Paper 4: Fundamentals of Business Mathematics & Statistic

3.50 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

Marginal Revenue : For any demand function p = f (q), the total revenue (TR) is the product of quantity

demand (q), and the price (p) per unit of output.

TR = q x p = q x f (q) (as p = f(q))

Now the marginal revenue represents the change in TR for each additional unit of sale, so

Marginal revenue ( )

MR d TR( )

= dq i.e., derivative of TR w.r.t. quantity demanded.

Here TR = Revenue = pq, AR (average revenue)

pq

= q

For TR maximum we may also have to find the output (q), making MR = 0 i.e.,

d(TR) 0

dq = (i.e., first derivative

w.r.t. output equal to zero) and hence we can estimate the price (p) and finally the maximum revenue.

Note : For profit maximisation, MR = MC.

Example 107 :A radio manufacturer produces ‘x’ sets per week at a total cost of Rs.

x^2

(^25) +3x 100 .+ He is
monopolist and the demand for his market is x = 75 – 3p ; where p is the price in rupees per set. Show that
the maximum net revenue is obtained when about 30 sets are produced per week. What is the monopoly
price?
Solution:
Net revenue (NR) = Sale – total cost = x × p – TC
75 x x^2
=x 3 − − 25 +3x 100+^
For max. net revenue, we have d(dxNR)= 0 or,
25 2 x 2x 3 0
3 25

− −^ +^ =

(^)
or,
2x 1 1
3 25

−^ +^

= 3 – 25 = – 22

or,

x^2811

75

(^) =
or, x=11 75 28 × =^30 (app.)
Now, p=^753 −x 75 30 45= − 3 = 3 =^15
∴ monopoly price = ` 15 per set.
Example 108 :A manufacturer can sell x items per month at a price p = 300 – 2x rupees. Produced items cost
the manufacturer y rupees y = 2x + 1000. How much profit will yield maximum profits?