BBA 2421

Business Mathematics

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 UNIT 2

Applications of Linear Equations and Non-Linear Functions

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 Equations and functions
• An equation is a mathematical statement setting two algebraic
expression equal to each other.
• An equation in which all variables are raised to the first power is known
as a linear equation. Examples of linear equations are:
𝑦 = 2𝑥 + 1 & 2𝑦 − 4𝑥 = 7
• Linear equations are generally expressed as 𝑎𝑥 + 𝑏𝑦 = 𝑐.
• A linear equation can be solved by moving the unknown variable to the
left-hand side of the equal sign and all the other terms to the right-
hand side. To solve the above two equations they will look like this:
𝑦 − 2𝑥 = 1
2𝑦 − 4𝑥 = 7

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• Another type of equation is what is known as a quadratic equation which takes
the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 ≠ 0. Such an
equation is solved by factoring or using the quadratic formula:
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
In-class exercise
• Solve the following:
𝑥 𝑥
i. 4
−3= +1
5
ii. 𝑥 2 + 13𝑥 + 30 = 0
iii. 8𝑃𝑏 − 3𝑃𝑚 = 7 and -𝑃𝑏 + 7𝑃𝑚 = 19

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Functions
• A function has the following components:
i. Independent variable(s) called the argument of the function
ii. Dependent variable called the value of the function
• The equation 𝑦 = −2𝑥 + 7 is a function where 𝑥 is the argument of the
function and 𝑦, also denoted as 𝑓(𝑥), is the value of the function at
any unique 𝑥.
• The set of all possible values of the argument of a function (𝑥) is called
the domain of a function while the set of all possible values of the
value of the function [𝑓(𝑥)] is called the range.
• A function is therefore an algebraic formula such that for each value of
the independent variable 𝑥, there is one and only one value of the
dependent variable 𝑓(𝑥).

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• There are basically five (5) types of functions and these are:
i. Linear function
ii. Quadratic function
iii. Polynomial function of degree 𝑛
iv. Rational function
v. Power function

Reading assignment: Read on the various types of functions listed

above.
• In business we have revenue and profit functions.

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Revenue, cost and profit functions
• We define revenue as the number of units sold multiplied with the price
per unit.
• That is, Total revenue (TR)=price× 𝑞𝑢𝑎𝑛𝑡𝑖𝑦 or simply 𝑇𝑅 𝑄 = 𝑃𝑄.
• Total cost represents money spent in producing the goods.
• While profit is the difference between Total Revenues (TR) and Total Cost
(TC), and the profit function is denoted by π.
• That is;
𝜋 = 𝑇𝑅 – 𝑇𝐶
• For instance, if I go to Soweto and order tomato at 𝐾106 per box. If the
box contains 35 tomatoes and I sell each tomato at 𝐾10. Then my revenue
is 35 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 × 𝐾10 𝑃𝑟𝑖𝑐𝑒 = 𝐾350. if all I spent is 𝐾106 then my
total cost is 𝐾106 and thus 𝜋 = 𝑇𝑅 – 𝑇𝐶 = 𝐾350 − 𝐾106 = 𝐾244.

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 Example using functions
Suppose the demand function is given by P = 100 – 2Q. Find TR as the
function of Q.
Solution
𝑇𝑅 = 𝑃 × 𝑄 = 100 – 2𝑄 × 𝑄 = 100𝑄 − 2𝑄 2
• We can also transform the demand function P=100-2Q into an
inverse demand function by making Q subject of the formula as
follows:
𝑃 = 100 – 2𝑄
2𝑄 = 100 − 𝑃
• If we divide both sides of the equation by 2 we get
2𝑄 100 𝑃
= − ⇨ 𝑄 = 50 − 0.5𝑃
2 2 2
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• The Total Cost function relates the production cost to the level of
output Q.
• The production function is an increasing function owing to the fact
that total cost increase as output increases.
• Total costs can be decomposed into fixed costs and variable costs. i.e.
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐹𝑖𝑥𝑒𝑑 𝐶𝑜𝑠𝑡 + 𝑇𝑜𝑡𝑎𝑙 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐶𝑜𝑠𝑡
𝑇𝐶 = 𝑇𝐹𝐶 + 𝑇𝑉𝐶
• Fixed costs are costs such as rent, land cost and cost of equipment
that are constant whatever the amount of goods produced.
• On the other hand, variable costs are costs that vary with output e.g.
the cost of raw materials, energy costs, cost of unskilled labour etc.
• Therefore, TVC = Variable cost x Output (TVC = VC x Q)

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• AVERAGE TOTAL COST OR AVERAGE COST FUNCTION
• The average total cost or average cost (AC) function is obtained by dividing the
total cost function by output.
𝐓𝐂 TFC + TVC 𝐓𝐅𝐂 𝐓𝐕𝐂 𝐓𝐅𝐂 𝐕𝐂 𝐱 𝐐
𝐴𝐶 𝑜𝑟 𝐴𝑇𝐶 = = = + = + ,
𝐐 𝐐 𝐐 𝐐 𝐐 𝐐
Where 𝑇𝑉𝐶 = 𝑉𝐶 𝑥 𝑄
𝐓𝐅𝐂
𝐴𝐶 = + 𝑉𝐶
𝐐
Example
• Assume the fixed costs of a firm are K100, while its variable costs are
K10 per unit. Find the average cost function as a function of output.

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 Solution

𝐓𝐂
𝐴𝐶 =
𝐐

𝑏𝑢𝑡 𝑇𝐶 = 𝑇𝐹𝐶 + 𝑇𝑉𝐶

= 100 + 𝑉𝐶 𝑥 𝑄

= 100 + 10𝑄

100 + 10Q
𝐴𝐶 = 𝑇𝐶/𝑄 = ( )
Q

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 Break-even analysis
• Break-Even Point is the point at which total cost and total revenue are
equal. In other words, it is the point at which the total revenue curve
intersects with the total cost curve. The diagram below illustrates
break-even analysis.
M
TR TC
TC B
TR

0 Q
𝑄𝐴 𝑄𝑀 𝑄𝐵

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• From the above, we note that the two curves intersect at points A and B, where
Total revenue =Total cost.
• We therefore, say that the firm breaks even at point A and B and that the quantities
𝑄𝐴 and 𝑄𝐵 are the quantities at which the firm breaks even. For Q< 𝑄𝐴 and Q> 𝑸𝑩 ,
the firm makes a loss because the total cost is relatively higher than the total
revenue.
• Thus, the firm makes increases its profits by increasing its output towards Q_A and
reducing its output towards 𝑄𝐵 .
• However, if Q is greater than 𝑄𝐴 but less than 𝑄𝐵 (i.e 𝑄𝐴 < Q > 𝑄𝐵 ), the firm makes a
profit because TR is greater than TC.
• NOTE: A firm wishing to maximise profit would produce 𝑄𝑀 because it is the output
that maximises the firm’s profit.

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Basic calculus rules
• If 𝑦 = 𝐴𝑥 𝑛 , where 𝐴 is a constant and 𝑛 is any real number then the
derivative of 𝑦 will be given by:
𝑑𝑦
= 𝑛𝐴𝑥 𝑛−1
𝑑𝑥
Example
Find the derivative of the following:
i. 𝑦 = 2𝑥 0
1
ii. 𝑦 = 2
𝑥
iii. 𝑦 = 3𝑥 3 + 4

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 Solution
𝑑𝑦
i. = 0 × 2𝑥 0−1 = 0
𝑑𝑥
𝑑𝑦 1 1−1 1
ii. 𝑑𝑥
=1×
2
𝑥 =
2
𝑑𝑦
iii. =3× 3𝑥 3−1 + 0 = 9𝑥 2
𝑑𝑥

Task: Read on the rules of differentiation.

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 Optimization

• This can be divided in two types: that is unconstrained and

constrained optimization.

• Recall that firms always want to maximise profits and levels of output
as well as minimise costs and the use of scarce natural resources.

• The objective of maximization and minimization can be attained using

differential calculus or differentiation techniques.

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 Unconstrained optimization
• Unconstrained optimization encompasses the simplest form of
optimization problems without conditions (constraints).
• The set-up is such that the analyst only has an objective function,
which is sufficient to derive the optimal solution using the approach
discussed below.
• Let’s assume we want to maximise utility given by the following
function 𝑈=𝑈 (𝑋1 ,𝑋2 ).
• Under unconstrained optimization, a consumer can keep taking ever
increasing amounts of 𝑋1 and 𝑋2 and therefore, lead to an infinite
solution with regard to the total utility.
• The following steps will be used to solve unconstrained optimization
problems.

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• Case 1: Functions of one variable.
• Assume 𝑦 = 𝑓(𝑥)
• Step 1
• Solve for the derivative of y which is 𝑦 ′ or 𝑓 ′ (𝑥) and equate to zero and to
find the stationary points, let’s say x = z.
• Step 2
i. If f´´ (z) > 0. Then the function has a minimum at x=z
ii. If f´´ (z) < 0. Then the function has a maximum at x=z
iii. If f´´ (z) =0. Then the point cannot be classified using the available
information.

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 Example
• Maximise 𝝅 for a firm with 𝑇𝑅 = 4000𝑄 − 33𝑄 2 and 𝑇𝐶 = 2𝑄 3 −
3𝑄 2 + 400𝑄 + 5000 assuming 𝑄 > 0.
Solution
• Set up the 𝝅 function
𝝅 = 𝑇𝑅 − 𝑇𝐶
= 4000𝑄 − 33𝑄 2 − 2𝑄 3 − 3𝑄 2 + 400𝑄 + 5000
−2𝑄 3 − 30𝑄 2 + 3600𝑄 − 5000
• Find the critical points/stationary points by finding the derivative of
π(Q) and equating it to zero.
𝜋 ′ = −6𝑄 2 − 60𝑄 + 3600 = 0
−6𝑄 2 − 60𝑄 + 3600 = 0

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 −6(𝑄 2 + 60𝑄 − 600) = 0
𝑄 2 + 60𝑄 − 600 = 0
𝑄 + 30 𝑄 − 20 = 0
• Solving the above, we have
• Q = -30 and Q = 20, as the critical points
• Since Q cannot be negative, we ignore the negative output (Q = -30)
because it has no economic significance.
• Therefore, find the second derivative and evaluate it at Q = 20 to
verify that Q = 20 is at the maximum.

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 𝜋 ′′ 𝑄 = −12𝑄 − 60
𝜋 ′′ 20 = −12 20 − 60 = −300 < 0
• Confirming that Q=20 is indeed at the maximum (i.e. Concave). Thus
the maximum profit is:
𝜋 20 = −2(20)3 − 30 20 2 + 3600 20 − 5000 = 39000.

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• Case 2: Functions of two variables
• Assume 𝑧 = 𝑓(𝑥, 𝑦)
• Step 1
• Find the stationary points by solving the simultaneous equations below:
𝜕𝑧
= 𝑓𝑥 𝑥, 𝑦 = 0
𝜕𝑥
𝜕𝑧 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑡𝑤𝑜 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ 𝑡𝑤𝑜 𝑢𝑛𝑘𝑛𝑜𝑤𝑛𝑠 𝑥 𝑎𝑛𝑑 𝑦
= 𝑓𝑦 𝑥, 𝑦 = 0
𝜕𝑦
• Step 2
• If (a, b) are the stationary/critical points obtained from solving the system of
equations above, then;
• The point (a, b) is a minimum iff;

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 2 2 2 2 2 2
𝜕 𝑧 𝜕 𝑧 𝜕 𝑧 𝜕 𝑧 𝜕 𝑧
2
> 0, 2 > 0, 2 2
− >0
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥𝜕𝑦
2
𝑓𝑥𝑥 𝑎, 𝑏 > 0, 𝑓𝑦𝑦 𝑎, 𝑏 > 0 𝑎𝑛𝑑 𝑓𝑥𝑥 𝑎, 𝑏 × 𝑓𝑦𝑦 𝑎, 𝑏 − 𝑓𝑥𝑦 𝑎, 𝑏 >0
• The point (a, b) has a maximum iff;
2 2 2 2 2 2
𝜕 𝑧 𝜕 𝑧 𝜕 𝑧 𝜕 𝑧 𝜕 𝑧
2
< 0, 2 < 0, 2 2
− >0
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥𝜕𝑦
2
𝑓𝑥𝑥 𝑎, 𝑏 < 0, 𝑓𝑦𝑦 𝑎, 𝑏 < 0 𝑎𝑛𝑑 𝑓𝑥𝑥 𝑎, 𝑏 × 𝑓𝑦𝑦 𝑎, 𝑏 − 𝑓𝑥𝑦 𝑎, 𝑏 >0

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 Example
• Find and classify the stationary points of the function
𝑍 = 3𝑥 2 − 𝑥𝑦 + 2𝑦 2 − 4𝑥 − 7𝑦 + 12
Solution
• Step 1
• Find the stationary points by solving the simultaneous equations below:
𝜕𝑧
= 𝑧𝑥 = 0
𝜕𝑥
𝜕𝑧 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
= 𝑧𝑦 = 0
𝜕𝑦
𝑍𝑥 = 6𝑥 − 𝑦 − 4 = 0
𝑍𝑦 = −𝑥 + 4𝑦 − 7 = 0

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• Solving gives us;
• x = 1, y = 2, i.e (1, 2) is the critical point
• Step 2
• Find the second derivative and evaluate them at the critical points
𝑍𝑥𝑥 = 6, 𝑍𝑦𝑦 = 4,
𝑍𝑥𝑥 1,2 = 6 > 0 𝑎𝑛𝑑 𝑍𝑦𝑦 1,2 = 4 > 0
• The cross partial derivatives are:
𝑍𝑥𝑦 = −1, 𝑍𝑦𝑥 = −1,
𝑍𝑥𝑦 1,2 = −1 𝑎𝑛𝑑 𝑍𝑦𝑥 1,2 = −1
𝑍𝑥𝑦 × 𝑍𝑦𝑥 = −1 × −1 = 1 > 0
• Since 𝑍𝑥𝑥 > 0 , 𝑍𝑦𝑦 > 0 and 𝑍𝑥𝑦 × 𝑍𝑦𝑥 > 0, the function has a minimum at
the point (1,2)

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 Constrained optimisation
• Constrained optimization includes optimization problems
that maximize or minimize a function subject to constraints.
• Given a function 𝑓(𝑥, 𝑦) subject to a constraint 𝑔 𝑥, 𝑦 =
𝑘 (a constant), a new function ℒ can be formed by
i. Setting the constraint equal to zero
ii. Multiplying the rearranged constraint by 𝜆 (the Lagrange
multiplier)
iii. Adding the product to the original function.

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• This leads to a new function called the Lagrangian function.
• The Lagrangian equation is a mathematical expression that combines
the objective function and the constraint to help solve for the optimal
choice of a given optimization problem.
• With the Lagrangian equation, we introduce a third-choice variable
called the Lagrange multiplier which is denoted by the Greek letter,
lamda.
• The Lagrange multiplier approximates the marginal impact of the
objective function caused by a small change in the constant of the
constraint.
• Constrained optimization can have several constraints including
equality and inequality constraints. In this unit, we will only focus on
constrained optimization problems with one equality constraint
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 Example
• Let’s assume that a firm wants to maximise its output Q=f (K, L).
• Let’s also assume that the unit costs of capital and labour are 𝑃𝑘 and
𝑃𝑙 respectively. The cost of the firm is given by;
𝑃𝑘 𝐾 + 𝑃𝑙 𝐿 = 𝑀
• This is the fixed cost the firm is willing to spend
• Thus, the problem the firm faces is that it wishes to:
𝑀𝑎𝑥𝑖𝑚𝑖𝑠𝑒 𝑄 = 𝑓 𝐾, 𝐿
𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 𝑃𝑘 𝐾 + 𝑃𝑙 𝐿 = 𝑀

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 • On a diagram the above will look
like this.
• The curves are known as
isoquants and along each
isoquant the output is the same
even though using different
combinations of capital and
labour.
• The optimal combination occurs
at point A.

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Solving constrained optimisation
• Differential calculus can be used to solve the constrained
maximization problem.
• Given the objective function 𝑓(𝑥 , 𝑦) subject to a constraint
𝒈(𝒙 , 𝒚) = 𝒌, the Lagrangian function can be obtained by adhering
to the following steps:
• Set the constraint equal to zero i.e. 𝒌 − 𝒈(𝒙, 𝒚) = 𝟎
• Then multiply it by Lagrange multiplier (𝝀) 𝒊. 𝒆. 𝝀[𝒌 − 𝒈(𝒙, 𝒚)], and then
• Add the product to the objective or original function to obtain the lagrangian
function F.
𝐹(𝑥, 𝑦, 𝜆) = 𝑓(𝑥, 𝑦) + 𝜆[𝑘 − 𝑔(𝑥, 𝑦)]

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• The stationary points/critical values 𝑥0 , 𝑦0 and 𝜆 at which the
function is optimized are found by taking the derivative of
𝐹(𝑥, 𝑦, 𝜆) with respect to all the independent variables, set them
equal to zero and solve simultaneously.
𝐹𝑥 𝑥, 𝑦, 𝜆 = 0, 𝐹𝑦 𝑥, 𝑦, 𝜆 = 0 𝑎𝑛𝑑𝐹𝜆 𝑥, 𝑦, 𝜆 = 0

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 Example
• Use the Lagrange multiplier method to optimise the objective function
below subject to the given constraint.
𝑍 = 4𝑥 2 – 2𝑥𝑦 + 6𝑦 2
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜: 𝑥 + 𝑦 = 72
Solution
• Set the constraint to zero, multiply it by λ and add it to the objective
function to obtain
𝐹(𝑥, 𝑦, 𝜆) = 4𝑥 2 – 2𝑥𝑦 + 6𝑦 2 + 𝜆(72 – 𝑥 – 𝑦)
• F.O.C
𝐹𝑥 = 8𝑥 – 2𝑦 – 𝜆 = 0 … … . . (𝑖 )
𝐹𝑦 = −2𝑥 + 12𝑦 – 𝜆 = 0 … … . . (𝑖𝑖)

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 𝐹𝜆 = 72 – 𝑥 – 𝑦 = 0 . . . (𝑖𝑖𝑖) • Substituting 𝑥 and 𝑦 into equation (i)
• Subtract equation (ii) from or (ii) gives us
equation (i) to eliminate λ, we 𝜆 = 276
have • The critical point is;
10𝑥 – 14𝑦 = 0 𝑥 = 42, 𝑦 = 30, 𝜆 = 276
𝑥 = 1.4𝑦 … … (𝑖𝑣) • Thus, the optimum point is;
• Substitute equation (iv) into 𝑭(𝒙, 𝒚, 𝝀)
equation (iii) = 𝟒(𝟒𝟐)𝟐 – 𝟐(𝟒𝟐)(𝟑𝟎)
1.4𝑦 + 𝑦 = 72 + 𝟔(𝟑𝟎)𝟐 + 𝟐𝟕𝟔(𝟕𝟐 – 𝟒𝟐 – 𝟑𝟎)
𝑦 = 30 = 𝟗𝟗𝟑𝟔
• Using 𝑥 = 1.4𝑦, substituting y
gives us
𝑋 = 1.4𝑥30 = 42
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 Interpretation of the lagrangian multiplier

• The Lagrangian multiplier λ approximates the marginal impact on the

objective function caused by a small change in the constant of the
constraint.

• E.g. a 1 unit increase (decrease) in the constant of the constraint

causes F to increase (decrease) by approximately λ units.

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 End of Unit 2

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