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.

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 cost function of producing x units of a product is given by M C = 3 x 2 - 10 x + 3 . If the total cost of producing 1 unit of the product is 6 , find the total cost function and the average cost.

Given, M C = 3 x 2 - 10 x + 3 C ( x ) = ∫ M C d x C ( x ) = ∫ ( 3 x 2 - 10 x + 3 ) d x C ( x ) = x 3 - 5 x 2 + 3 x Let fixed cost = K 6 = 1 3 - 5 ( 1 2 ) + 3 ( 1 ) + K K = 7 Therefore C ( x ) = x 3 - 5 x 2 + 3 x + 7 A C = Total cost Number of units A C = x 3 - 5 x 2 + 3 x + 7 x A C = x 2 - 5 x + 3 + 7 x

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