DSC1520 2023 S1 Assignment 5 (Mock Exam)
Solutions
NS Nkomo1 and Prof. S Mukeru2
1,2
University of South Africa
Department of Decision Sciences
Email ids: 1 nkomons@unisa.ac.za, 2 mukers@unisa.ac.za
1,2
Question 1
The surfing club provides swimming lessons to boost their funds. The club has
fixed costs of R1 035 per day, when offering these classes-mostly for insuranceand it costs them R195 for each lesson given. How many lessons did they provide
if total costs amounted to R3 000 ?
Solution 1
The surfing club’s total costs function is
T C = F C + V C = 1 035 + 195q,
where q is the number of lessons given per day.
When the club’s total costs is R3 000 then the number of lesson will be
3 000
=
195q
3 000 − 1 035 = 1 965
1 965
=
195
= 10, 0769 ≈ 10.
q
q
1 035 + 195q
=
Question 2
A firm’s fixed cost to produce cups amount to R1 260 per week and it costs R22
to produce each cup. These cups are sold for R65 each. What would be the
firm’s profit if they produce and sell 160 cups?
1
Solution 2
prof it(P ) = T R − T C
where
TR
= price per cup × number of cups sold
=
65q,
where q is the number of cups produced (or sold) per week.
TC
=
FC + V C
=
1 260 + 22q.
Hence, the profit function becomes
P
=
65q − (1 260 + 22q)
=
65q − 1 260 − 22q
=
43q − 1 260.
Also, when they sell 160 cups, their profit is R5 620, which can be calculated as
P (160) = 43(160) − 1 260 = 5 620.
Question 3
The demand and supply functions for pans are
q = 500 − 5p
and
p = 0, 8q + 30,
respectively, with the price in rand and q the quantity in boxes. Find the
point(s) where the demand and supply function intersect.
Solution 3
The demand function q = 500 − 5p can be transformed to
p=
500 q
− = 100 = 0, 2q.
5
5
Now, that both the demand and supply function are in the form p(q), we can
equate them to find the point where they intesect.
100 = 0, 2q
=
0, 8q + 30
100 − 30
=
0, 2q + 0, 8q
q
=
70.
To find the value of p, we substitute q = 70 on one of the given equations. Then
p = 100 − 0, 2(70) = 86.
2
Question 4
The demand function for a certain kind of cellphones is given by
q = 90 000 − 5p.
Determine the point elasticity of demand when p = 5 000.
Solution 4
When p = 5 000, then
q = 90 000 − 5(5 000) = 65 000,
that is (p0 ; q0 ) = (5 000; 65 000)
and
p0
q0
5 000
= −5 ×
65 000
= −0, 385.
= −b ×
ϵd
|ϵ| = 0, 385 < 1, the demand is inelastic at this price. 1% increase (or decrease)
in price will cause a 0, 385% decrease (or increase) in demand.
Question 5
Consider the supply function for a kind of computer, namely
p = 10 + 0, 5q.
Determine the arc elasticity if price increases from R1 000 to R3 000.
Solution 5
The supply function p = 10 + 0, 5q can be transformed to q = 2p − 20.
We need to find the quantities demanded at these prices.
If p = 1 000, then q = 2(1 000) − 20 = 1 980, to get (p1 ; q1 ) = (1 000; 1 980).
If p = 3 000, then q = 2(3 000) − 20 = 5 980, to get (p1 ; q1 ) = (3 000; 5 980).
The arc price elasticuty is
ϵs
=
=
≈
p1 + p2
q1 + q2
1 000 + 3 000
2×
1 980 + 5 980
1.
q×
|ϵ| = 1, the supply is unit elastic. The 1% change in supply is equal to 1%
change in price.
3
Question 6
The demand and supply functions for a certain product are given as
qd = 300 − 2pd
and
qs = −200 + 3ps ,
respectively. Determine the price (p) and quantity (q) at equilibrium.
Solution 6
To find the equilibrium price and quantity, we need to solve the demand and
supply functions simultaneously. Since the functions are both given in the form
q(p), we set them equal and solve for p (the equilibrium price), that is
qd
=
qs
300 − 2p
=
−200 + 3p
300 + 200
=
2p + 3p
500
=
5p
p
=
100.
To find the equlilbrium quantity, we substitute p = 100 into either one of the
given functions, to find
q = 300 − 2(100) = 100
or
q = −200 + 3(100).
Then we have (p; q) = (100; 100).
Question 7
Find the slope of the curve
y = 3x3 + 4x2 − 7x + 9
at x = 10.
Solution 7
The slope of the curve
y = 3x3 + 4x2 − 7x + 9
is called its derivative and is given by
dy
= f ′ (x) = 9x2 + 8x − 7.
dx
The slpe at the point x = 10 is
f ′ (x) = 9(10)2 + 8(10) + 7 = 973.
4
Question 8
Find H ′′ (x) where
H(x) = xe2x
Solution 8
H ′ (x)
dxe2x
dx
de2x
dx
= x
+ e2x
dx
dx
= 2xe2x + e2x · 1
=
=
(2x + 1)e2x
Using a product rule a second time, we get
H ′′ (x)
=
d(2x + 1)ex
dx
de2x
d(2x + 1)
(2x + 1)
+ e2x
dx
dx
2(2x + 1)e2x + 2e2x
=
2(2x + 2)e2x
=
=
Question 10
Find the stationary points of
Q = p3 − p2 − 5p.
Solution 10
The stationary points are when the derivative is zero, so we need to find Q′ and
set it equal to zero.
dQ
= Q′ = 3p2 − 2p − 5
dp
Now, let it equal to zero
3p2 − 2p − 5 = 0.
We need to solve this quadratic equation:
(3p − 5)(p + 1) = 0
Solving this gives p = 35 , p = −1.
By substituting our p values 35 and −1 into
Q = p3 − p2 − 5p,
we get 53 , − 175
27 and −1, 3, which they are stationary points.
5
Question 11
Given the function
g(x) =
1 1000
x
− x3 + 1,
5
what is the third derivative of g?
Solution 11
To get to the third derivative of g, we first need to find the first and then the
second derivative of g. Where the fist derivative of g is given by:
g ′ (x)
=
=
dg(x)
dx
200x999 − 3x2
and the second derivative of g is given by
g ′′ (x)
dg ′ (x)
dx
199 800x998 − 6x
=
=
Then, the third derivative of g is:
d3 g
dx3
=
=
dg ′′ (x)
dx
199 400 400x997 − 6.
Question 12
Find the turning point(s) of the following function and determine their nature:
M (n) = n2 − 4n + 4.
Solution 12
We start by differentiating the given function and then set the result equal to
zero, that is
M ′ (n) = 2n − 4 = 0.
When we solve this equation, we find n = 2 with
M (2) = (2)2 − 4(2) + 4 = 0.
The turning point is therefore at (2; 0). We now need to determine whether the
point is a maximum or minimum point, by using the second derivative, namely
M ′′ (n) =
d(2n − 4)
= 2.
dn
Since M ′′ (2) = 2 > 0, we have a local minimum point at n = 2.
6
Question 13
Determine the intervals along which the following function increases or decreases:
F (x) = −x3 + 9x2 − 24x + 26.
Solution 13
To find the stationary points of F , we find the derivative of F , set it equal to
zero and solve for x.
The derivative is
dF
dx
= −3x2 + 18x − 24.
F ′ (x)
=
Now, set it equal to zero and solve
−3x2 + 18x − 24
=
0
x − 6x + 8
=
0
(x − 4)(x − 2)
=
0
2
x = 2 or
x = 4.
Therefore, F has stationary points at x = 2 and x = 4. We now need to
determine whether these points are maximum or minimum points, by using the
second derivative, namely
F ′′ (x) =
d(−3x2 + 18x − 24)
= −6x + 18.
dx
Since F ( 4) = −6(4) + 18 = −6 < 0, we have a local maximum point at x = 4.
Since F ( 2) = −6(2) + 18 = 6 > 0, we have a local minimum point at x = 2.
With this information at hand, we can easily draw a rough graph of F and
specify that the function F (p) decreases over (−∞; 2) and (4; ∞), while increases
over (2; 4)
Question 14
Suppose we have the demand function given as
p = 72 − 7, 5q,
where q is the number of units to be produced and sold. Determine the marginal
revenue after 4 units have been sold.
7
Solution 14
From the given demand function
p = 72 − 7, 5q,
we find the total revenue function to be
T R(q)
= pq
=
(72 − 7, 5q)q
=
72q − 7, 5q 2 .
By differentiation, we find the marginal revenue function to be
MR
dT R
dq
d
=
(72q − 7, 5q 2 )
dq
= 72 − 15q.
=
From the M R function, we find that when 4 units have been sold, the revenue
from the next unit sold will be
M R(3) = 72 − 15(4) = R12
Question 15
Given total cost function (T C) as
T C = q 3 − q 2 + 120q,
where q is the number of units to be produced. Determine the cost to produce
the 17t h unit.
Solution 15
Given Total cost function
T C = q 3 − q 2 + 120q,
MC
dT C
dq
d 3
=
(q − q 2 + 120q)
dq
= 3q 2 − 2q + 120.
=
When q = 16, then
M C(16) = 3(16)2 − 2(16) + 120 = 856.
The cost to produce the 17th unit, is R856.
8
Question 16
Determine the intervals along which the given function m(n) incraeses or decreases, where
m(n) = 30n2 − n3 .
Solution 16
To find the stationary points of m, we find the derivative of m, set it equal to
zero and solve for n.
The derivative is
m′ (n) = 60n − 3n2 .
Now, set it equal to zero and solve
60n − 3n2
=
0
2
=
0
n(20 − n)
=
0
20n − n
n = 0 or
n = 20.
Therefore, m has stationary points at n = 0 and n = 20.
We now need to determine whether these points are maximum or minimum
points, by using the second derivative, namely
m′′ (n)
=
=
d
(60n − 3n2 )
dn
60 − 6n.
Since m′′ (0) = 60 > 0, we have a local minimum point at n = 0.
Since m′′ (20) = −60 < 0, we have a local maximum point at n = 20.
With this information at hand, we can easily draw a rough graph of m and
specify that m is decreasing over (−∞; 0) and (20; ∞), while increases over
(0; 20).
Question 17
The demand function of a firm is
Q = 900 − 1, 5P,
where P and Q represent the price and quantity, respctively. At what price is
revenue T R at maximum?
Solution 17
TR
= Q×P
=
900 − 1, 5P P
=
900P − 1, 5P 2 .
9
To find the price at maximum T R, we need to first find T R′ . let T R′ = 0 and
solve to P
T R′ = −3P + 900
−3P + 900
=
0
3P
=
900
P = 300
Therefore, P = 300 when the revenue is at maximum.
Question 18
If the production function is given by
√
Q = 400 L − 5L,
where Q denotes output and L denotes the size of workforce, calculate the value
of marginal production of labour if L = 16.
Solution 18
Marginal product M P
MP
400
dQ
= √ −5
dL
2 L
200
√ − 5.
L
=
=
If L = 16, then
MP
=
=
200
√ −5
16
45.
Question 19
A firm has the following total and cost functions, respectively.
TR
=
10Q − 4Q2
TC
=
16 − Q2 ,
where Q is the number of unites produced and sold (in thousands). How many
unites should be produced to maximise the profit?
10
Solution 19
When the profit is at maximum: Marginal revenue (M T ) = Margigal cost (M C)
MR
dT R
dQ
10 − 8Q
=
=
and
MC
dT C
dQ
−2Q
=
=
now we have
MR
= MC
10 − 8Q = −2Q
Q =
1, 67
Question 9
The demand for seats at a mini soccer match is given by
Q = 150 − P 2 ,
where Q is the number of seats and P is the price per seat. Find the price
alasticity ϵ of demand if seats cost R5 each. What does this value mean?
Solution 9
Q(5)
ϵd
=
150 − (5)2
=
150 − 25
=
125
=
△Q P
△P Q
=
−2 ×
=
−0.08
|ϵd | = 0.08 < 1
11
5
125
elastic
Question 21
The demand function Q and total cost function T (q) of a commodity are given
by the equations
P
=
100 − 4q,
TC
=
144 + 52q,
where P and q are the price and quantity, respectively.
Determine the quantity that must be produced and sold at breaks even.
Solution 21
From the given functions, we find the total revenue function to be
T R = pq = (100 − 4q)q = 100q − 4q 2 .
We know that break-even occurs when total revenue is equal to total cost. We
can therefore find the break-even points for the firm by setting T R = T C, that
is
100q − 4q 2
=
144 + 52q
2
−4q + 48q − 144
=
0
q 2 − 12q + 36
=
0
(q − 6)(q − 6)
=
0
q
=
6.
The firm therefore breaks even when 6 units of the good are produced and sold.
Question 22
The total cost of producing a certain good is given by
T C = 30 ln (q + 300) + 150.
Find the marginal cost (M C) and the avarage cost (AC) functions.
Solution 22
MC
=
=
=
=
dT C
dq
d
30 ln (q + 300) + 150
dq
30
(1)
q + 300
30
q + 300
12
and
AC
TC
q
30 ln (q + 300) + 150
q
=
=
Question 23
Evaluate
Z
1
9
√
3t2 + t2 t − 1
dt.
t2
Solution 23
We start by simplifying,
√
Z 9 2
Z 9
1
3t + t2 t − 1
dt
=
2 + t 2 − t−2 dt
2
t
1
1
#9
"
3
t−1
t2
=
3t + 3 −
−1
2
1
#9
"
−1
t
2 3
=
3t + t 2 −
3
−1
1
3
3
2
(9)−1 2
= 3 · 9 + (9) 2 −
− 3 · 1 + (1) 2 + (1)−1
3
−1
3
1
2
= 27 + 18 + − 3 − − 1
9
3
= 40
Question 24
Find
Z √
2x
+ 5e5x dx
1 − 4x2
Solution 24
Z √
2x
+ 5e5x dx
1 − 4x2
Z
=
√
x
dx +
1 − 4x2
let
u = 1 − 4x2 . Then du = −8xdx, so xdx = − 81 du,
and let
13
Z
e5x dx
v = 5x. Then dv = 5dx, so dx = 15 dv.
Therefore,
Z 2x
5x
√
+ 5e dx =
1 − 4x2
Z
Z
x
2 √
dx + 5 e5x dx
1 − 4x2
Z
Z
1
1
√ du + ev dv
= −
4
u
Z
Z
1
1
u− 2 du + ev dv
= −
4
1 √
= − (2 u) + ev + c
4
1 p
= − ( 1 − 4x2 ) + e5x + c
2
Question 25
Find the area under the curve
F (x) =
x2
x
,
−1
covered between x = 2 and x = 4.
Solution 25
The area between the curve F (x) and the x-axis from x = 4 and x = 8 is
determine as
Z 4
x
dx.
2−1
x
2
Let u = x2 − 1, then du = 2xdx, so xdx = 12 du.
If x = 4 then u = 15.
If x = 8 then u = 63.
Therefore,
Z 4
Z
x
1 63 1
dx
=
du
2
2 15 u
2 x −1
i63
1h
ln u
=
2
15
1
=
ln 63 − ln 15 .
2
Question 26
The marginal cost for a certain product is given by
100
MC =
,
q + 30
where q is the number of units produced. Given that the total cost of producing
10 units is R500. Determine the total cost function.
14
Solution 26
Total cost (T C) can be calculated by integrating the marginal cost function,
that is
Z
TC =
M Cdq
Z
100
=
dq
q + 30
Z
1
= 100
dq
q + 30
= 100 ln (q + 30) + c.
Given that T C = 500 when q = 10, we therefore find the constant, c, by setting
100 ln (40) + c = 500
to get
c = 500 − 100 ln (40) = 131.
The total cost function is therefore
T C = 100 ln (q + 30) + 131.
Question 27
At a small tool-hire company, the estimated rate of increase in the maintenance
cost of power drills is given by
C(t) = 2e2t + 2t + 19,
where C is in rand and t is time in weeks. Give the total maintenance cost
during the first 15 weeks.
Solution 27
Total maintenance costs during the first 10 weeks (0 ≤ t ≤ 15) are given by
Z
TC
=
15
2e2t + 2t + 19 dt
0
=
h
i15
e2t + t2 + 19t
=
e30 + 225 + 285 − 1
=
509 + e30
0
15
Question 28
The demand and supply functions for bikes are
pd = 900e−0.1q
and
ps = 3e0.9q ,
respectively. Where p is the price, and q the quantity. What is the total surplus
at equilibrium market?
Solution 28
We find the equilibrium price and quantity by setting pd = ps , that is
900e−0.1q
=
3e0.9q
−0.1q
=
e0.9q
300e
300
q
= eq
=
ln 300
To get p, we can substitute q = ln 300 into any of the given functions.
p = 3e0,9 ln 300 = 508, 7804134 ≈ 509.
Now, we have (p0 ; q0 ) = (509; ln 300), we can then use the consumer and concumer surplus formula, then
Z ln 300
CS =
900e−0.1q dq − (509)(ln 300)
0
900 −0,1q ln 300
e
− 2901
0
(−0, 1)
=−0,1 ln 300
= −9 000 e
− 1 − 2901
=
=
1 009
and
Z
PS
=
ln 300
(509)(ln 300) −
3e0.9q dq
0
=
2 901 −
=
2 341
i
3 h 0,9 ln 300
e
−1
0, 9
We can not find the total surplus (T S), that is
TS
= CS + P S
=
1 009 + 2 341
=
3 350
16
Question 29
The marginal cost of a good is given by
M C = −10q 4 + 3q 2 + 80q,
where q is the number of units produced. It is also given that the fixed cost of
production is R1 500. Determine the total cost function.
Solution 29
We can find the total cost by integrating the marginal cost, that is
Z
TC =
M Cdq
Z 4
2
=
− 10q + 3q + 80q dq
= −2q 5 + q 3 + 40q 2 + c,
given that F C = 1 500, this means T C(0) = 1 500, that is
1 500 = −2(0)5 + (0)3 + 40(0)2 + c
therefore, c = 1 500, then
T C = −2q 5 + q 3 + 40q 2 + 1 500
Question 30
Suppose the marginal revenue function for a good is
M R = 100 − 13q 12 + 16q 3 ,
where q is the number of units sold, and that total revenue from selling one unit
is R3 316. Find the total revenue function.
Solution 30
We can find the total revenue by integrating the marginal reveue, that is
Z
TR =
M Rdq
Z =
100 − 13q 12 + 16q 3
=
300q − q 13 + 4q 2 + c.
Given that total revenue from selling one unit is R3 316, meaning T R(1) = 3 316,
that is
3 316 = 100(1) − (1)13 + 4(1)2 + c
therefore, c = 3 213, then
T R = 100q − q + 4q 2 + 3 213.
17