A product can be manufactured at a total cost C ( x ) = x 2 100 + 100 x + 40 , where x is the number of units produced. The price at which each unit can be sold is given by P ( x ) = 200 - x 400 . Determine the production level x at which the profit is maximum. What is the price per unit and total profit at this level of production?

Given, cost function C ( x ) = x 2 100 + 100 x + 40 Price P ( x ) = 200 - x 400 Revenue function R ( x ) = P ( x ) × x = ( 200 - x 400 ) × x = 200 x - x 2 400 Profit G ( x ) = R ( x ) - C ( x ) = 200 x - x 2 400 - x 2 100 - 100 x - 40 = 100 x - 5 x 2 400 - 40 For maxima/minima d d x G ( x ) = 0 d d x G ( x ) = d d x ( 100 x - 5 x 2 400 - 40 ) 0 = 100 - x 40 x = 4000 d 2 d x 2 G ( x ) = - 1 40 < 0 Hence, x = 4000 is point of maxima. P ( 4000 ) = 200 - 4000 400 = 190 G ( 4000 ) = 100 ( 4000 ) - 5 ( 4000 ) 2 400 - 40 = 199960 Hence, price per unit is ₹ 190 and profit is ₹ 199960 .

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.