FOUNDATIONS OF MATHEMATICS
LECTURE 3
Graphical Presentation in Business Analysis
1. Graphs of linear equations of cost, profit, demand, supply, depreciation
Example 1.1 (Profit) Each Sunday a newspaper agency sells x copies of a certain newspaper for 1€
per copy. The cost to the agency of each newspaper is 0.50€. The agency pays a fixed cost (for
storage, delivery and so on) of 100€ per Sunday. Write an equation that relates the profit P in Euros to
the number x of copies sold. Graph this equation.
The cost equation is C  FC  vc  x , where FC = 100 and vc = 0.5€. The profit is evaluated as
P  px  C  px  FC  vc  x , where p = 1.
Thus the profit equation is P = 0.5x – 100. We can graph this linear equation by using Excel (see
Fig1) or by hand as a straight line passing through 2 points, for example (0,100) and (500,150).
Example 1.2. (Cost) The management of a company manufacturing surfboards has fixed costs of
200€ per day and total costs of 1,400€ per day at a daily output of twenty boards.
a) Assuming the total cost per day is linearly related to the total output per day, write an equation
relating these two quantities. ( Hints: Find the equation of the line that passes through
(0, 200) and (20,1400) .); b) Graph the equation for 0  x  20 .
(a) The cost equation has a coefficient in front of x equal to the slope k 
1400  200
 60 .
20  0
Then the cost equation is C  60x  200 . (b) The equation can be graphed using Excel or by hand
through the above 2 points.
Fig. 1
Fig. 2
2. Graphical presentation of systems of equations in Business Analysis
Use Excel to solve the problems (See: Problems with Excel Part 1, Lecture 3 in Learning Resources)
2.1 Break-even Analysis
The cost of producing x boxes of chocolates is given by the equation C  2.8x  600 and each box
sells for 4€.
a) Find the break-even point graphically;
b) If it is known that at least 450 units will be sold, show graphically what should be the price charged
for each item to guarantee no loss?
2.2 Supply-Demand Management
Imagine that we want to find the market equilibrium point for the plasma TV sets. The task is: At a
price (p) of 2400€ the supply (s) of the plasma TV sets is 120 units while their demand (d) is 560
1
units. If the price is raised to 2700€ per unit the supply and demand will be 160 units and 380 units,
respectively. The following supply and demand equations have been derived on the base of this data:
s
2
p  200,
15
d 
3
p  2000
5
Graph the obtained equations and determine the market equilibrium point graphically. Find the
revenue at the equilibrium point.
We can use Excel to graph these equations and determine the market equilibrium point from the table.
Solve the following problems by traditional methods:
1. (Depreciation) Office equipment was purchased for 20,000€ and is assumed to have a scrap value of
2,000€ after 10 years. If its value is depreciated linearly from 20,000€ to 2,000€:
a) Find the linear equation that relates value (V) in Euros to time (t) in years.
b) Graph the equation for 0  t  10
c) What would be the value of the equipment after 6 years?
2. (Supply) At a price of 2.5€ per unit, a firm will supply 8000 shirts per month; at 4€ per unit, the
firm will supply 14,000 shirts per month. Determine the supply equation, assuming it to be linear.
3. (Market Equilibrium) The following supply and demand equations have been proposed after some
observations of the process of producing and selling (weekly) cars:
s  2 p  10,
d
8000
 370 ,
p
where p is the price per unit of thousands of Euros and s and d are number of units supplied and
demanded per month.
a) Find the equilibrium point (algebraically or by using Excel).
b) Determine the total revenue received by the manufacturer at the equilibrium point.
Mathematical Tools
1. Linear Equation in Two Variables. Different Forms of a Straight Line Equation.
The standard form of a linear equation in two variables is Ax  By  C  0 ,
provided A 2  B 2  0 . The solution set is formed by all ordered pairs of real numbers ( x, y ) that
satisfy the equation. In a rectangular coordinate system the graph of this equation is a straight line
drawn through the points whose coordinates satisfy the equation. To graph a linear equation in two
variables we plot any two points of their solution set and draw the line through these two points.
The points where the line crosses the axes – called the intercepts, are often the easiest to find
when graphing the equation. To find y-intercept, we let x = 0 and solve the eq. for y; to find xintercept, we let y = 0 and solve for x.
Example 1. Graph the equation 2 x  y  1  0 .
Solution: The graph of this eq. is a straight line passing through the points M(0,b) and N(a,0), where b
is the y-intercept and a is the x-intercept. Let x = 0. Then after replacing x by 0 in the eq. we obtain y =
-1, e.g. b = -1. When y = 0 we obtain from the eq. x = 0.5, e.g. a = 0.5. Hence the straight line passes
through the points M(0,-1) and N(0.5,0).
Different Forms of a Straight Line Equation.
Every straight line in a Cartesian coordinate system is a graph of a linear equation in two
variables The general form of the equation of a straight line g is:
g: Ax  By  C  0
(1)
The notation means the correspondence between the straight line g and the equation. This equation
could be transformed to different particular forms, which are more convenient for application.
 If the line does not pass through the origin ( 0,0) , i.e. C  0 and both A  0 and B  0 it has the
2
intercept form equation:
x y
  1,
a b
where a stands for x - intercept point and b - for y - intercept point.
g:
(2)
 If the line is horizontal, then its slope is 0, there is no x-intercept point and its equation gets the y intercept form:
g: y  b (all points have one and the same ordinate)
(3)
 If the line is vertical, there is no y-intercept point and its equation is transformed to the
x - intercept form:
g: x  a
(all points has one and the same abscissa).
(4)
 If the line passes through the point P( x1 , y1 ) with a slope k then it has the point slope form
equation:
g: y  y1  k ( x  x1 ) .
(5)
 If the line is not vertical, i.e. B  0 its equation is transformed to the slope-intercept form:
g: y  kx  n ,
(6)
where k is the slope of the line, n is its y - intercept point.
Definition. The slope (or gradient) of a line passing through the points P( x1 , y1 ) and Q( x2 , y2 ) is
y 2  y1
(7)
, x2  x1 ,
x2  x1
where y2  y1 is the so called rize and x1  x2  the run . The slope of a rising line is positive but
given by the formula: k 
the slope of a falling line is negative. Its meaning is the rate of change of y in respect to x.
Example 2. Find the intercept, slope-intercept and point-slope forms of the straight line equation
(presented in its general form): l : 3 x  2 y  6  0 .
Solution: We can transform the given equation as follows:
l : 3x  2 y  6  0  3x  2 y  6 :(6) 
x
y
  1 (intercept form of the eq.)
2 3
We can find the slope of the line replacing the coordinates of the y- and x- intercept points P(0,3) and
Q(-2,0) in (7): k 
03
3 3

  1.5 .
20 2 2
The other forms of the equation can be obtained by replacing k = 1.5, n = 3 in (5) and x1  0, y1  3
l : y  3  1.5 x (point-slope form) .
in (6): l : y  1.5 x  3 (slope-intercept form);
Theorem:
Let two straight lines are given by the equations l1 : y  k1 x  n1 and l 2 : y  k 2 x  n 2 . Then
* l 1 is parallel to l 2 if and only if k1  k 2 ;
* l 1 is perpendicular to l 2 if and only if k1 k 2  1 .
Answer the question:
Given the cost equation C  16 x  7000 . Which of the following statements is not correct?
A. The variable cost per unit is 16€;
B. The slope of the line corresponding to this equation is 16;
C. If the price per unit is 20€ the Profit equation is P  4x  7000 ;
D. The profit and cost straight lines are parallel;
E. None of the above.
Methods for solving systems of linear equations in two variables
Substitution method.
The following steps have to be performed:
1) Solve one of the equations with respect to one of the variables;
2) Substitute this variable by the expression obtained in (1) in the remaining equation;
3
3) Solve the equation in one variable obtained in (2).
4) Find the value of the remaining variable by back substitution into the expression obtained in 1).
5) Check the solution.
Addition method.
According to this method :
1) Make the coefficients of one of the unknown variables in both equations exactly the same
magnitude and opposite in sign;
2) Add the two equations to eliminate one of the variables;
3) Solve the obtained equation in one variable;
4) Substitute the obtained variable in one of the given equations and solve this equation in respect to
the other variable;
5) Check the solution.
Example 1.1. Solve the following system:
4 x  7 y  3
3x  5 y  49
(i)
(ii )
Solution: Applying the substitution method, we solve (i) for x : 4 x  7 y  3
Substituting this value of x into (ii), we obtain
or
x
1
7 y  3
4
3
(7 y  3)  5 y  49 . We multiply both sides by 4
4
and solve for y :
3(7 y  3)  20 y  196  21y  9  20 y  196  41 y  196  9  205  y 
Putting y = 5 in (i) we get x 
205
 5.
41
1
(35  3)  8 .Thus the solution is x = 8 and y = 5 or the ordered pair (8,5).
4
x y 5
(i)
Example 1.2. Solve the system by using the addition method:
3x  2 y  1
(ii )
Solution: It is appropriate to multiply both sides of (i) by 2 and add the obtained equation to (ii):
x  y  5  2 x  2 y  10  (2 x  2 y )  (3x  2 y )  10  1  11  5 x  0. y  11  5 x  11 or x 
After replacing the obtained value of x in (11) we get
11
11 25  11 14
4
 y  5 y  5 

2
5
5
5
5
5
.
The solution is x  2
1
4
, y2
5
5
.
Graph of the Solution Set
To graph the solution set of system 1) we have to graph the solution sets of the two equations. These
are the straight lines l1 : a1 x  b1 y  c1 , l 2 : a 2 x  b2 y  c 2 .
Then the following three cases are possible:
a) l 1 intersects l 2 in one point - the solution set is only one point in the plane;
b) l 1 is parallel to l 2 - the solution set is empty set -  ;
c) l 1 = l 2 , i.e. the two lines coincide - infinitely many points belong to the solution set, namely all
the points on the line.
Exercises
I. Solve the following systems of linear equations:
1.
5.
3x  2 y  1
x y0
40 s  50t  10
100s  200t  30
2.
x y2
x  4y  7
6.
2x  y  2
3.
x  3y  1
120 p  20q  10
7.
120 p  15q  35
4
4.
x  2y  2
2x  4 y  1
2 C  3 R  1
22 C  13Ry  12
8.
s  2t  2
2s  4t  4
11
1
2
5
5