demand and revenue functions

Question

Suppose that the total cost of production of a certain item is given by C(x) = 0.04x^3 − 0.6x^2 + 32x + 96 with x the number of items produced. Suppose also that the demand function is p(x) = 132−2.6x.

(a) Determine the marginal revenue and marginal cost when x = 3.

(b) Find the profit function P(x).

(c) Find the production level that will maximize profit.

Answer 1

R(x) = xp(x) = 132x - 2.6x²; this is revenue function.

(a) Merginal revenue R'(x) = 132 - 5.2x; merginal cost is C'(x) = 0.12x² - 1.2x + 32.;

R'(3) = $116.4 per item; C'(3) = $29.48 per item

(b) P(x) = R(x) - C(x) = 132x - 2.6x² - 0.04x³ + 0.6x² - 32x - 96 = - 0.04x³ - 2x² + 100x - 96.

P'(x) = - 0.12x² - 4x + 100 = 0; 3x² + 100x - 2500 = 0; x = (-50 + 100)/3 = 50/3.

P''(x) = - 0.24x - 4; P''(50/3) = - 8 <0. By second derivative test we have maximum profit at x = 50/3

P_max = P(50/3) = -0.04·(50/3)³ - 2·(50/3)² + 100·(50/3) - 96 = $829.93