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Application of Calculus in Commerce and Economics
OPTIONAL - II
Mathematics for
Commerce, Economics
and Business
41
Notes
APPLICATION OF CALCULUS IN
COMMERCE AND ECONOMICS
æö
We have learnt in calculus that when 'y' is a function of 'x', the derivative of y w.r.to x i.e. çdy÷
ç ÷
ç ÷÷
èø
dx
measures the instantaneous rate of change of y with respect to x. In Economics and commerce
we come across many such variables where one variable is a function of the other. For example,
the quantity demanded can be said to be a function of price. Supply and price or cost and
quantity demanded are some other such variables. Calculus helps us in finding the rate at which
one such quantity changes with respect to the other. Marginal analysis in Economics and
Commerce is the most direct application of differential calculus. In this context, differential calculus
also helps in solving problems of finding maximum profit or minimum cost etc., while integral
calculus is used to find he cost function when the marginal cost is given and to find total revenue
when marginal revenue is given.
In this lesson, we shall study about the total, average or marginal functions and the optimisation
problems.
OBJECTIVES
After studying this lesson, you will be able to :
· define Total Cost, Variable Cost, Average Cost, Marginal Cost, Total Revenue, Marginal
Revenue and Average Revenue;
· find marginal cost and average cost when total cost is given;
· find marginal revenue and average revenue when total revenue is given;
· find optimum profit and minimum total cost under given conditions; and
· find total cost/ total revenue when marginal cost/marginal revenue are given, under given
conditions.
212
MATHEMATICS
Application of Calculus in Commerce and Economics
EXPECTED BACKGROUND KNOWLEDGE OPTIONAL - II
Mathematics for
· Derivative of a function Commerce, Economics
Integration of a function and Business
·
41.1 BASIC FUNCTIONS
Before studying the application of calculus, let us first define some functions which are used in
business and economics. Notes
41.1.1 Cost Function
The total cost C of producing and marketing x units of a product depends upon the number of
units (x). So the function relating C and x is called Cost-function and is written as C = C (x).
The total cost of producing x units of the product consists of two parts
(i) Fixed Cost
(ii) Variable Cost i.e. C (x) = F + V (x)
Fixed Cost : The fixed cost consists of all types of costs which do not change with the level of
production. For example, the rent of the premises, the insurance, taxes, etc.
Variable Cost : The variable cost is the sum of all costs that are dependent on the level of
production. For example, the cost of material, labour cost, cost of packaging, etc.
41.1.2 Demand Function
An equation that relates price per unit and quantity demanded at that price is called a demand
function.
If 'p' is the price per unit of a certain product and x is the number of units demanded, then we
can write the demand function as x= f(p)
or p = g (x) i.e., price (p) expressed as a function of x.
41.1.3 Revenue function
If x is the number of units of certain product sold at a rate of Rs. 'p' per unit, then the amount
derived from the sale of x units of a product is the total revenue. Thus, if R represents the total
revenue from x units of the product at the rate of Rs. 'p' per unit then
R= p.x is the total revenue
Thus, the Revenue function R (x) = p.x. = x .p (x)
41.1.4 Profit Function
The profit is calculated by subtracting the total cost from the total revenue obtained by selling x
units of a product. Thus, if P (x) is the profit function, then
P(x) = R(x) − C(x)
41.1.5 Break-Even Point
Break even point is that value of x (number of units of the product sold) for which there is no
MATHEMATICS 213
Application of Calculus in Commerce and Economics
OPTIONAL - II profit or loss.
Mathematics for i.e. At Break-Even point P (x ) = 0
Commerce, Economics
and Business or R(x)-=C(x)0 i.e. R(x)= C(x)
Let us take some examples.
Example 41.1 For a new product, a manufacturer spends Rs. 1,00,000 on the infrastructure
Notes and the variable cost is estimated as Rs.150 per unit of the product. The sale price per unit was
fixed at Rs.200.
Find (i) Cost function (ii) Revenue function
(iii) Profit function, and (iv) the break even point.
Solution : (i) Let x be the number of units produced and sold,
then cost function C ( x) = Fixed cost + Variable Cost
= 1,00,000 + 150 x
(ii) Revenue function = p.x = 200 x
(iii) Profit function P ( x ) = R(x)-C(x)
=200x-+(100,000150x)
=-50x100,000
(iv) At Break-Even point P(x)0=
50x-=100,0000
100,000
x==2000
50
Hence x = 2000 is the break even point.
i.e. When 2000 units of the product are produced and sold, there will be no profit or loss.
Example 41.2 A Company produced a product with Rs 18000 as fixed costs.
The variable cost is estimated to be 30% of the total revenue when it is sold at a rate of Rs. 20
per unit. Find the total revenue, total cost and profit functions.
Solution : (i) Here, price per unit (p) = Rs. 20
Total Revenue R ( x ) = p. x = 20 x where x is the number of units sold.
30
(ii) Cost function C(x)=+18000R(x)
100
30
=18000+´20x
100
=+180006x
(iii) Profit function P(x)=-R(x)C(x)
214
MATHEMATICS
Application of Calculus in Commerce and Economics
=20x-+(180006x) OPTIONAL - II
Mathematics for
=-14x18000 Commerce, Economics
and Business
Example 41.3 A manufacturing company finds that the daily cost of producing x items of a
product is given by C(x)=+210x7000
(i) If each item is sold for Rs. 350, find the minimum number that must be produced and sold
daily to ensure no loss. Notes
(ii) If the selling price is increased by Rs. 35 per piece, what would be the break even point.
Solution :
(i) Here, R(x)=350x and C(x)=+210x7000
\ P(x)=--350x210x7000
=-140x7000
For no loss P(x)0=
Þ 140x-=70000 or x=50
Hence, to ensure no loss, the company must produce and sell at least 50 items daily.
(ii) When selling price is increased by Rs. 35 per unit,
R(x)=+=(35035)x385x
\ P(x)=385x-+(210x7000)
=-175x7000
At Break even point P(x)0=
Þ 175x-=70000
7000
∴ x==40
175
CHECK YOUR PROGRESS 41.1
1. The fixed cost of a new product is Rs. 18000 and the variable cost per unit is Rs. 550. If
demand function p(x)=-4000150x, find the break even values.
2. A company spends Rs. 25000 on infrastructure and the variable cost of producing one
item is Rs. 45. If this item is sold for Rs. 65, find the break-even point.
3. A television manufacturer find that the total cost of producing and selling x television sets
is 2 . Each product is sold for Rs. 6000. Determine the
Cx=50x++3000x43750
( )
break even points.
4. A company sells its product at Rs.60 per unit. Fixed cost for the company is Rs.18000
and the variable cost is estimated to be 25 % of total revenue. Determine :
(i) the total revenue function (ii) the total cost function (iii) the breakeven point.
MATHEMATICS 215
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