Business Calculus

Question

1. Given the cost function C(x)=1/4x^3-2x^2+6x+20, where x is hundreds of items produced and C(x) is thousands of dollars, find the following.

a. The level of production which minimizes marginal cost

b. The first derivative of the average cost function

c. Show that when the average cost is minimized, the marginal cost and average cost are equal

Answer 1

C(x) = ¼x³ – 2x² + 6x + 20

a. The level of production which minimizes marginal cost

marginal cost = C'(x) = ¾x² – 4x + 6 is minimized when C''(x) = 0

C''(x) = ³⁄₂x – 4 = 0 when x = ⁸⁄₃

b. The first derivative of the average cost function

C_avg(x) = C(x)/x = ¼x² – 2x + 6 + 20/x is the average cost function

C'_avg(x) = ½x – 2 – 20/x²

c. Show that when the average cost is minimized, the marginal cost and average cost are equal.

Average cost is minimized when

C'_avg(x) = ½x – 2 – 20/x² = 0

Multiplying both sides by x² gives eqn (1)

(1) ½x³ – 2x² – 20 = 0

When marginal cost and average cost are equal C'(x) = C(x)/x

This means ¾x² – 4x + 6 = ¼x² – 2x + 6 + 20/x

Canceling the 6's and collecting like terms gives

½x² – 2x – 20/x = 0

Multiplying both sides by x also gives eqn (1)

(1) ½x³ – 2x² – 20 = 0

This shows that when the average cost is minimized, the marginal cost and average cost are equal.