Question

Q.4. In Shraddha's farm the cost function for output x is given by
C =x³÷3- 20x² + 70x Find the output for which i) Marginal cost (Cm) is
minimum. ii) Average cost (CA) is minimum.​

Answer 1

Answer:

average cost is minimum for x>30

Answer 2

Answer:

(i) The Marginal Cost is minimum at x = 20

(ii) The Average Cost is minimum at x = 30.

Step-by-step explanation:

C = $\frac{x^3}{3}$ - 20x² + 70x

(i) We know that Marginal Cost = C'(x)

where C(x) is the cost function.

Therefore, here, Marginal Cost (CM) = $\frac{1}{3}$ * 3x² - 20 * 2 * x + 70

                                                             = x² - 40x + 70

We have to find the output where CM is minimum

Therefore, differentiating with respect to x

=> $\frac{dCM}{dx}$ = 2x - 40 + 0                               --(i)

Differentiating again with respect to x

=> $\frac{d^2CM}{dx^2}$ = 2

As 2 > 0 therefore, minima.

Therefore, putting equation (i) = 0

=> 2x - 40 = 0

=> 2x = 40

=> x = 20

Therefore, the Marginal Cost is minimum at x = 20.

(ii) We know that Average Cost = C(x)/x

where C(x) is the cost function.

Therefore, here, Average Cost (CA) = ( $\frac{x^3}{3}$ - 20x² + 70x ) / x

                                                             = $\frac{x^2}{3}$ - 20x + 70

We have to find the output where CA is minimum

Therefore, differentiating with respect to x

=> $\frac{dCA}{dx}$ = $\frac{1}{3}$* 2x - 20 + 0                               --(ii)

Differentiating again with respect to x

=> $\frac{d^2CA}{dx^2}$ = 2/3

As 2/3 > 0 therefore, minima.

Therefore, putting equation (ii) = 0

=> 2x/3 - 20 = 0

=> 2x/3 = 20

=> 2x = 60

=> x = 30

Therefore, the Average Cost is minimum at x = 30.