Question

A store has been selling calculators at ₹ 350 each. A market survey indicates that a reduction in price (p) of calculator increases the number of units (x) sold. The relation between the price and quantity sold is given by the demand function `p=450-1/2x.`

Based on the above information, answer the following questions:

1. Determine the number of units (x) that should be sold to maximise the revenue R(x) = x p(x). Also, verify the result.
2. What rebate in price of calculator should the store give to maximise the revenue?

Answer 1

(i) The Demand function

`P = 450-1/2x`

Revenue function R(x) = Px

`= (450-1/2x)xx x`

`R = 450x - 1/2 x^2`

`(dR)/dx = 450-x`

`(dR)/dx=0`

450 − x = 0

x = 450

`(d^2R)/dx^2 = -1`

`(d^2R)/dx^2<0`

when x = 450

To find number of unit x for R is maximum,

We should find 'x' where `(dR)/dx = 0, (d^2R)/dx^2<0`

`P= 450-1/2xx450`

p = 450 − 225

P = ₹ 125

∴ The revenue is maximum when P = 125, the price per unit is ₹ 125 and 450 unit are demanded.

(ii) Rebate in price of calculator

= 350 − 125 = ₹ 125

P  = ₹ 125