 = 36.
R8
3
8
CS = 0 (100 − q 2 ) dq − 36 × 8 = (100q − q3 ) 0 − 288 = 629,33 − 288 = 341,33.
q0 = 9, p0 =
100
9+1
= 10. The graph
p0 = 10.
of
pd
is shown in Figure 9.20(b).
CS
is the area between the
is the area under the demand
urve and above the line at
(a) pd = 100 − q 2 at q0 = 8
(b) pd =
100
q+1
at q0 = 9
Figure 9.20: Consumer surplus
CS =
2a
At
R9
100
0 q+1
9
dq − 10 × 9 = 100 ln(q + 1) 0 − 90 = 100(ln 10 − ln 1) − 90 = 230,26 − 90 = 140,26.
q0 = 4, p0 = 10 −
above the supply
P S = 9,8 × 4 −
2b
At
urve
1
4+1
= 10 − 0,2 = 9,8. The graph
p0 = 9,8.
of
ps
is shown in Figure 9.21(a).
PS
is the area
urve and under the line at
R4
0
(10 −
1
q+1 ) dq
= 39,2 − (10q − ln(q + 1))
q0 = 5, p0 = 52 + 4 = 29. The graph
and under the line at p0 = 29.
of
ps
4
0
= 39,2 − [(40 − ln 5) − (0 − ln 1)] = 0,81.
is shown in Figure 9.21(b).
1
(a) ps = 10 − q+1
at q0 = 4
PS
is the area above the supply
(b) ps = q 2 + 4 at q0 = 5
Figure 9.21: Produ er surplus
138
DSC1520/SG001
P S = 29 × 5 −
3a
R5
0
3
(q 2 + 4) dq = 145 − ( q3 + 4q)
5
0
= 145 − [(41,67 + 20) − 0] = 83,33.
Equilibrium is where demand and supply are equal,
this equation gives
3b CS =
p0 = 55
and
pd = ps ,
that is when
q0 = 90.
100 − 0,5q = 10 + 0,5q .
90
R 90
(100 − 0,5q) dq − p0 q0 = (100q − 0,25q 2 ) 0 − 55 × 90 = (6 975 − 0) − 4 950 = 2 025.
R 90
90
P S = 55 × 90 − 0 (10 + 0,5q) dq = 4 950 − (10q + 0,25q 2 ) 0 = 4 950 − (900 + 2 025) = 2 025.
0
Total surplus is
2 025 + 2 025 = 4 050.
Figure 9.22:
pd = 100 − 0,5q
and
139
ps = 10 + 0,5q
at equilibrium
Solving
PART V
MATHEMATICAL BACKGROUND ASSUMED
TO BE IN PLACE
140
Appendix A: Numbers and variables
A.1 Priorities
Simple operations in mathemati s are performed in the following order:
1. parenthesis (bra kets)
2. exponentiation
3. multipli ation/division
4. addition/subtra tion
Consider the following:
2 × 3 + 4 × 5 = 6 + 20 = 26
8 ÷ 2 − 6 ÷ 3 = 4 − 2 = 2.
5 × 3 + 15 ÷ 3 − 4 × 5 = 15 + 5 − 20 = 0.
Parenthesis (or bra kets) in mathemati s are mainly used for grouping and
larity. For example, we know
that
4 × 3 + 7 × 5 = 12 + 35 = 47
when we apply the normal priority rules. However if we rst need to add 3 and 7 before multiplying by 4
and 5, we use bra kets and write
4 × (3 + 7) × 5,
whi h indi ates that the expression in bra kets must rst be
al ulated before the other operations, so that
4 × 10 × 5 = 40 × 5 = 200.
We also use bra kets to indi ate fra tional exponents, for example
82/3 = 8(2÷3) .
Also note that
−22 = −4,
but
(−2)2 = 4.
A square root sign a ts as bra kets, so
When an expression
p
22 + 5 + 3 =
p
(4 + 5) + 3 =
√
9 + 3 = 3 + 3 = 6.
ontains bra kets within bra kets, as in
(42 − (6 × 3 − 80 ÷ 5)2 ) ÷ (92 − 77),
141
DSC1520/SG001
we rst need to rst
al ulate the value in the innermost bra kets, namely
get
6 × 3 − 80 ÷ 5 = 18 − 16 = 2,
to
(42 − (2)2 ) ÷ (92 − 77) = (16 − 4) ÷ (81 − 77) = 12 ÷ 4 = 3.
A tivity
Evaluate ea h of the following expressions:
1.
2.
4 × 3 ÷ 2 + 5 × 6 − 20 ÷ 4 × 2
√
32 × 7 − 9 × 8 + 22 ÷ 4 × 3
A.2 Variables
If you need to nd the area of a 4 m by 3 m re tangle, you multiply 4 by 3. But if you do not have spe i
values for the re tangle, you
by
L × W.
an
all the length of the re tangle
This will be true for any value of
width, in this
ase, are
alled
L
and
W.
L
and its width
W,
and the area is given
Su h letters or symbols that represent length and
variables.
The rule to express one quantity as a per entage of another is to divide the rst quantity by the se ond and
multiply by 100. If the rst number is denoted by
x
y
x
and the se ond by
y,
these words
an be expressed as
× 100.
This is the great advantage of using variables to represent numbers. They enable us to write down rules or
expressions that are
ompletely general. If we then need to nd the answer to a parti ular
to do is to substitute the parti ular values of the variables (x,
A
variable
is simply a label that represents something and
In printed text it is represented by either itali
a, b, c, . . . , x, y, z )
y , L, W , et
an assume any one of several numeri
letters of the Roman alphabet (A, B, C,
or other symbols like Greek letters (α, β, σ, χ, et .).
A tivity
1. Determine the numeri al value of ea h of the following:
(a)
12x + 17
(b)
x2 − 3
( )
(d)
2. If
if
x+7
if x = 5
4
(x + 3)(x − 2)
a = 2, b = 1
(a)
x=2
x=4
if
and
if
c = 7,
x=6
then nd the value of the following:
2(a + b − c) + c(b − a)
(b)
a2 + b2 + c2
( )
(a + b)(b − c)
(d)
2b −
c+3
a
+ b2
3. Write ea h of the following statements as a mathemati al expression:
(a) The sum of
x
and
y.
(b) Subtra t the sum of
( ) Three times
x
a
and
b
from eight.
added to two times
ase, all we need
.) into the appropriate expression.
y.
142
values.
. . . , X, Y, Z
or
DSC1520/SG001
(d) Robert's age in seven years' time if he is
y
years old at the moment.
A.3 Laws of operations
There are three laws appli able to
tipli ation (×)
1.
and
division (÷).
ombinations of the basi
These are the
operations
ommutative, asso iative
addition (+), subtra tion (−), muland distributive laws.
Commutative laws
The two most basi
laws are the
ommutative laws of addition and multipli ation.
They say that it
does not matter in whi h order numbers are added or multiplied.
(a) The
ommutative law of addition
The sum of two numbers is unique, meaning that it does not matter whi h number is pla ed rst
or se ond, the result is the same.
For example,
6+2 = 2+6
(= 8)
or, in terms of variables,
a + b = b + a.
(b) The
ommutative law of multipli ation
The produ t of two numbers is unique, meaning that it does not matter whi h number is pla ed
rst or se ond, the result is the same.
For example,
5×2=2×5
(= 10)
and, in terms of variables,
a × b = b × a.
But, what about subtra tion and division? Does the
ommutative law apply to them too? The following
examples show that it does not:
⊲
For subtra tion,
7 − 3 6= 3 − 7.
Here,
ommutation (i.e.
of 4).
⊲
(The sign 6= stands for is not equal to.)
hanging the order) leads to the negative of the initial result (−4 instead
For division,
10 ÷ 2 6= 2 ÷ 10.
Here,
2.
ommutation leads to the re ipro al of the initial result (1/5 instead of 5).
Asso iated laws
Asso iated laws say that the order in whi h three (or more) numbers are added (or multiplied) does
not have an ee t on the result.
⊲
The asso iated law of addition
The sum of three (or more) numbers is unique it does not depend on whi h two are added rst,
the result is the same.
Consider the following example, where we use bra kets to indi ate the sequen e in whi h the
operations are to be performed:
7 + (3 + 2) = (7 + 3) + 2
143
(= 12)
DSC1520/SG001
In terms of variables,
(a + b) + c = a + (b + c).
This implies that when adding long
olumns of numbers, it does not matter whi h numbers are
added rst and whi h last.
⊲
The asso iated law of multipli ation
The produ t of three (or more) numbers is unique it does not depend on whi h two are multiplied
rst, the result is the same.
Consider the following examples. Bra kets again indi ate the sequen e in whi h the operations
are to be performed:
5 × (4 × 2) = (5 × 4) × 2
(= 40)
In terms of variables,
(a × b) × c = a × (b × c).
Again, when multiplying many numbers, it does not matter whi h numbers are multiplied rst
and whi h last.
When it
omes to subtra tion and division, however, we must be very
areful, as the following examples
show:
10 − (5 − 2) 6= (10 − 5) − 2 and (8 ÷ 4) ÷ 2 6= 8 ÷ (4 ÷ 2).
This means that an expression like
8÷4÷2
is highly ambiguous, be ause the result depends on the
order in whi h the operations are performed. One should always indi ate and enfor e a spe i
order
by using bra kets.
So far, we have stated laws for expressions
multipli ation. Let's take a look at a
3.
ontaining one type of operation only, namely addition or
ombination of these two operations.
Distributive law of multipli ation over addition
The produ t of a number with the sum (or dieren e) of two other numbers is equal to the sum (or
dieren e) of the produ ts of the rst number and ea h of the other two numbers.
Consider the following example:
6 × (2 + 3) = 6 × 2 + 6 × 3
In terms of variables,
a(b + c) = ab + ac.
144
(= 30)
DSC1520/SG001
Note:
⊲
Sin e multipli ation obeys the
ommutative law, the fa tors involved in the statements above may be
reversed, for example
6 × (2 + 3) = 6 × 2 + 6 × 3 = 2 × 6 + 3 × 6 = (2 + 3) × 6
and, in terms of variables,
x × (y + z) = xy + xz = yx + zx = (y + z)x.
⊲
Although we
annot easily formulate similar laws for subtra tion and division, we
transform expressions
ontaining subtra tion and division to ones
an frequently
ontaining only addition and multi-
pli ation.
For the
In the
ommutitative law:
ase of subtra tion,
hange the sign of the number being subtra ted and then add, for example
11 − 4 = 11 + −4 = −4 + 11
In the
(the commutative law).
ase of division, repla e the number being divided by its inverse and multiply, for example
9 ÷ 3 = 9 × 3−1 = 3−1 × 9
For the distributitative law, we
(the commutative law).
ombine these, for example
(17 − 4) ÷ 5 = (17 + −4) × 5−1 = 17 × 5−1 + −4 × 5−1 .
Although these laws are elementary almost to the extent of being self-evident they
tools for rearranging expressions when used in
an be very powerful
onjun tion.
A.3.1 Addition and subtra tion
Only terms of the
same type
an be added or subtra ted. For example,
2 + 3 + 5 = 10
but
2 + 3x + 5 = 7 + 3x.
Likewise,
5x + 8x − 3x = 10x
but
5x + 8x − 3xy = 13x − 3xy.
A.3.2 Multipli ation and division
When terms with the same sign (+ or
−)
are multiplied or divided, the result is positive. However, when
the signs dier, the result is negative. Examples are as follows:
7 × 5 = 35
but
7 × −5 = −35,
35 ÷ 5 = 7
but
35 ÷ −5 = −7.
It is important to remember that zero times any real number is zero and zero divided by any real number is
zero. In symbols
0 × (any real number) = 0
Note the following important multipli ation
and
0 ÷ (any real number) = 0.
onventions:
2(x + y) = 2x + 2y,
145
(A.1)
DSC1520/SG001
(x + y)2 = (x + y)(x + y) = x2 + xy + yx + y 2 = x2 + 2xy + y 2 ,
(A.2)
(x − y)2 = (x − y)(x − y) = x2 − xy − yx + y 2 = x2 − 2xy + y 2
(A.3)
(x + y)(x − y) = x2 − xy + yx − y 2 = x2 − y 2 .
(A.4)
A tivity
1. Apply the distributive law of multipli ation over addition to expand the following expressions to
ontain
the sum of four terms, ea h term being the produ t of three numbers. (You do not need to a tually
al ulate the result.)
(a)
7 × (6 × (5 + 4) + 3 × (2 + 1))
(b)
A(B(C + D) + E(F + G))
2. Whi h of the
ommutative or asso iative laws are asso iated with the following expressions?
(a)
2 + (5 + 4) = (2 + 5) + 4
(b)
(ab)c = c(ab)
( )
(3 + 7) + 4 = (7 + 3) + 4
(d)
2(7 + 4) = 2 × 7 + 2 × 4
3. Add (subtra t) the following terms:
(a)
(b)
8x + 6xy − 12x + 6 + 2xy
3x2 + 4x + 7 − 2x2 − 8x + 2
4. Simplify the following expressions:
(a)
(b)
(a + b)(6a − 3b)
(x + 2)2 − x(x + 2)
( )
4x2 +2x
2x
(d)
x2 −2xy+y 2
x−y
A.4 Fra tions
We
all the top part of a fra tion the
numerator
and the bottom part the
fraction =
To illustrate what a fra tion a tually is,
that is
numerator
.
denominator
onsider the fra tion
10
2 . This number indi ates that ten must be
split into groups of size two ea h, that is
2 + 2 + 2 + 2 + 2 = 10.
This means that ve su h groups of two
denominator,
an be formed from ten, giving
10
= 5.
2
146
DSC1520/SG001
Note that any number
an be written as a fra tion with denominator one, for example
100
1 000
5
, 1 000 =
,
5 = , 100 =
1
1
1
A.4.1 Multipli ation and division
To multiply fra tions, we multiply the numerators with ea h other and the denominators with ea h other,
for example
2 1
2×1
2
× =
= .
3 2
3×2
6
This nal fra tion
an be simplied by dividing both the numerator and denominator by two to get the
1
answer to be .
3
Whatever is done to the numerator of a fra tion, must also be done to the denominator to
ensure that the number that the fra tion represents does not
Consider the following example of multiplying fra tions
hange.
ontaining variables:
3 x+1
3(x + 1)
3x + 3
×
=
= 2
.
x x−3
x(x − 3)
x − 3x
To divide by a fra tion, we simply
multiply by the invert of the fra tion.
Remember that
1
x
= x−1 .
Examples of this operation are
5
3
4
and
2x
x+y
3x
2(x−y)
=
=5÷
5 4
20
3
= × =
,
4
1 3
3
2x
3x
2x
2(x − y)
4x(x − y)
4(x − y)
÷
=
×
=
=
.
x + y 2(x − y)
x+y
3x
3x(x + y)
3(x + y)
A.4.2 Addition and subtra tion
To add or subtra t fra tions, you need to transform the fra tions to have the same denominator.
This is
done as follows:
⊲
Find a
ommon denominator.
This is a number (or term) that is divisible by ea h of the fra tions'
denominators (often the produ t of all the denominators).
⊲
For ea h fra tion, divide its denominator into the
ommon denominator and multiply the numerator
by the result.
⊲
Now, add (or subtra t) and simplify the answer.
Consider the following numeri al example:
1(3)(5) + 2(2)(5) − 3(2)(3)
15 + 20 − 18
17
1 2 3
+ − =
=
= .
2 3 5
(2)(3)(5)
30
30
Also, when working with variables,
1
3
x + 3(x + 2)
x + 3x + 6
4x + 6
+ =
=
=
.
x+2 x
(x + 2)(x)
x(x + 2)
x(x + 2)
147
DSC1520/SG001
A tivity
Simplify ea h of the following expressions:
1.
2
3
2.
7
2x
3.
x−3
5
1
5
÷
−
x
9
+
2
x
−
x
5
A.5 Fa torisation
To simplify an expression like
2x(x + 1),
multiply the
2x
into the bra ket to nd
2x2 + 2x.
Also, by
multipli ation we nd that
(x + 2)(x + 3) = x2 + 3x + 2x + 6 = x2 + 5x + 6.
When this pro ess is reversed, we fa torise
1 to nd that
2x2 + 2x = 2x(x + 1)
and
x2 + 5x + 6 = (x + 2)(x + 3).
A tivity
1. Simplify the following expressions:
(a)
2(3x + 5) + x(4x2 + 1)
(b)
100(1 − x)(1 + x)
( )
4x2 + 7x + 2x(4x − 5)
(d)
(5x + 1)2
(e)
(5x2 + 4)(x + 1)
2. Fa torise the following expressions:
(a)
63 − 7x2
(b)
f s − f r + qr − qs
( )
12ax + 3ay + 8bx + 2by
(d)
2a2 + 11a + 12
(e)
4x2 − 13x + 3
(f )
2x2 − 8x − 24
A.6 Fun tions
A fun tion is a rule that relates how one quantity depends on other quantities.
For example, if ea h item of a
ertain produ t is sold for R8, then the in ome when ten items are sold, is
R80; if 20 items are sold the in ome is R160; et . The number of items sold is a variable and we denote it
by
q.
1
Finding what to multiply together to get an expression.
148
DSC1520/SG001
The relationship between the in ome,
I,
and the number sold,
q,
is given by the fun tion,
f
and denoted by
I = f (q) = 8Q.
Variable
I
If we use
is
x
alled the
dependent variable
and
for the independent variable and
relationship between
x
y
and
q
y
the
independent variable.
for the dependent variable, we
an write the fun tional
as
y = f (x).
In mathemati s, we often
ome a ross fun tions like
y = x2 + 5x + 2,
whi h
an also be written as
y = f (x) = x2 + 5x + 2,
to show that
y
This notation
x
is a fun tion of
(depends on
x).
an be used to determine the value of the dependent variable for
pendent variable. For example, if
x = 3,
ertain values of the inde-
then by substitution
y = f (3) = (3)2 + 5(3) + 2 = 9 + 15 + 2 = 26.
A tivity
1. Evaluate the fun tion
f (x) = 3x + 20
2. Given that the height (in
x = −4.
when
entimetres) of a parti ular obje t at time
t
(in years) is
nd the height after two years.
3. If
Note:
⊲
F (t) = 3t − t2
Any fun tion
the fun tion
⊲ y = 12 x + 3
for
t ≤ 2,
nd
F (2)
and
h(t) = 50t − 4,9t2 ,
F (3).
an also be written as an equation. For example,
y = f (x) = 2x + 6
an be written as
⊲ y = f (x) = x2 + 6x − 10
an be written as the equation
2y − x − 3 = 0,
an be written as
y = 2x + 6,
or
y − 2x − 6 = 0,
and
y − x2 − 6x + 10 = 0.
A.7 Polynomials
A polynomial is a fun tion involving only
non-negative, integer
powers of
x.
In general, we dene a polynomial as a fun tion of the form
f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x1 + a0 ,
a0 , a1 , a2 , . . . , an are real numbers
degree of a polynomial is the highest
where
alled the
The
power of
This general formula might look quite
not really the
ompli ated, but
oe ients of the polynomial.
x in its expression.
onsider the following examples to see that this is
ase:
⊲ f (x) = 2x2 − x − 2 is a polynomial
This is alled a quadrati fun tion.
of degree 2, sin e 2 is the highest power of
149
x.
DSC1520/SG001
⊲ f (x) = 4x3 − 3x2 + 2 is a polynomial
This is alled a ubi fun tion.
⊲ f (x) = x7 − 4x5 + 1
Fun tions are
in the formula.
is a polynomial of degree 7.
Note that we only need the highest power of
powers of
x
of degree 3, as 3 is the highest power of
x
to determine the degree of the polynomial.
lassied as polynomials if they
ontain only
non-negative, integer
x.
The following examples are not polynomials:
∗ f (x) = x3 −
√
x + 10
∗ f (x) = 5x2 − 4x +
is not a polynomial sin e it
3
x is not a polynomial sin e it
ontains
√
1
x = x2 .
ontains a negative power of
x,
namely
3
x
= 3x−1 .
The following table shows a summary of polynomials of degrees 0, 1, 2 and 3:
Degree
Fun tion name
0
Constant
1
Linear fun tion
2
Quadrati
3
Cubi
fun tion
fun tion
General form
a, b, c, . . .
y = ax + b
y = ax2 + bx + c
y = ax3 + bx2 + cx + d
A tivity
For ea h of the following fun tions, state whether it is a polynomial and, if so, give the degree thereof. If it
is not a polynomial, say why not.
1.
f (x) = x3 + 3
2.
f (x) = −x2 + πx
√
f (x) = x − 1
3.
4.
f (x) = 23x5 − 100 + 3x17
5.
f (x) = 4x3 − 3x−3
A.8 Per entages
As we know,
We
per ent
means
per hundred.
al ulate 5% of 250 as follows:
When we say a number in reases by a
So, 5% stands for 5 per every hundred, or
5
100 .
1 250
5
× 250 =
= 12,5.
100
100
ertain per entage, say
Number ×
x%,
then the number in reases by
x
,
100
and the in reased number is given by
Number + the increase = Number + Number ×
Consider the following examples:
150
x x
.
= Number 1 +
100
100
DSC1520/SG001
1. We
al ulate 23% of 1 534 as
1 534
35 282
23
×
=
= 352,82.
100
1
100
2. If you earn a salary of R55 240 per month and you get a 12% in rease, then your salary in reases by
12
662 880
× 55 240 =
= R6 628,80.
100
100
Your new monthly salary will be
55 240 + 6 628,80 = R61 868,80,
whi h
an also be
al ulated in one step as
New salary = Old salary(1 + 12%) = 55 240 × 1,12 = R61 868,80.
3. An apartment was valued at R63 400 in 2008. If we know that this is 5% higher than the pri e paid
for the apartment in 2006, then we
an nd the value of the apartment in 2006 (let's
all it the basi
pri e) as follows:
5
63 400 = basic price × 1 +
100
whi h leads to
basic price = 63 400 ×
= basic price ×
105
,
100
100
= R60 380,95.
105
A tivity
1. Cal ulate (i) 12% of 360,20
(ii) 11,5% of R523,00.
2. The pri e of a new washing ma hine is R4 950, whi h in ludes 15% value added tax (VAT). Cal ulate
the pri e without VAT.
3. A
ompany plans to phase out a
the present output is 500
ertain model of
ars per week,
ar by redu ing their output by 20% per week. If
al ulate the number of
ars to be produ ed in the next three
weeks.
4. In 2005, a house was valued at R450 000 and in 2011 it was valued at R700 000.
(a) Find the per entage in rease in the value of the house between 2005 and 2011.
(b) If house pri es are predi ted to rise by 8% per year between 2011 and 2015,
al ulate the proje ted
value of the house in 2012 and 2013.
( ) The number of rst-year BCom students registered at a
ertain university is 1 540 females and
2 211 males. What per entage of students is (i) male and (ii) female?
151
Appendix B: Equations and inequalities
B.1 Solving equations
To solve an equation, one needs to nd the value(s) of the variable(s) for whi h the left-hand side (LHS) of
the equation is equal to the right-hand side (RHS).
Note that an equation may have one of the following three types of solutions:
1. a
2.
3.
unique solution,
as was the
ase in the examples above
innitely many solutions, for example,
(x = 4, y = 6), (x = 1, y = 9), . . .
no solution,
for example, the equation
5
x
the equation
=0
x + y = 10
has solutions (x
= 5, y = 5),
has no solution, be ause there is no value of
x
that
an
be multiplied by 0 to obtain 5
B.1.1 Linear equations
To solve a simple equation like
x + 2 = 5,
we need to nd the value of
if
x = 3.
An equation like
and the
x
for whi h the LHS is equal to the RHS. It is
The solution to the equation is therefore,
2x − 6 + 5x = x + 3,
lear that this will be true only
x = 3.
is solved by manipulating it to get the terms
ontaining
x
on the LHS
onstant terms at the RHS:
2x − 6 + 5x = x + 3
7x − 6 = x + 3
(Add like terms on both sides.)
7x = x + 3 + 6
(Add 6 on both sides.)
7x − x = 9
(Subtract x on both sides.)
6x = 9
9
x = = 1,5.
6
(Divide by 6 on both sides.)
Another kind of equation to be solved, may be the produ t of two linear expressions, for example:
(x + 3)(x − 6) = 0
To solve this, we see that if
if
x−6=0
(or
x = 6),
x+3= 0
(or
x = −3),
the LHS is zero and the equation has been solved. Also,
the equation is true.
The solution is therefore
x = −3 or
152
x = 6.
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B.1.2 Quadrati equations
The general form of a quadrati
equation is
ax2 + bx + c = 0,
where
a, b
and
c
are
onstants.
The solution to this quadrati
Of
x that makes
√
−b ± b2 − 4ac
.
x=
2a
equation is the value of
ourse there are also other ways to solve quadrati
tion B.1. If you
the LHS equal to zero, given by
(B.1)
equations it is not always ne essary to use Equa-
an fa torise the fun tion, the roots are found easily by setting ea h of the fa tors equal to
zero.
Consider for instan e the equation
x + 10 = 11x2 − x + 1.
Here, we rst simplify to get the equation in standard form as
11x2 − 2x − 9 = 0,
with
a = 11, b = −2
and
c = −9.
Using Equation B.1, we nd the solution to be
−(−2) ±
x=
2±
p
p
(−2)2 − 4(11)(−9)
2(11)
4 − (−396)
√ 22
2 ± 400
=
22
2 ± 20
=
22
−18
22
or
=
22
22
=1
or
− 0,8181.
=
1
By using fa torisation , we nd
11x2 − 2x − 9 = (11x + 9)(x − 1) = 0.
The solution is therefore either
11x + 9 = 0,
whi h gives
x=
−9
11
= −0,8181,
or
x − 1 = 0,
whi h gives
x = 1.
Examples
1. The equation
(x − 5)(x + 5) = 0
an be solved by setting ea h of the fa tors equal to zero, or we
expand the LHS to nd
2
(x − 5)(x + 5) = 0
x − 5x + 5x − 25 = 0
x2 − 25 = 0
x2 = 25
√
x = ± 25
1
See page 148.
x = 5 or x = −5.
153
an
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2. The equation
(x + 2)2 − (x − 2)2 = 4
is solved by expanding the LHS, that is
(x + 2)(x + 2) − (x − 2)(x − 2) = 4
x2 + 4x + 4 − (x2 − 4x + 4) = 4
x2 + 4x + 4 − x2 + 4x − 4 = 4
8x = 4
1
x= .
2
3. The equation
3x2 + 25x − 18 = 0
an be solved by using fa torisation:
3x2 + 25x − 18 = 0
(3x − 2)(x + 9) = 0
3x − 2 = 0 or x + 9 = 0
2
or x = −9
x=
3
A tivity
1. Solve the following linear equations:
(a)
(b)
( )
(d)
14 + 9x = 95
16 + y
= 25
4
x
+ 2 = 2x
3
3
x−3 2 x
+ − =
5
x 5
10
2. Solve the following quadrati
equations:
(a)
(x − 2)(x + 4) = 2x
(b)
x(x2 − 2) = 0
( )
4x2 + 7x − 2x(2x − 5) = 17
3y y
−12y
+
= 480
y
2
2
(d)
(e)
5Q
P +2
1
P +2
= 20
(f )
3
2
−
=1
x 2x
(g)
2x(y + 2) − 2y(x + 2) = 0
(h)
2
=0
Q
154
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B.2 Inequalities
Equations are statements where the expression at the LHS is
equal to the expression at the RHS. Inequalities,
>) or
on the other hand, are statements where the expression at the LHS is either greater than (denoted by
less than (denoted by
<)
the expression at the RHS. For example:
9 > 5,
2<5
or
9x > 2x, 2x < 3x + 3
A handy way to visualise inequalities, is by using a number line su h as
Numbers in reasing in value.
−5
Note that zero is the
−4
−3
−2
−1
0
1
2
3
4
entral point with numbers in reasing positively to the right and negatively to the left.
It is important to realise that when numbers in rease to the negative side, they
means that
−5 < −1
5
and
de rease
in magnitude. This
−10 > −100.
An inequality statement su h as
x > −3
is indi ated on the number line as follows:
All numbers greater than −3.
−5
−4
−3
−2
−1
0
1
2
3
4
When we add or subtra t values on both sides of an inequality, it does not
−8 on both
−3 > −6, whi
8 or
sides of the inequality
is
h both are true.
5 > 2,
we nd
5 + 8 > 2 + 8,
that is
5
hange. For instan e, if we add
13 > 10
and
5 − 8 > 2 − 8,
However, when we multiply (or divide) by a negative number, the dire tion of the inequality
example, if we multiply both sides of the inequality
5(−2) > 2(−2)
or
To make the inequality true, we need to
5>2
by
−2,
x+3 = 5
hanges. For
we nd
− 10 > −4, which of course is not true.
hange the inequality symbol from > to < to have
is true ( he k on the number line).
When an equation like
that
is solved, the solution is
x=2
−10 < −4,
whi h
a unique number. The solution to an
inequality, however, is a range of values for whi h the inequality holds.
Examples
1. The solution to
x−2 > 3
is
x > 5.
This represents all numbers greater than, but not in luding, ve.
On a number line, this is
x>5
−2
2. We solve the inequality
−1
0
25
> 10
x
1
2
3
4
5
by multiplying on both sides by
6
x,
7
that is
8
25 > 10x
Illustrated on a number line, we have
x < 2,5
−2
−1
0
1
2
3
155
4
5
6
7
8
or
x < 2,5.
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3. Solving the inequality
2
2x − 6 ≤ 12 − 4x,
with
2x − 6 ≤ 12 − 4x
2x + 4x ≤ 12 + 6
x > 0,
gives
(Add 6 and 4x on both sides.)
6x ≤ 18
x ≤ 3 and x > 0.
Note that
below as a
x ≤ 3 indi ates that three is in luded in the solution. This is indi ated on the
bla k dot at x = 3, while the dot at x = 0 is open, sin e zero is not in luded:
0<x≤3
−5
−4
−3
−2
−1
0
4. Consider the following solution to the inequality
3x − 29 ≤ 7x + 11
3x − 7x ≤ 11 + 29
2
3
4
5
3x − 29 ≤ 7x + 11:
(Add 29 and subtract 7x on both sides.)
−4x ≤ 40
x ≥ −10.
5. The solution to the inequality
1
(Divide by − 4 and swap the inequality.)
(x − 3)(−x + 5) > 0
x > 3 or
is
− x > −5, that is x < 5.
On the number line, this is shown as
3<x<5
−2
−1
0
1
2
3
4
5
6
7
8
A tivity
Find the range of values for whi h the following inequalities are true, assuming that
1.
2.
3.
2
x − 12 > 10
−75
> 15
x
2x − 6 ≤ 12 − 4x
The symbol ≤ stands for less than or equal to, and ≥ for greater than or equal to.
156
x > 0.
number line
Appendix C: Linear fun tions
C.1 Graphs of linear fun tions
The general form of a linear fun tion is
y = f (x) = ax + b,
where
a
and
b
are
onstants. Here,
a
The graph of su h a linear fun tion
the line and
b
is
alled the
x.
oe ient of
onsists of a straight line drawn on a system of axes, with
a
the slope of
the inter ept on the verti al axis.
Consider the graph of su h a straight line in Figure C.1:
y
y
y
b
x
inter ept
x
=a
+b
inter ept
b
−b
a
x
b
a
Figure C.1: Graph of
The
⊲
x
and
The
y
y
inter epts are
⊲
The
x
learly indi ated in the gure. These
inter ept is found where
therefore gives the
y
y = f (x) = ax + b
x = 0,
that is where
y = 0,
that is where
with
a>0
an be determined algebrai ally as follows:
y = a(0) + b = b.
The
b
in the fun tion
y = ax + b
inter ept.
inter ept is found where
The slope of a linear fun tion is given by the
⊲
the
⊲
the number of units by whi h
⊲
the ratio of the
0 = ax + b,
oe ient of
x,
namely
giving
a.
ax = −b
or
x=
−b
a , if
a 6= 0.
The slope is dened as
hange in height per unit in rease in horizontal distan e; or
hange in
y
y
hanges when
values and the
x
in reases by one unit; or
hange in
Slope =
x
values, that is
∆y
y2 − y1
=
.
∆x
x2 − x1
157
(C.1)
DSC1520/SG001
y
From Figure C.1 we see that
in reases by
b
units when
Slope =
The form of a linear fun tion
Figure C.2 shows the dierent
b
b
a
x
=b×
in reases by
b
a units. The slope is therefore
a
= a.
b
an be anti ipated by simply looking at its slope (a) and the
y
inter ept (b).
ases:
y
y
a > 0, b > 0
a < 0, b > 0
x
x
(a) Positive slope; positive y inter ept
(b) Negative slope; positive y inter ept
y
y
a > 0, b < 0
a < 0, b < 0
x
x
( ) Positive slope; negative y inter ept
(d) Negative slope; negative y inter ept
y
a = 0, b < 0
x
(e) Zero slope; negative y inter ept
Figure C.2: Straight line graphs with dierent slopes and
y
inter epts
C.2 Plotting linear fun tions
In this se tion we dis uss two ways to plot linear fun tions, namely by using
⊲
⊲
the slope and
y
axis inter ept as given by the fun tion; and
the inter epts on the
x
and
y
axes.
C.2.1 Using the slope and y axis inter ept
For the general linear fun tion
Consider the fun tion
y = ax + b,
the slope of the line is given by
y = 2x − 1.
158
a
and the
y
axis inter ept by
b.
DSC1520/SG001
The slope of this line is
a = 2.
This means that for every unit in rease in
x, y
in reases by two
units.
y
The
axis inter ept is at
y = b = −1.
The fun tion is plotted in Figure C.3(a), with
Maxima to show that it is indeed the
∗
at the
y
inter ept. We in lude the graph from
orre t line.
y
1
Slope =
0
−1
∗
−1
1
2 units up
1 unit right
2
=2
x
3
−2
(a) Using slope and y inter ept
(b) Maxima
Figure C.3: Graph of
y = 2x − 1
C.2.2 Using the inter epts on the x and y axes
Any linear fun tion
an be graphed if two points on the line are known.
It is important to realise that everywhere on the
Consider the fun tion
We know that
Likewise,
the point
x=0
y = 0 on
1
2; 0 .
A straight line
y
axis,
x = 0.
Also, everywhere on the
x
axis,
y = 0.
y = 2 − 4x.
on the
the
x
y
axis. The
axis. The
x
y
inter ept is therefore at
y = 2 − 4(0) = 2,
inter ept is therefore found where
an now be drawn through the points
Again, we in lude the graph by Maxima as
(0; 2)
and
onrmation.
1
2;
0
0 = 2 − 4x
that is the point
or
x=
−2
−4
=
(0; 2).
1
2 , that is
to get the graph in Figure C.4(a).
A tivity
1. Plot ea h of the following fun tions by using the inter epts on the
(a)
y =x+1
(b)
y = −5x − 25
2. Graph ea h of the following fun tions by using the slope and the
(a)
y = −6 + 9x
(b)
y = −2x + 4
y
x
and
y
axes:
inter ept:
C.3 Determining formulæ of linear fun tions
It is often ne essary to nd the fun tion (or equation) representing a line from available data. We may only
have two points on the line, or we may have the slope of the line and a point that it passes through.
159
DSC1520/SG001
y
6
4
2
b
(0; 2)
0
−1
−2
b
1
2;
0
1
x
−4
(a) Using slope and y inter ept
(b) Maxima
Figure C.4: Graph of
y = 2 − 4x
C.3.1 Given two points on the line
As noted before, any line
an be graphed if two of the points on the line are known. This also means that if
we have two points that fall on a line, we
an determine the linear fun tion representing the line.
For example, if we know that the points (1; 3) and (3; 7) are on a line, we
with
a
the slope and
b
an nd the fun tion
the verti al inter ept of the line.
From Equation C.1 on page 157 we know that
a=
If we take
(x1 ; y1 ) = (1; 3)
and
∆y
y2 − y1
=
.
∆x
x2 − x1
(x2 ; y2 ) = (3; 7),
a=
then
7−3
4
= = 2,
3−1
2
giving the general linear fun tion to be
y = 2x + b.
Now, to nd the value of
b,
we use one of the given points, say (1; 3), to nd
3 = 2(1) + b
or
b = 3 − 2 = 1.
The fun tion of the line is therefore
y = 2x + 1.
160
y = ax + b
DSC1520/SG001
Note:
⊲
(x1 ; y1 ) and whi h as (x2 ; y2 ).
(x2 ; y2 ) = (1; 3), we would have
It does not matter whi h point we label as
(x1 ; y1 ) = (3; 7)
way around with
and
a=
If we have taken it the other
3−7
−4
=
= 2,
1−3
−2
as before.
⊲
We
ould also have used the other point (3; 7) to nd the value of
7 = 2(3) + b
or
b,
namely
b = 7 − 6 = 1,
whi h is the same!
C.3.2 Given the slope of the line and any point on the line
Sin e the slope of a line is the same when measured between any two points, we
be
(x1 ; y1 )
and
(x; y),
an
onsider the points to
and write (from Equation C.1)
a=
y − y1
.
x − x1
Simplifying this gives
y − y1 = a(x − x1 )
or
y = a(x − x1 ) + y1 ,
whi h is the fun tion of the line with slope
Consider for example a line with slope
a
that passes through the point
a = 2
and a point
(x1 ; y1 ).
(x1 ; y1 ) = (3; 5)
on the line.
The equation
representing this line is
y = a(x − x1 ) + y1
= 2(x − 3) + 5
= 2x − 6 + 5
= 2x − 1.
Sometimes data about the line in question are not given expli itly, and you need to nd data from given
information. Consider for example the following:
Miriam bakes vetkoek to be sold at the station. If she bakes 20 vetkoek, it
40 vetkoek, her
ost is R13. Assuming that there is a linear relationship between the
of vetkoek baked, nd the linear
ost and the number
ost fun tion.
Let's take the independent variable
be the
osts her R8,00 and if she bakes
x
to be the number of vetkoek baked and the dependent variable
ost to bake them. We write the given information in a table to summarise it for ourselves:
x
y
20
8
40
13
161
y
to
DSC1520/SG001
Two data points that satisfy the linear relationship are therefore
P1 = (x1 ; y1 ) = (20; 8)
and
P2 = (x2 ; y2 ) = (40; 13).
As before, we nd the slope of the line to be
a=
13 − 8
5
1
y2 − y1
=
=
= ,
x2 − x1
40 − 20
20
4
so the fun tion of the line is given by
y=
1
x + b.
4
We now use any one of the given points to determine the value of
1
8 = (20) + b
4
The fun tion is therefore
y=
or
1
x+3
4
b.
Say we use
P1 = (20; 8)
to get
b = 8 − 5 = 3.
or
y = 0,25x + 3.
A tivity
1. Determine the fun tion of the straight line that passes through the points
(−2; 8)
and
(4; 1).
2. Find the equation of a line with slope 1,5 that passes through the point (2; 4).
3. Determine the expression that represents the straight line through the points
4. Find the inter epts on the
x
and
y
axes for the linear fun tion
5. A bus agen y has room for 60 people on a bus tour. If they
(1; 2)
and
(3; 3).
y = f (x) = −4x + 3.
harge R600 per person, they will be able
to ll the bus. They know from experien e that if they in rease the pri e of the tour by R50 they will
lose three
ustomers. Determine the demand fun tion if demand
fun tion of the pri e
p
d
(number of
ustomers) is a linear
(in rand).
6. Mr Walsh sells Brighter-and-Better washing powder. If he sells the washing powder for R190 per box,
demand is 26 000 boxes per week. If he
p
is the pri e per box,
d
harges R210 per box, the weekly demand is 16 000 boxes. If
is the weekly demand and we assume there is a linear relationship between
pri e and demand, derive an expression for weekly demand.
C.4 Plotting linear inequalities
The fun tion
y = f (x) > 3x − 2
represents an inequality and
y > 3x − 2
To plot the inequality
y > 3x − 2,
or
y − 3x + 2 > 0.
we start by plotting the line
whi h side of the line the inequality holds.
First, we plot the fun tion
an be written as
y = 3x − 2
with slope
To nd where the inequality holds, we
a=3
y = 3x − 2
and verti al inter ept
as before and then de ide on
b = −2
as in Figure C.5(a).
onsider the origin (i.e. the point (0; 0)) and nd
0 > 3(0) − 2
or
0 > −2,
whi h is true. The origin therefore falls within the inequality and we
162
an indi ate it as shown in Figure E.3.
DSC1520/SG001
y
y
1
1
0
−1
−1
0
−1
−1
x
1
−2
−2
−3
−3
−4
−4
(a) The fun tion y = 3x − 2
Note that the inequality is
stri tly
1
x
(b) Inequality y > 3x − 2
larger than (>).
It therefore does not in lude the line itself, and we
indi ate it as a dotted (or dashed) line.
The shaded area in Figure E.3 is
alled the feasible area (or region) of the inequality. If we have two (or
more) inequalities and need to determine where both of them hold, the feasible area is the area where the
feasible areas overlap.
C.5 Simultaneous equations/inequalities
C.5.1 Solving simultaneous linear equations
To solve simultaneous linear equations, one may use either the method of elimination or the method of
substitution. In the examples below we apply either of these methods as and when they are appropriate.
Two equations in two variables
To illustrate, we work through a few examples of solving systems of simultaneous equations.
1. Consider the following system of simultaneous equations:
x + 3y = 4
−x + 2y = 6.
To solve the system algebrai ally, we try to eliminate one of the variables and nd the value of the
remaining variable. Sin e we have an
x
in one equation and a
adding the equations to get
0 + 3y + 2y = 10
−x
in the other, we
or
5y = 10,
giving
x = −2,
whi h results in
y = 2.
By substituting
y=2
into the rst equation, we nd
x + 3(2) = 4
and into the se ond
−x + 2(2) = 6
giving
163
−x=2
or
x = −2.
an eliminate
x
by
DSC1520/SG001
We may therefore use either equation for the substitution.
The solution to the system is therefore at the point
(x; y) = (−2; 2),
that is where the lines interse t.
2. Solve the following equations simultaneously:
2x + 3y = 12,5
(1)
−x + 2y = 6
(2)
From equation (2) we nd that
x = 2y − 6.
When we repla e the
x
(3)
in equation (1) with
2y − 6,
we nd
2(2y − 6) + 3y = 12,5
4y − 12 + 3y = 12,5
7y = 24,5
y = 3,5.
When we now substitute
y = 3,5
into equation (3), we get
x = 2(3,5) − 6 = 1.
The solution to this system of simultaneous equations is therefore
(x; y) = (1; 3,5).
3. Solve the following system of simultaneous equations:
2x + 3y = 0,75
(1)
5x + 2y = 6
(2)
From equation (1) we nd
2x = −3y + 0,75
x = −1,5y + 0,375.
(3)
Substituting this into equation (2) gives
5(−1,5y + 0,375) + 2y = 6
−7,5y + 1,875 + 2y = 6
−5,5y = 6 − 1,875
4,125
y=
= −0,75.
−5,5
When we substitute
y = −0,75
into equation (3), we nd
x = −1,5(−0,75) + 0,375 = 1,5.
The solution to this system of simultaneous equations is the point where the lines representing the
equations interse t, that is the point
(x; y) = (1,5; −0,75).
Three equations in three variables
In the
ase of three equations
ontaining three variables (unknowns), the same methods
Consider the following set of three simultaneous equations in three variables:
2x + y − z = 4
(1)
x+ y−z = 3
(2)
2x + 2y + z = 12
(3)
164
an be applied.
DSC1520/SG001
y
and
y − z = 2,
and
When we subtra t equation (2) from equation (1), we eliminate both
When we substitute
x=1
giving
2 + 2y + z = 12
giving
Adding equations (4) and (5) eliminates
x = 1.
z,
(4)
2y + z = 10.
(5)
and we nd
3y = 12 or
y=4
and get
into equations (2) and (3) we nd
1+ y−z = 3
We now substitute
z,
y = 4.
into equation (4) and nd
4 − z = 2 or
z = 2.
The solution to this system of simultaneous equations is the point where all three equations pass through,
that is where
x = 1, y = 4
z=2
and
in the three-dimensional spa e.
C.5.2 Solving simultaneous equations graphi ally
When two simultaneous linear equations are plotted on the same axes, a solution to the system is found
where the equations interse t.
Consider again the system
x + 3y = 4
−x + 2y = 6.
Writing the equations in standard fun tional format, gives
1
4
y =− x+
3
3
and
y = 0,5x + 3.
Graphi ally, the solution of the system is found at the point where the lines interse t, as shown in Figure C.5.
y
y=
− 1x
3
+
3
4
3
2
b
0
y=
−6
−5
,5 x
(−2; 2)
+3
−4
1
−3
−2
−1
Figure C.5: Simultaneous equations
1
x + 3y = 4
2
and
3
4
x
−x + 2y = 6
Depending on the equations, the solution to a system of equations may be one of the following:
⊲
a
unique solution
⊲ no solution
when the lines
when they never
⊲ innitely many solutions
ross (as above)
ross (e.g. when the lines are parallel)
when the lines overlap
165
DSC1520/SG001
C.5.3 Systems of inequalities
To solve a system
onsisting of inequalities, we need to nd the feasible area of ea h inequality. The solution
is then given by the area where the individual feasible areas
overlap.
Consider, for instan e, the following system of inequalities:
x + 3y ≤ 4
(1)
−x + 2y < 6.
(2)
The lines representing these equations have been shown in Figure C.5 (above), where we found the point of
(−2; 2).
interse tion to be at
We now need to nd the feasible area of ea h inequality. As before, we
onsider the origin to determine on
whi h side of ea h line the equation holds.
For inequality (1), at (0; 0) we nd that
(0) + 3(0) = 0 ≤ 4,
whi h is true. The feasible area therefore in ludes the origin. This area is therefore below the line, as shown
in Figure C.6(a).
For inequality (2), at (0; 0) we nd that
−(0) + 2(0) = 0 < 6,
whi h is also true.
So, again the feasible area
ontains the origin and is below the line as shown in Fig-
ure C.6(b).
y
x+
3y ≤
4 (1
)
y
3
3
2
Feasible area
1
−x
b
−6
−5
−4
−3
−2
−1
1
2
3
4
=6
2
(2)
Feasible area
1
b
x
−6
−5
(a) Feasible area of x + 3y ≤ 4
−4
−3
−2
−1
1
2
3
4
x
(b) Feasible area of −x + 2y < 6
Figure C.6: Inequalities
Note:
y
+2
x + 3y ≤ 4
and
−x + 2y < 6
For the < inequality, the line representing the equation is not in luded and is indi ated by a dashed
line not a solid line as for the ≤ inequality.
The feasible area representing the solution to the system of inequalities is found where the individual feasible
areas overlap, that is the
ross-hat hed area in Figure C.7.
A tivity
1. Solve the following system of equations:
2y + 10x = 580
(1)
y − 10x = −40
(2)
2. Find the solution to the following set of simultaneous equations:
x
+1
2
y = 0,5x + 2
y=
166
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y
x+
3y =
4
+
−x
−6
−5
=6
2y
3
(1)
2
(2)
Feasible area
−4
−3
−2
1
−1
Figure C.7: Simultaneous inequalities
1
x + 3y ≤ 4
2
and
3
−x + 2y < 6
3. Solve the following system of simultaneous equations:
4x − 3y + 1 = 13
0,5x + y − 3 = 4
4. Find the solution to the following simultaneous equations:
p=
18 − 10q
5
2=
3q + 5p
2
5. Solve the following simultaneous equations:
7
5
q − 3p =
2
2
and
3p = 3(q − 3).
6. Solve the following system of inequalities:
2x − 3y ≤ 12
(1)
x + 5y ≤ 20
(2)
x ≥ 0.
(3)
167
4
x
Appendix D: Non-linear fun tions
In this appendix, we dis uss quadrati
and hyperboli
and
ubi
fun tions in detail and we tou h on exponential, logarithmi
fun tions. The graphs of these fun tions have distin t features.
In Figure D.1 examples of quadrati
and
1
ubi
fun tions are shown :
f (x)
f (x)
100
30
50
20
0
−4 −2
−50
10
−3 −2 −1
1
2
x
3
2
4
6
8
10
x
−100
(a) f (x) = 3x2 + x + 3
(b) f (x) = 0,5x3 − 5x2 + 1,5x + 25
Figure D.1: Graphs of quadrati
and
ubi
fun tions
D.1 Quadrati fun tions
The fun tion
y = ax2 + bx + c
is
x,
alled a quadrati
fun tion. Here,
a, b
and
c
are
onstant values, with
a
and
b
the
oe ients of
x2
and
respe tively.
Solving a quadrati fun tion implies that we need to
uts the x axis, that is where y = 0. These values are
2 the standard quadrati
same as solving
nd the values of
x
where the graph of the fun tion
alled the roots of the quadrati
fun tion. This is the
equation
ax2 + bx + c = 0.
The equation
x2 = 3−2x, for example, is written in standard form as either x2 +2x−3 = 0 or −x2 −2x+3 = 0.
These equations are the same, sin e we
sign by
Note:
−1,
We may multiply by
fun tions
f (x) = x2 + 2x − 3
graphs in Figure D.6.
1
2
an simply multiply either one of them on both sides of the equal
to nd the other.
−1
and
equations, but not when working with fun tions. The
f (x) = −x2 − 2x + 3 are obviously not the same, as an be seen from their
when working with
These graphs were generated by Maxima. We en ourage you to use this software to visualise non-linear fun tions.
See Se tion B.1.2 on page 153 for information on solving quadrati equations.
168
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f (x)
f (x)
4
4
2
2
0
−4 −3 −2 −1
−2
1
2
0
−4 −3 −2 −1
−2
x
−4
1
2
x
−4
(a) f (x) = x2 + 2x − 3
(b) f (x) = −x2 − 2x + 3
Figure D.2: Quadrati
fun tion multiplied by
−1
(using Maxima)
D.1.1 Properties and graphs
The properties that follow, are given in terms of the quadrati
fun tion in
standard form,
namely
f (x) = ax2 + bx + c.
(a)
The general form of the graph
From Figure D.6(a) (above) we see that when the
oe ient of
x2
is positive, the turning point of the
graph is at the bottom, and the legs of the graph point upwards.
On the other hand, as
an be seen from Figure D.6(b), when the
oe ient of
x2
is negative, the turning
point is at the top and the legs point downwards.
When the
oe ient of
and when the
(b)
x2
oe ient of
is positive (>
x2
0)
is negative (<
the fun tion has a minimum turning point,
0)
the graph has a maximum turning point.
The turning point
The
The
oordinates of the turning point are important when plotting the graph of a quadrati
x
3
value of the turning point is alled the vertex of the graph and it is given by
xm =
The
fun tion.
−b
.
2a
oordinates of the turning point of a quadrati
fun tion are
given by
( )
−b
2a
.
The y axis inter ept
The
y
inter ept of a quadrati
verti al inter ept of
f
f is the point where it
f (0) = a(0)2 + b(0) + c = c.
fun tion
is therefore
The
3
−b
;f
2a
y
inter ept of
f (x) = ax2 + bx + c
uts the
axis, that is where
is given by
When we dis uss dierentiation later on, you will see how this point is derived.
169
y
c.
x = 0.
The
DSC1520/SG001
(d)
The x axis inter ept(s)
The points where a quadrati
f (x) = ax2 + bx + c = 0
fun tion
and solve for
uts the
x
x
axis, are found where
y = 0.
We nd these roots by either fa torisation or by using the `minus
x=
The kind of roots of a quadrati
square root sign, namely
⊲
When
an
−b ±
b'
formula,
√
b2 − 4ac
.
2a
fun tion
an be determined by looking at the expression under the
whi h is
alled the
b2 − 4ac,
b2 −4ac > 0, we
We therefore need to set
to nd the roots of the fun tion.
al ulate
√
dis riminant.
b2 − 4ac and the solution
onsists of two dierent, real-valued
roots. Figure D.3(a) shows su h a fun tion.
⊲
When
b2 − 4ac = 0,
the `minus
b2 − 4ac < 0,
we
fun tion therefore tou hes the
⊲
When
b' formula be omes −b
2a , whi h is the turning point of the graph. The
x axis at its turning point. Figure D.3(b) shows su h a fun tion.
annot evaluate the square root, sin e the square root of a negative number
4
is not a real number . In this
ase the fun tion does not tou h the
x
axis and has no
x
inter epts.
Figure D.3( ) shows su h a fun tion.
These dierent kinds of roots are visualised in Figure D.3.
indi ates the type of roots
Che k for yourself that the dis riminant
orre tly.
f (x)
f (x)
25
4
0
−5 −4 −3 −2 −1
−2
2
−4
15
−6
10
−8
5
f (x)
0
−4 −3 −2 −1
−2
1
2
x
x
1
20
−10
−4
(a) f (x) = x2 + 2x − 3
−3 −2 −1
(b) f (x) = −x2 − 2x + 3
1
2
3
x
( ) f (x) = 3x2 + x + 3
Figure D.3: Dierent kinds of roots the value of the dis riminant
Example
Use the properties dis ussed above to sket h the graph of the quadrati
fun tion
y = f (x) = −x2 − 2x + 3.
(a) We see that
(b) The
y
a = −1 < 0
and we know the parabola will have a maximum turning point.
inter ept is 3. (If we set
x = 0,
we nd
y = 3.)
b2 − 4ac = (−2)2 − 4(−1)(3) = 16 > 0,
x2 + 2x − 3 = 0) has two distin t roots.
( ) Sin e the dis riminant is
is the same as
4
No real value exists that gives a negative value when squared.
170
the equation
−x2 − 2x + 3 = 0
(whi h
DSC1520/SG001
By fa torising the LHS, we nd
the
x
axis.
(d) The vertex of
f
the turning point is
The graph of
f
x = −b
2a =
at (−1; 4).
is at
(x + 3)(x − 1) = 0,
−(−2)
2(−1)
= −1.
At
whi h gives
x = −1
is shown in Figure D.4. The points that were
x = −3
we nd
and
x=1
as the inter epts on
f (−1) = −(−1)2 − 2(−1) + 3 = 4,
so
al ulated above are indi ated as bla k dots.
(−1; 4) f (x)
4
3
2
1
−3
−2
Figure D.4:
0
−1
−1
1
2
x
f (x) = −x2 − 2x + 3
A tivity
For ea h of the following quadrati
fun tions, determine the inter epts on the axes and the turning point.
Indi ate the shape of ea h by means of a rough graph.
Solutions are available on page 194.
1.
f (x) = 2x2 − x − 3
2.
f (x) = 4x2 − 16x + 16
3.
f (x) = −3x2 + 3x − 2
D.2 Cubi fun tions
A
ubi
fun tion in general notation is
f (x) = ax3 + bx2 + cx + d,
where
a, b, c and d
are
onstants.
Unfortunately there are no general rules (as for quadrati
sin e, depending on the
fun tions) to analyse and plot
oe ients of the fun tion, they may behave so dierently.
ubi
fun tions
As shown in Figure
D.5, they may have no turning points and a single root (Figure D.5(a)), or two turning points with one root
(Figure D.5(b)), two roots (Figure D.5( )) or three roots (Figure D.5(d)):
To illustrate, we work through a few examples. You will see that we
estimates for the roots of
ubi
an easily nd the general form and
fun tions by employing mathemati al software like Maxima.
171
DSC1520/SG001
f (x)
60
f (x)
40
4
20
3
0
−4 −2
−20
2
1
−1.0
−0.5
0.5
1.0
2
4
6
x
8
−40
x
(a) f (x) = 3x3 + 2
(b) f (x) = 0,5x3 − 5x2 + 8,5x + 27
f (x)
40
f (x)
20
40
0
−12 −10 −8 −6 −4 −2
−20
20
−4
−3
0
−1
−20
−2
1
x
4
−40
x
2
2
−60
−80
−40
−100
( ) f (x) = 3x3 + 9x2
(d) f (x) = −0,5x3 − 5x2 + 8,5x + 27
Figure D.5: Dierent kinds of roots the value of the dis riminant
Examples
1. Consider the fun tions
f (x) = x3
f (x) = y = x3 are found by
3
same applies for f (x) = −x .
The roots of
root. The
and
setting
y = 0,
f (x) = −x3 .
that is
f (x)
f (x)
20
20
10
10
0
−3 −2 −1
−10
1
2
3
1
2
3
x=0
as the only
x
−20
(a) f (x) = x3
Figure D.6: Graphs of
whi h gives
0
−3 −2 −1
−10
x
−20
It is
x3 = 0,
(b) f (x) = −x3
f (x) = x3
and
f (x) = −x3
lear that these fun tions are ree tions of one another, while ea h has only one root at
no turning points.
172
x = 0 and
DSC1520/SG001
2. Consider the
ubi
fun tion
f (x) = x3 − 9x + 5.
By simply looking at the fun tion, we
To nd the roots of a
ubi
an say that the inter ept on the
y
axis is at
fun tion like this, we need to solve the equation
5
not a simple task and we will not attempt it at this stage .
y = 5.
x3 − 9x + 5 = 0,
whi h is
f (x)
30
20
10
0
−4 −3 −2 −1
−10
1
2
3
4
x
−20
Figure D.7:
f (x) = x3 − 9x + 5
A tivity
Graph ea h of the following
1.
2.
ubi
f (x) = 0,5x3 − 8x2 + 200
Y = 3x3 + 9x2
fun tions and nd their roots (use a mathemati al tool like Maxima):
over the interval
over the interval
−5 < x < 15.
−4 < x < 2.
D.3 Exponential fun tions
D.3.1 Denition
Exponential fun tions have the general form
y = f (x) = ax ,
where
base
⊲ a
is a
⊲ x
is the power (or index) of the exponential fun tion.
Let's
onstant, referred to as the
onsider
a=4
and evaluate
4x
of the exponential fun tion, and
as shown in Table D.1:
The graph of this fun tion is shown in Figure D.8:
Figure D.9 shows the following exponential fun tions plotted on the same axes:
y = 2−x , y = 2x , y = ex
5
and y = 3,5x .
You will nd the tools to determine the turning points of ubi fun tions when we get to dierentiation in Study unit 4.
173
DSC1520/SG001
x
4x
−3
−2
−1
0
1
2
3
0,016
0,06
0,25
1
4
16
64
Table D.1:
f (x) = 4x
f (x)
60
50
40
30
20
10
−3 −2 −1
1
Figure D.8:
Note that
e
e
in the third fun tion is the number
is the base of the
natural logarithms
2
x
3
y = 4x
e = 2,7182818284590452353602874713527 . . . .
6
and is very important in mathemati s
f (x)
y = 3,5x
5
y = 2−x
The number
y = ex
y = 2x
4
3
2
1
−2
−1
1
Figure D.9: Graphs of
x
2
2−x , 2x , ex
and
3,5x
From this graph, we see the following properties of exponential fun tions:
⊲
All
urves are
ontinuous (no breaks) and pass through the point
the power zero is one ((real
6
(0; 1).
number)0 = (variable)0 = 1).
Read more about this at https://www.mathsisfun. om/numbers/e-eulers-number.html.
174
Remember that anything to
DSC1520/SG001
2x ,
⊲
When the index (or power) is positive, as in
⊲
When the index (or power) is negative, as in
⊲
Exponential fun tions with larger bases in rease more rapidly for
x<
⊲
0. Compare 2x and
From Figure D.10 it is
y > 0),
the
2−x ,
urve in reases and is
the
alled a growth
urve de reases and is
x>0
urve.
alled a de ay
urve.
and de rease more rapidly for
3,5x in the graph.
y = ax and y = a−x
y = −a−x are always below
lear that the graphs of
while the graphs of
y=
−ax and
(a) y = 1,5x and y = −1,5x
x
y < 0).
are always above the
the
x
axis (i.e.
(b) y = 1,5−x and y = −1,5−x
Figure D.10: Graphs of positive and negative exponential fun tions (Maxima)
D.3.2 Working with exponentials
The rules below must be followed when multiplying, dividing and raising exponentials to a power:
1. To multiply two exponentials with the same base, add the indi es:
am × an = a(m+n)
Example:
43 × 45 = 4(3+5) = 48
2. To divide two exponentials with the same base, subtra t the indi es:
am ÷ an =
am
= a(m−n)
an
Example:
43 ÷ 45 =
43
1
= 4(3−5) = 4−2 = 2
45
4
3. To raise an exponential to a power, multiply the indi es:
(am )k = am×k
Example:
(52 )3 = 52×3 = 56
175
axis (i.e.
DSC1520/SG001
Examples
1. Simplifying
25
23 2−4 , gives
25
25
=
23 2−4
23−4
25
= −1
2
= 25−(−1)
= 26 = 64.
2. Simplifying
√
35
, gives
3−4 34
√
5
35
32
= −4+4
3−4 34
3
5
32
= 0
3
= 32,5
√
1
( a = a2 )
(30 = 1)
= 15,588.
3. Simplifying
7eα eβ
, gives
e3
7eα eβ
7eα+β
=
e3
e3
α+β−3
= 7e
.
4. Simplifying
2L0,5
L−2
2
(Add when multiplying.)
(Subtract when dividing.)
, gives
2L0,5
L−2
2
2
= 2L0,5−(−2)
= 2L2,5
(Subtract when dividing.)
2
= 22 L2×2,5
(Multiply when raising to power.)
5
= 4L .
D.3.3 Solving equations
To solve an equation
ontaining exponentials, start by writing ea h side of it as
(base)power ,
with the bases
the same throughout. Then, sin e the bases are identi al, the powers must also be equal for the LHS to be
equal to the RHS. For example,
2x = 24
leads to
x = 4.
Examples
1. To solve the equation
2x =
1
16 , we write both sides in the form
2x =
whi h leads to
1
1
= 4 = 2−4 ,
16
2
x = −4.
176
2? ,
that is
DSC1520/SG001
2. To solve
e2x
e4+x
= 1,
we need to write it with both sides in the form
e2x
=1
e4+x
e2x = e4+x
whi h leads to
3. To solve
2
4t
=
2x = 4 + x,
e? ,
that is
(Multiply both sides by e4+x .)
resulting in
x = 4.
1
2 , we need to write both sides as
4? ,
that is
1
2
=
4t
2
2 × 2 = 4t
4 = 4t ,
resulting in
t = 1.
4. In the following equation we have
form
e? ,
t
in the exponent of
e.
We therefore need to write both sides in the
that is
2
+5
e5 et+1
1
et = 5+t+1
e
et = e−t−6 ,
5 + 2et =
whi h results in
t = −t − 6,
or
2t = −6,
(Subtract 5 and divide by 2 on both sides.)
giving
t = −3.
A tivity
1. Simplify ea h of the following expressions:
(a)
(b)
( )
(d)
4L2
L−2
xy
3y 2
2Q
√
3P P
2
−
xy 2
2y
ex (1+e1−x )
e
2
2. Solve ea h of the following equations:
√1
16
(a)
2x =
(b)
(1 + 2t)0,5 = 12
( )
e5x =
1
xx−5
(d)
16 −
(e)
2
5t
1 −2t
8
=2
=0
177
DSC1520/SG001
D.4 Logarithmi fun tions
D.4.1 Denition
In the previous se tion we solved equations by manipulating them to have the same base throughout and
then equating the powers. But, what if it is not possible to do this? For example, take the equation
The RHS
annot be written as two to some power, and we
For su h problems we use logarithms.
2x = 10.
annot solve the equation.
We say that to take the logarithm of a number is the inverse of
exponentiation in the same way as subtra tion is the inverse of addition and division is the inverse of
multipli ation.
The logarithm of
x
is the power to whi h base
words, the logarithm of
x
to base
b
b
x. In other
by = x, written
must be raised to yield
is the solution
y
to the equation
as
logb x = y.
We know that
3
sin e 2
100 = 102 ,
therefore we
an write
log10 100 = 2.
In the same way we
an write
log2 8 = 3,
= 8.
In general, we say that if
Number = BasePower ,
then
logBase Number = Power.
Of
ourse, the base
exponentiation.
(D.1)
an be any number su h as 2, 3, 5 or 10, as we have seen in previous examples of
The most frequently used bases are 2 (used in
base, used in s ien e and engineering) and
e
omputer s ien e), 10 (the most
(the base of the natural logarithm, with
in mathemati s and physi s).
On your
base
al ulator you should nd the
e (loge x).
b ≈ 2,718,
ommon
used widely
log key for logarithms to base 10 (log10 x) and the ln key for logs to
We will use only these two bases in this module.
D.4.2 Graphs and properties
In Figure D.11
as
ln x
log x and ln x are graphed on the same axes. Note that Maxima does not have a fun tion su h
log(x), whi h represents ln(x). We therefore have to enter log10 (x) as log(x)/ log(10).
it only has
Figure D.11: Graphs of
ln x
(blue) and
log10 (x) =
From the graphs, we note the following properties of algorithms:
178
ln(x)
ln(10) (red)
DSC1520/SG001
⊲
The logarithm of values
≤0
⊲
The logarithm of one is zero for any base (log 1
⊲
Logarithms of values less than one is negative.
⊲
Logarithms of values greater than one is positive.
do not exist.
= ln 1 = 0).
D.4.3 Working with logarithms
The rules below must be followed when
1. To add two logarithmi
ombining and manipulating logarithmi
fun tions:
values to the same base, nd the logarithm of the produ t of the values:
logb (M ) + logb (N ) = logb (M N )
Example:
ln(14) + ln(9) = ln(14 × 9) = ln(126) = 4,8363
Che k if this is indeed true:
2. To subtra t two logarithmi
ln(14) = 2,6391, ln(9) = 2,1972
and the sum is
2,6391 + 2,1972 = 4,8363
values to the same base, nd the logarithm of the quotient of the values:
logb (M ) − logb (N ) = logb
M
N
Example:
ln(40) − ln(9) = ln
Che k if this is indeed true:
40
9
= ln(4,4444) = 1,4917
ln(40) = 3,6889, ln(9) = 2,1972
and
3,6889 − 2,1972 = 1,4917
3. To nd the logarithm of a number raised to a power, multiply the power by the logarithm of the
number:
logb (M z ) = z logb (M )
Example:
log(43 ) = 3 log(4) = 3 × 0,6021 = 1,8063
Che k if this is indeed true:
log(43 ) = log(64) = 1,8062
(The dieren e in the fourth de imal is due to
rounding.)
4. To
hange the base of the logarithm of a value, divide the logarithm to the new base of the value by
the logarithm to the original base:
logb (N ) =
logx (N )
logx (b)
Example:
log(24) =
Che k if this is indeed true:
log(24) = 1,3802
and
179
ln(24)
ln(10)
ln(24)
ln(10)
=
3,1781
2,3026
= 1,3802
(Use your
al ulator.)
DSC1520/SG001
Examples
1. We simplify the expression
logx (25) + logx (5) − logx (45)
logs are to the same base, we
by employing the rules above. Sin e all the
an say that
25 + 35
logx (25) + logx (35) − logx (45) = logx
45
= logx (1,3333).
If
x = 5,
(Rules 1 and 2)
then the answer is evaluated as
log5 (1,3333) =
If
x = 10,
the answer is
ln(1,3333)
= 0,1787.
ln(5)
log(1,3333) = 0,0125;
and if
2. Using the rules above to simplify the expression
(Rule 3)
x = e,
then it is
ln(1,3333) = 0,2877.
4 log(7) − 3 log(0,85) + log(10):
4 log(7) − 3 log(0,85) + log(10) = log(74 ) − log(0,853 ) + log(10)
4 7
= log
+ log(10)
0,853
= log(3 909,6276 × 10)
(Rule 3)
(Rule 2)
(Rule 1)
= log(39 096,276)
= 4,592
D.4.4 Solving equations
In the following examples, the variable for whi h we need to solve the equations, is in an exponent and we
need to use logarithms.
1. To solve an equation like
21 = 3(2x ),
we rst need to write it in the form
number = basepower .
We
an simplify the equation by dividing on both sides by three to nd
21
= 2x
3
2x = 7.
or
Now, to nd the solution, we take logarithms at both sides either to base 10, or
available on our
al ulators. Let's use base
e
to nd
ln 2x = ln 7.
By using Rule 3 above, we
an write
x ln 2 = ln 7
or
x=
ln 7
= 2,807.
ln 2
2. We follow the same pro edure to solve the equation
15 + 5,12x = 24,2.
By subtra ting 15 on both sides, we nd
5,12x = 9,2.
180
e,
sin e they are
DSC1520/SG001
Now, by taking the logarithm to base 10 on both sides, we nd
log(5,1)2x = log(9,2)
2x log(5,1) = log(9,2)
log(9,2)
2x =
= 1,362
log(5,1)
1,362
= 0,681.
x=
2
When an equation
(Rule 3)
ontains a logarithm, on the other hand, we have to reverse the pro ess. We know from
Equation D.1 (page 178) that
log (Base)Power = Power,
for example
log(10x ) = x
Consider the equation
e(x+3) = 1.
If we
and
ln(ex ) = x.
an take the logarithm to base
e
on both sides, then
ln e(x+3) = ln(1)
x+3=0
The equation
2x 2x+1 = 7
x = −3.
an be solved by taking logarithms to base two on both sides:
2x 2x+1 = 7
2x+x+1 = 7
(Rule 1)
2x+1
log2 (2
) = log2 (7)
ln(7)
= 2,8074
2x + 1 =
ln(2)
2,807 − 1
= 0,9037.
x=
2
A tivity
1. Simplify ea h of the following expressions:
(a)
log(x) + log(x − 5)
(b)
2 ln(x − 2) − ln(2x)
x+1
1
log 1−x
+ log x+1
( )
2. Solve ea h of the following equations:
(a)
5 = 3(10)x
(b)
5 = 10x−4
( )
125 = 230et
(d)
3 log(x + 5) = 3,6
(e)
2 + ln(x − 4,5) = 4
(f )
et = 9 − 2et
(g)
20 = 40(1 − e−t )
(h)
2 ln(5x) − 2 ln(2x) = 2,5
181
(Rule 4)
DSC1520/SG001
D.5 Hyperboli fun tions
Consider the hyperboli
fun tion with the following formula in standard form
a
+ q,
x−p
y=
with
a, p
and
q
onstants.
Properties that
a>0
an be derived dire tly from the standard form of the hyperboli
⊲
If
⊲
There is a verti al asymptote at
the graph falls in quadrants one and three, while it falls in quadrants two and four if
The simplest hyperboli
x=p
and a horizontal asymptote at
fun tion is
y=
p=0
where both
fun tion, are the following:
and
q = 0.
a < 0.
y = q.
1
,
x
Figure D.12 shows the graph of this fun tion:
f (x)
8
6
Verti al
asymptote
4
2
−2
0
−1−2
1
x
2
Horizontal
asymptote
−4
−6
−8
Figure D.12: The hyperboli
From the graph, it is
the
x
axis as
The value of
x
lear that there is a horizontal asymptote at
be omes very big (x
y=
1
x
fun tion
→ ∞)
and as
annot be determined at
verti al asymptote at
x = 0.
x
x=0
y=
y = 0,
1
x
meaning that the graph approa hes
be omes very small (x
→ −∞).
sin e division by zero is not allowed, therefore there is a
In the general formula, the verti al asymptote is where
x = p.
In the standard formula,
given by
y=q
y =
a
+ q,
x−p
x − p = 0,
that is where
the horizontal asymptote is
and the verti al asymptote is given by
x = p.
Examples
1. Consider the hyperboli
fun tion
y=
Here,
p=0
y = 3.
The
and
x
q = 3,
1
+ 3.
x
so the graph has a verti al asymptote at
axis inter ept is where
1
+ 3 = 0 so that
x
y = 0,
x=0
and a horizontal asymptote at
that is where
1
= −3 or 1 = −3x giving
x
182
1
x = − = −0,333.
3
DSC1520/SG001
Figure D.13 shows the graph of
1
x
y=
+ 3.
Note the asymptotes at
x=0
and
y = 3.
f (x)
8
6
4
2
0
−1−2
x
1
−4
−6
−8
Figure D.13: Graph of
2. Figure D.14 shows the graph of the hyperboli
p = −1
and
+3
200
x+1
−1 ≤ x ≤ 20.
We see that there are asymptotes at
with
1
x
fun tion
f (x) = y =
over the interval
y=
q = 0.
x = −1 and y = 0,
whi h
an be derived from the general formula
f (x)
200
150
100
50
5
10
15
Figure D.14: Graph of
3. Let's solve the hyperboli
20
y=
x
200
x+1
fun tion
x+5
= x + 1.
x−5
We start by multiplying on both sides by
x−5
to get
x + 5 = (x + 1)(x − 5) = x2 − 4x − 5.
Simplifying this gives
x2 − 5x − 10 = 0.
183
DSC1520/SG001
We
an now nd the solution by using the quadrati
x=
−(−5) ±
= 6,53
Alternatively, you
or
formula (see page 170) to nd
p
(−5)2 − 4(−10)
2
− 1,53.
an use Maxima to nd the roots of the fun tion to be
√
( 65 − 5)
x=−
, or
2
whi h is simplied to
x = −1,5311
or
√
65 + 5
x=
,
2
x = 6,5311.
A tivity
1. Graph ea h of the following fun tions:
(a)
y =2+
(b)
y=
1
x+1
4
1−x
2. Solve ea h of the following hyperboli
(a)
(b)
fun tions:
5
= 10
q − 3,5
x
=3
x+4
D.6 Composite fun tions
Two fun tions
g
and
f
may be
ombined to get a new fun tion. For example, suppose
f (x) = x2
then the fun tions
an be
and
g(x) = x + 3,
ombined to nd
f [g(x)] = (g(x))2 = (x + 3)2
(Replace x by x + 3 in f.)
or
g[f (x)] = f (x) + 3 = x2 + 3.
Su h a
ombined fun tion is
fun tions are
alled a
(Replace x by x2 in g.)
omposite fun tion (a
ombined are obviously very important, as
fun tion of a fun tion).
The order in whi h the
an be seen from the following examples.
Examples
1. Consider the fun tions
f (x) = 2x
These fun tions may be
and
g(x) = ex .
and
g(f (x)) = e(f (x)) = e2x .
ombined to nd
f (g(x)) = 2(g(x)) = 2ex
184
DSC1520/SG001
2. Consider the fun tions
f (x) = 3x
We
an
and
g(x) = 2x2 − 5.
ombine these fun tions to get
f (g(x)) = 3(g(x)) = 3(2x2 − 5) = 6x2 − 15
and
g(f (x)) = 2(f (x))2 − 5 = 2(3x)2 − 5 = 2(9x2 ) − 5 = 18x2 − 5.
3. The fun tions
f (x) = 2x + 3
an be
and
g(x) =
2
x
ombined to get
4
2
+3= +3
f (g(x)) = 2(g(x)) + 3 = 2
x
x
and
g(f (x)) =
2
2
=
.
(f (x))
2x + 3
A tivity
Find the
f (g(x))
√
g(x) = x
omposite fun tions
1.
f (x) = e4x
2.
f (x) = x − 8
3.
f (x) =
5
2x2
and
and
and
and
g(f (x))
for the following fun tions:
g(x) = x2
g(x) = 10x
185
Appendix E: Solutions to a tivities (Appendix)
E.1 Se tion A.1 (Priorities)
1 4 × 3 ÷ 2 + 5 × 6 − 20 ÷ 4 × 2 = 12 ÷ 2 + 30 − 5 × 2 = 6 + 30 − 10 = 26
√
2 32 × 7 − 9 × 8 + 22 ÷ 4 × 3 = 9 × 7 − 3 × 8 + 1 × 3 = 63 − 24 + 3 = 42
E.2 Se tion A.2 (Variables)
1a 12(2) + 17 = 41
1b 42 − 3 = 16 − 3 = 13
1
5+7
4
=
12
4
=3
1d (6 + 3)(6 − 2) = 9 × 4 = 36
2a 2(2 + 1 − 7) + 7(1 − 2) = 2(−4) + 7(−1) = −8 − 7 = −15
2b 22 + 12 + 72 = 4 + 1 + 49 = 54
2 (2 + 1)(1 − 7) = 3(−6) = −18
2
2d 2(1) − 7+3
2 + 1 = 2 − 5 + 1 = −2
3a x + y
3b 8 − (a + b)
3 3x + 2y
3d y + 7
E.3 Se tion A.3 (Laws of operations)
1a 7 × (6 × (5 + 4) + 3 × (2 + 1)) = 7 × (6 × 5 + 6 × 4 + 3 × 2 + 3 × 1) = 7 × 6 × 5 + 7 × 6 × 4 + 7 × 3 × 2 + 7 × 3 × 1
1b A(BC + BD + EF + EG) = ABC + ABD + AEF + AEG
2a
Asso iative law of addition.
2b
Commutitative law of multipli ation
2
Commutitative law of addition
2d
Distributive law of multipli ation over addition
3a 8x + 6xy − 12x + 6 + 2xy = (8x − 12x) + (6xy + 2xy) + 6 = −4x + 8xy + 6
3b 3x2 + 4x + 7 − 2x2 − 8x + 2 = (3x2 − 2x2 ) + (4x − 8x) + 7 + 2 = x2 − 4x + 9
186
DSC1520/SG001
4a (a + b)(6a − 3b) = 6a2 − 3ab + 6ab − 3b2 = 6a2 + 3ab − 3b2
4b (x + 2)2 − x(x + 2) = (x + 2)(x + 2) − x(x + 2) = x2 + 4x + 4 − x2 − 2x = 2x + 4;
or
(x + 2)2 − x(x + 2) = (x + 2)(x + 2 − x) = 2(x + 2) = 2x + 4
4
4x2 +2x
4d
x2 −2xy+y 2
x−y
2x
2x(2x+1)
2x
=
=
by fa torisation (see Se tion A.5)
= 2x + 1
(x−y)2
x−y
=x−y
E.4 Se tion A.4 (Fra tions)
1
2
3
2
7
2x
3
x−3
5
÷
1
5
−
=
x
9
+
2
3
=
2
x
×
5
1
=
10
3
7(9)−x(2x)
18x
−
x
5
=
=
63−2x2
18x
(x−3)x+2(5)−x(x)
5x
=
x2 −3x+10−x2
5x
=
10−3x
5x
E.5 Se tion A.5 (Fa torisation)
1a 2(3x + 5) + x(4x2 + 1) = 6x + 10 + 4x3 + x = 4x3 + 7x + 10
1b 100(1 − x)(1 + x) = 100(1 + x − x − x2 ) = 100 − 100x2
1 4x2 + 7x + 2x(4x − 5) = 4x2 + 7x + 8x2 − 10x = 12x2 − 3x
1d (5x + 1)2 = (5x + 1)(5x + 1) = 25x2 + 5x + 5x + 1 = 25x2 + 10x + 1
1e (5x2 + 4)(x + 1) = 5x3 + 5x2 + 4x + 5
2a 63 − 7x2 = 7 × 9 − 7x2 = 7(9 − x2 ) = 7(32 − x2 ) = 7(3 − x)(3 + x)
(See Equation A.4.)
2b f s − f r + qr − qs = f (s − r) + q(r − s) = f (s − r) − q(s − r) = (s − r)(f − q)
2 (12ax + 3ay) + (8bx + 2by) = 3a(4x + y) + 2b(4x + y) = (4x + y)(3a + 2b)
2d 2a2 + 11a + 12 = (2a + 3)(a + 4)
2e 4x2 − 13x + 3 = (4x − 1)(x − 3)
2f 2x2 − 8x − 24 = 2(x2 − 4x − 12) = 2(x − 6)(x + 2)
E.6 Se tion A.6 (Fun tions)
1 f (−4) = 3(−4) + 20 = −12 + 20 = 8
2 h(2) = 50(2) − 4,9(2)2 = 100 − 19,6 = 80,4
m.
3 F (2) = 3(2) − (2)2 = 6 − 4 = 2 and F (3) is undened,
sin e
E.7 Se tion A.7 (Polynomials)
1
A polynomial of degree 3
2
A polynomial of degree 2
3
Not a polynomial it
4
A polynomial of degree 17
ontains
x0,5
(only integers allowed).
187
t>2
DSC1520/SG001
5
Not a polynomial it
ontains a negative power of
x
(only positive powers allowed).
E.8 Se tion A.8 (Per entages)
1
(i)
2
Pri e without VAT is
3
Week 1:
0,12 × 360,20 = 43,22
Week 3:
4 950 × (1 − 15%) = 4 950 × 0,85 = R4 207,50
The pri e in reased by
pri e in 2005.
4b
700 000 − 450 000 = 250 000.
This is an in rease of
250 000
450 000
= 0,56
or 56% on the
700 000(1 + 0,08) = 756 000
756 000 × 1,08 = 816 480
In 2012:
In 2013:
4
0,115 × 523,00 = 60,15
500(1 − 0,20) = 500 × 0,8 = 400
400 × 0,8 = 320
320 × 0,8 = 256
Week 2:
4a
(ii)
In total there are 3 751 students. Of these, (i)
0,41 or 41%
are female.
2 211 ÷ 3 751 = 0,59 or 59% are male and (ii) 1 540 ÷ 3 751 =
E.9 Se tion B.1 (Solving equations)
1a 14 + 9x = 95 giving 9x = 81, resulting in x = 9.
1b
1
in
1d
16+y
4
= 25
From
x
3
From
x−3
5
gives
16 + y = 100
that results in
+ 2 = 2x we nd x + 6 = 6x
−5x = −6 or x = 65 .
10x).
+
2
x
−
x
5
This results in
(multiply by 3) or
From
2b
The solution to
2
Simplifying results in
2d
Simplifying gives
2e
From
√ (x − 2)(x
√ + 4) = 2x
x = + 8 or − 8.
we nd
x(x2 − 2) = 0
x2 + 4x − 2x − 8 − 2x = x2 − 8 = 0,
is either
follows that
5Q
P +2
2f
From
2
x
2g
From
2x(y + 2) − 2y(x + 2) = 0
3
2x
=1
x=0
= 20
we nd
4 − 3 = 2x
or
1
P +2
x2 − 2 = 0,
that is
whi h results in
(multiply by
follows that
solution has an innite number of solutions.
2h
t
6x
and add 6). This results
ommon denominator
whi h gives
x=0
or
x2 = 8,
√
x=+ 2
or
resulting in
√
x = − 2.
4x2 + 7x − 2x(2x − 5) = 4x2 + 7x − 4x2 + 10x = 17x = 17, giving x = 1.
−12y
3y
4y
y
=
12
= −12(2y) = 480, resulting in 2y = −40 or y = −20.
+
y
2
2
2
P 6= −2.)
−
x − 6x = −6 (subtra
3
follows that 2x(x − 3) + 10(2) − 2x(x) = 3x (multiply by
= 10
2x2 − 6x + 20 − 2x2 = 3x or −6x − 3x = −20, giving x = 20
9 .
2a
( P5Q
+2 )
= 20
1
( P +2
)
y = 84.
5Q = 20
ommon denominator
or
2x).
Q = 4.
This results in
2xy + 2x − 2xy − 4y = 2x − 4y = 0,
There is no solution to this equation, be ause there is no number that
E.10 Se tion B.2 (Inequalities)
1 x − 12 > 10 gives x > 10 + 12 = 22.
188
(This is only true if
resulting in
x=
1
2
x = 2y .
an be divided to get zero.
This
DSC1520/SG001
x > 22
−10
2
−75
> 15
x
3
From
gives
−75 > 15x
2x − 6 ≤ 12 − 4x
Assuming that
0
x>0
or
10
x < −5.
follows that
20
But, if we assume that
2x + 4x ≤ 12 + 6
gives the solution to be
or
30
x > 0,
6x ≤ 18,
40
no solution exists.
giving
x ≤ 3.
0 < x ≤ 3.
0<x≤3
−3
0
3
6
E.11 Se tion C.2 (Plotting linear fun tions)
1a
The x
(−1; 0).
inter ept is on the
Likewise, the
The graph of
1b
The
The
x
y
x
axis, where
y = 0.
This gives
0 = x + 1,
or
y inter ept is where x = 0, giving y = 0 + 1 = 1. The inter
y = x + 1 is drawn through the points (0,1) and (−1; 0) to
x = −1.
The inter ept is therefore
ept is therefore (0; 1).
give the following:
y axis where x = 0. This gives y = −5(0) − 25 = −25, giving the point (0; −25).
y = 0, that is at 0 = −5x − 25 or x = −25
5 = −5, that is the point (−5; 0).
inter ept is on the
inter ept is where
The graph through these points is as follows:
2a
Here, written in the form
we have
y = 9x − 6
with
a=9
and
b = −6.
b, whi h is −6, so the graph goes through the point (0; −6).
a = 9, meaning that for ea h unit we move to the right (on the x axis) we need to move 9 units
up. So, from (0; −6), if we move one unit to the right, we need to move 9 units up to y = −6 + 9 = 3. The
new point is then (1; 3).
The
y
y = f (x) = ax + b,
inter ept is given by
The slope is
Verify that the following graph represents this information:
189
DSC1520/SG001
2b
For the fun tion
y = f (x) = −2x + 4,
goes downwards from left to right.
y = b = 4 (i.e. the point (0; 4)). The slope a = −2 indi ates
in rease in the x value, y de reases by 2. From the point (0; 4), if we move one unit
we must de rease y by 2 (from 4), to get y = 4 − 2 = 2. This gives the point (1; 2).
The
y
we see that the slope of the line is negative (−2). The line therefore
inter ept is at
that for every unit of
to the right to
x = 1,
Verify that the following is a graph of this fun tion:
E.12 Se tion C.3 (Determining formulæ of linear fun tions)
1
1
y = ax + b is a = xy22 −y
−x1 for points P1 = (x1 ; y1 )
P1 = (x1 ; y1 ) = (−2; 8) and P2 = (x2 ; y2 ) = (4; 1), we nd
The slope of a line
a=
and
P2 = (x2 ; y2 )
on the line. Taking
−7
1−8
=
,
4 − (−2)
6
giving
−7
x + b.
6
value of b,
y=
Now use either one of the given points to nd the
1=
−7
(4) + b
6
The line passing through the points
P1
or
and
P2
b=1+
Here, a = 1,5 and the given point is (x1 ; y1 ) = (2; 4),
1,5(x − 2) + 4 = 1,5x − 3 + 4 = 1,5x + 1.
If we take
(x1 ; y1 ) = (1; 2)
and
7(2)
3 14
17
= +
=
.
3
3
3
3
so the equation of the line is
y = ax + b.
(x2 ; y2 ) = (3; 3),
a=
Then,
17
−7
x+ .
6
3
2
The general expression of the line is
P2 = (4; 1).
is therefore given by the fun tion
y=
3
say
then the slope of the line is
y2 − y1
1
3−2
= = 0,5,
=
x2 − x1
3−1
2
190
y = a(x − x1 ) + y1 =
DSC1520/SG001
so the expression be omes
y = 0,5x + b.
Using the point
2 = 0,5(1) + b
or
(x1 ; y1 ) = (1; 2),
we nd the value of
b
as
b = 2 − 0,5 = 1,5.
The expression is therefore
y = 0,5x + 1,5.
4
The
5
y = 0,
inter ept is found where
y inter
y = 3.
The
at
x
ept is given by
b in
that is
0 = −4x + 3
the general expression
−4x = −3,
whi h gives
y = ax + b (where x = 0).
The
y
x=
3
4
= 0,75.
inter ept is therefore
d = ap + b. We are given two points on the line,
(p2 ; d2 ) = (600 + 50; 60 − 3) = (650; 57). The slope of the demand line is
The general demand fun tion is given by
(p1 ; d1 ) = (600; 60)
and
a=
The value of
b
(p1 ; d1 ) = (600; 60)
is found by using
namely
57 − 60
−3
d2 − d1
=
=
= −0,06.
p2 − p1
650 − 600
50
to nd
demand fun tion is therefore
6
or
60 = −0,06(600) + b
or
b = 60 + 36 = 96.
The
d = −0,06p + 96.
The general expression representing demand is
d = ap + b,
p
b.
sin e the independent variable is
need to nd the values of
a
and
(Mr Walsh
an de ide on the pri e) and the dependent variable is
The given information provides two points on the weekly demand line, namely
(p1 ; d1 ) = (190; 26 000)
The slope of the line is
a=
so the demand line be omes
To nd the value of
b,
and
(p2 ; d2 ) = (210; 16 000).
d2 − d1
16 000 − 26 000
−10 000
=
=
= −500,
p2 − p1
210 − 190
20
d = −500p + b.
we use one of the points on the line, say
(p2 ; d2 ) = (210; 16 000),
16 000 = −500(210) + b
or
b = 16 000 + 105 000 = 121 000.
The demand line is therefore given by
d = −500p + 121 000.
E.13 Se tion C.5 (Simultaneous equations/inequalities)
1
From equation (2) we nd that
y = 10x − 40.
When we substitute this into equation (1), we nd
2(10x − 40) + 10x = 580
20x − 80 + 10x = 580
30x = 660
x = 22,
191
to nd
d.
We
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and from equation (2) it follows that
y = 10(22) − 40 = 180.
This gives the point (22; 180) as solution to the system of simultaneous equations, whi h is
onrmed by the
plot from Maxima in Figure E.1:
Figure E.1: Equations
2
It is
2y + 10x = 580
and
lear that the two fun tions have the same slope, namely
and the system has no solution. This is
y − 10x = −40
a=
1
2 . The lines will therefore never
onrmed by the plot from Maxima in Figure E.2:
Figure E.2: Equations with the same slope
3
First, we simplify the equations to nd
4x − 3y = 12
(1)
0,5x + y = 7.
From (2) we nd that
y = 7 − 0,5x.
(2)
When we substitute this into (1) we nd
4x − 3(7 − 0,5x) = 12
4x − 21 + 1,5x = 12
5,5x = 33
x = 6,
and
4
y = 7 − 0,5(6) = 4.
Again we rst simplify and get
5p + 10q = 18
(1)
5p + 3q = 4.
(2)
When we substitute the rst equation as it is given
5
p=
18 − 10q
5
18 − 10q
+ 3q = 4
5
18 − 10q + 3q = 4
−7q = −14
q = 2.
192
into (2), we nd
ross
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Substituting this value into (1) gives
5p + 10 × 2 = 18
The solution is therefore
5
or
p=
−2
= −0,4.
5
(p; q) = (−0,4; 2).
Simplifying the equations gives
5q − 6p = 7
(1)
q − p = 3.
From (2) we nd
q = p + 3.
Substituting this into (1) gives
Substituting this into (2) gives
6
(2)
5(p + 3) − 6p = 7,
whi h results in
p = 8.
q = 11.
To graph the inequalities, we write the rst two in standard form to get
2x − 3y ≤ 12
x + 5y ≤ 20
−3y ≤ −2x + 12
2
y ≥ x−4
3
5y ≤ −x + 20
x
y ≤ − + 4.
5
Plotting ea h of the three inequalities gives the graphs as shown in Figure E.3.
2
4
Feasible area
Feasible area
2
0
2
4
6
8
10
2
−2
Feasible area
0
2
0
2
−4
4
6
8
4
6
8
10
10
−2
−2
(a) y ≥ 23 x − 4
(b) y ≤ − x5 + 4
( ) x≥0
Figure E.3: Individual feasible areas
The feasible area of the system
onsists of the area where these areas overlap, as shown in Figure E.4.
4
2
Feasible area
0
2
4
6
8
10
−2
−4
Figure E.4: Feasible area of the system of inequalities
193
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E.14 Se tion D.1 (Quadrati fun tions)
1
For
f (x) = 2x2 − x − 3, a = 2, b = −1
−(−1)
xm = −b
2a = 2(2) =
at (0,25; −3,125).
The vertex is at
turning point is
The
y
inter ept is at
1
4
and
c = −3.
= 0,25.
Sin e
a > 0, f
At the vertex,
f (0,25) = 2(0,25)2 − 0,25 − 3 = −3,125.
b2 − 4ac = (−1)2 − 4(2)(−3) = 25 > 0, f
x =
=
=
√
−b− b2 −4ac
2a
√
−(−1)− 25
2×2
−4
4
and
(2x − 3)(x + 1) = 0,
=
=
The
y
xm =
inter ept is at
−b
2a
=
−(−16)
2×4
f (x) = 2x2 − x − 3
and
c = 16.
Sin e
a>0f
For
giving the turning point as
f (x) = 4x2 − 16x + 16
f (x) = −3x2 + 3x − 2, a = −3, b = 3
point as
The
y
xm =
(0,5; −1,25).
inter ept is at
−b
2a
=
y = −2
−(3)
2×−3
= 0,5
and
and
c = −2.
axis. The
Sin e
hes the
x axis at its minimum
(Maxima)
a < 0, f
has a maximum turning point.
f (0,5) = −3(0,5)2 + 3(0,5) − 2 = −1,25,
giving the turning
(=c).
b2 − 4ac = (3)2 − 4(−3)(−2) = 9 − 24 = −15 < 0,
graph of f is shown in Figure E.7:
The dis riminant
x
(2; 0).
y = 16 (= c).
b2 −4ac = (−16)2 −4(4)(16) = 0. Therefore, the graph tou
(2; 0). The graph of f is shown in Figure E.6:
The vertex is at
is shown in Figure E.5:
has a minimum turning point.
= 2 and f (2) = 4(2)2 − 16(2) + 16 = 0,
Figure E.6:
3
f
(Maxima)
The dis riminant is
point at
inter epts, namely
−b+ b2 −4ac
2a
√
−(−1)+ 25
2×2
6
4
resulting in the same roots. The graph of
f (x) = 4x2 − 16x + 16, a = 4, b = −16
The vertex is at
x
√
= 1,5.
Figure E.5:
For
has two distin t
x =
= −1
2
The
y = −3 (= c).
Sin e the dis riminant
Fa torisation gives
has a minimum turning point.
194
therefore the graph does not tou h the
DSC1520/SG001
Figure E.7:
f (x) = −3x2 + 3x − 2
(Maxima)
E.15 Se tion D.2 (Cubi fun tions)
1
The graph of the fun tion is shown in Figure E.8:
Figure E.8:
Maxima gives the roots of
f
f (x) = 0,5x3 − 8x2 + 200
(Maxima)
as
[x=6.483089182511198,x=-4.425328095280196,x=13.942238912769℄.
2
The graph of the fun tion is shown in Figure E.9:
Figure E.9:
Maxima gives the roots of
f
as
f (x) = 3x3 + 9x2
[x=0.0,x=0.0,x=-3.0℄,
whi h
(Maxima)
orresponds with the graph.
E.16 Se tion D.3 (Exponential fun tions)
1a
1b
1
2
= (4L2−(−2) )2 = (4L4 )2 = 16L8
=
2Q
3P ×P 0,5
−
4L2
L−2
xy
3y 2
2Q
√
3P P
xy 2
2y
2
=
=
2Q
3P 1,5
x
3y
−
xy 2
2
=
x
3y
−
x2 y 2
4
=
4x−3x2 y 3
12y
195
=
x(4−3xy 3 )
12y
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1d
ex (1+e1−x )
e
2a 2x =
√1
16
= ex−1 (1 + e1−x ) = ex−1 + ex−1+1−x = ex−1 + e0 = ex−1 + 1
1
4
=
= 2−2 , which results in x = −2.
2b
From
(1 + 2t)0,5 = 12
2
From
e5x =
or
x=
1
we nd
ex−5
5
6.
2d
Simplify to nd
2e
We
1 + 2t = 144
we nd
16 −
e5x+(x−5) = 1 whi
1 −2t
8
an write the equation
by squaring both sides. This results in
h is simplied to
e6x−5 = e0 .
= 16 − 82t = 24 − (23 )2t = 24 − 23×2t = 0,
2
5t
=
2
1 as
5t = 1
or
5t = 50 ,
whi h results in
t=
144−1
2
= 71,5.
From this follows that
whi h gives
24 = 26t
or
6x − 5 = 0
t=
2
3.
t = 0.
E.17 Se tion D.4 (Logarithmi fun tions)
1a log(x) + log(x − 5) = log(x(x − 5)) = log(x2 − 5x)
1b 2 ln(x − 2) − ln(2x) = ln(x − 2)2 − ln(2x) = ln
1 log
x+1
1−x
+ log
1
x+1
= log
2a
If
5 = 3(10)x ,
2b
If
5 = 10x−4 ,
2
If
125 = 230et ,
2d
From
2e
From
2f
From
2g
et = 9 − 2et
From
40(1 − e−t ) = 20
2h
From
2 ln(5x) − ln(2x) = 2,5
then
then
10x =
then
et =
12,5x =
5
3 giving
1
x+1
125
230 giving
= 12,1825
= log
1
1−x
= 0,2218
= log(1 − x)−1 = − log(1 − x)
x = 4,6990
giving
log(x + 5) = 1,2
3et = 9
or
ln(x − 4,5) = 2
or
et = 3,
e−t =
follows that
or
5
3
(x−2)2
2x
125
t = ln 230
= −0,6098
follows that
follows that
i
x = log
follows that
2 + ln(x − 4,5) = 4
x = 11,889.
in
x+1
1−x
x − 4 = log 5 = 0,6990
3 log(x + 5) = 3,6
e2,5
h
follows that
x = 0,9746.
20−40
−40
10log(x+5) = 101,2 ,
or
eln(x−4,5) = e2 ,
whi h results in
= 0,5,
whi h results in
resulting in
x = 10,8489
x − 4,5 = 7,3891
or
t = ln(3) = 1,0986.
−t = ln(0,5) = −0,6931 or t = 0,6931.
2
ln(5x)2 − ln(2x) = 2,5 or ln 25x
= 12,5x = 2,5. This results
2x
resulting in
E.18 Se tion D.5 (Hyperboli fun tions)
1a
Comparing the fun tion
y = 2+
1
x+1
with the standard form
whi h indi ates that there is a verti al asymptote at
y inter ept is where x = 0, that is at y = 2 +
−2x − 2 = 1, resulting in x = 1+2
−2 = −1,5.
The
The graph is shown in Figure E.10:
4
Comparing the fun tion y =
1b
1−x
=
−4
x−1
1
1
x = −1
= 3.
y=
a
x−p
+ q , we see that p = −1 and q = 2,
and a horizontal asymptote at
The
x
inter ept is where
with the standard form, we see that
y =2+
y = 2.
1
x+1
= 0,
a < 0, p = 1
and
that is
q = 0,
whi h indi ates that the graph falls in the se ond and fourth quadrants, there is a verti al asymptote at
x = −1
and a horizontal asymptote at
There is no
x
inter ept and the
y
y = 0.
inter ept is at
y=
4
1+0
= 4.
The graph is shown in Figure E.11:
2a
Simplifying the equation gives
5
q−3,5
= 10,
whi h gives
196
40 = 10q ,
resulting in
q = 4.
DSC1520/SG001
f (x)
4
3
2
1
−3
−2
−1
1
2
y=
Figure E.10: Graph of
3
1
x
x
+3
f (x)
6
4
2
−3
−2
0
−1
−2
1
2
3
y=
4
1−x
x
−4
−6
Figure E.11: Graph of
2b
From
x
x+4
=3
follows that
x = 3x + 12,
whi h results in
x = −6.
E.19 Se tion D.6 (Composite fun tions)
√
1 f (g(x)) = e(g(x)) = e
x
2 f (g(x)) = f (x2 ) = x2 − 8
3 f (g(x)) = f (10x) =
and
5
2(10x2 )2
g(f (x)) =
p
f (x) =
√
e4x
g(f (x)) = g(x − 8) = (x − 8)2 = x2 − 16x + 64
5
1
= 2(100x
and
g(f (x)) = g 2x5 2 = 10 2x5 2 = 25x2
4 ) = 40x4
and
197