Question

The demand function is  `x = (24 - 2p)/(3)` where x is the number of units demanded and p is the price per unit. Find:
(i) The revenue function R in terms of p.
(ii) The price and the number of units demanded in which the revenue is maximum. 

Answer 1

`x =  (24 - 2p)/(3)`

3x = 24 - 2p
2p = 24 - 3x

⇒ p =` 12 - (3)/(2) x`

(i) R(x) = p.x =`(12 - (3)/(2) x) x`

= `12x - (3)/(2) x^2` 

(ii) let y = R(x)

`(dy)/(dx) = (dR)/(dx)` = 12 - 3x

`(d^2y)/(dx^2) = -3 < 0`

Hence, revenue is maximum
∴ The point of maxima is given by :

`(dy)/(dx) = 0`  ⇒ 3x = 12

⇒ x = 4

Hence, 4 units must be produced and the price demanded is :

R(4) = `12(4) - (3)/(2) (4)^2`

R(4) = `48 - (3)/(2) xx 4 xx 4`

R(4) = 48 - 24

R(4) = ₹ 24.