A manufacturer can produce x items at a total cost of Rs. 120 x + x 2 2 . If he sells the whole produce at p per item, where p = 2 100 - x 4 , then find the production level x at which the profit is maximum. What is the price per item and the total profit?

Given, cost function C ( x ) = x 2 2 + 120 x Price P ( x ) = 2 ( 100 - x 4 ) = 200 - x 2 Revenue function R ( x ) = P ( x ) × x = ( 200 - x 2 ) × x = 200 x - x 2 2 Profit G ( x ) = R ( x ) - C ( x ) = 200 x - x 2 2 - x 2 2 - 120 x = - x 2 + 80 x For maxima/minima d d x G ( x ) = 0 d d x G ( x ) = d d x ( 80 x - x 2 ) 0 = 80 - 2 x x = 40 d 2 d x 2 G ( x ) = - 2 < 0 Hence, x = 40 is point of maxima. P ( 40 ) = 200 - 40 2 = 180 G ( 40 ) = 80 . 40 - ( 40 ) 2 = 1600 Hence, price per unit is ₹ 180 and profit is ₹ 1600 .

A manufacturer can produce x items at a total cost of Rs. 120 x + x 2 2 . If he sells the whole produce at p per item, where p = 2 100 - x 4 , then find the production level x at which the profit is maximum. What is the price per item and the total profit?

Given, cost function C ( x ) = x 2 2 + 120 x Price P ( x ) = 2 ( 100 - x 4 ) = 200 - x 2 Revenue function R ( x ) = P ( x ) × x = ( 200 - x 2 ) × x = 200 x - x 2 2 Profit G ( x ) = R ( x ) - C ( x ) = 200 x - x 2 2 - x 2 2 - 120 x = - x 2 + 80 x For maxima/minima d d x G ( x ) = 0 d d x G ( x ) = d d x ( 80 x - x 2 ) 0 = 80 - 2 x x = 40 d 2 d x 2 G ( x ) = - 2 < 0 Hence, x = 40 is point of maxima. P ( 40 ) = 200 - 40 2 = 180 G ( 40 ) = 80 . 40 - ( 40 ) 2 = 1600 Hence, price per unit is ₹ 180 and profit is ₹ 1600 .

Find the profit maximising output level, given x = 200 - 10 p and A C = 10 + 7 30 x where x represents the units of output, p is the price per unit and A C is the average cost.

Given, average cost function A C = 10 + 7 30 x Cost function C ( x ) = A C × x = ( 10 + 7 30 x ) × x = 10 x + 7 30 x 2 x = 200 - 10 p p = 20 - x 10 Revenue R ( x ) = p × x = ( 20 - x 10 ) × x = 20 x - x 2 10 Profit P ( x ) = R ( x ) - C ( x ) = 20 x - x 2 10 - 10 x - 7 30 x 2 = 10 x - x 2 3 For profit to be maximum, d d x P ( x ) = 0 d d x ( 10 x - x 2 3 ) = 0 10 - 2 x 3 = 0 x = 15 Now, if profit is maximum then , d 2 d x 2 P ( x ) < 0 d 2 d x 2 ( 10 - 2 x 3 ) = - 2 3 < 0 Hence, for x = 15 , profit will be maximum.

The marginal revenue of selling x units of a commodity is given by M R = 20 e x / 10 1 + x 10 . If the revenue obtained on selling 10 units of the commodity is Rs 200 e , find the revenue function.

M R = 20 e x 10 ( 1 + x 10 ) Revenue function R ( x ) = ∫ M R d x R ( x ) = ∫ 20 e x 10 ( 1 + x 10 ) d x R ( x ) = 20 [ 10 e x 10 + x e x 10 - 10 e x 10 ] + k R ( x ) = 20 x e x 10 + k Also, given R ( 10 ) = 200 e 20 x e x 10 + k = 200 e 200 e + k = 200 e k = 0 Hence, R ( x ) = 20 x e x 10