Question

A restaurant sells x hamburger per week. Assume that the weekly cost and demand equations

C(x) = 1.1x + 300

p(x) = 5 - 0.03x 0 ≤ x ≤ 1,000

Find the number of hamburgers that the restaurant should sell and the price that the restaurant should charge for each hamburger to maximize the profit.

Answer 1

Note that profit = revenue - cost

= p(x)*x - c(x)

= 5x - 0.03x^2 - 1.1x - 300 subject to 0 ≤ x ≤ 1,000

Profit is maximum when its derivative is 0

5 - 0.06x - 1.1 = 0

x = 3.9/0.06 = 65 units

Hence profit maximizing quantity is x = 65 units and price P = 5 - 0.03*65 = 3.05 per unit

Profit = 3.05*65 - 1.1*65 - 300 = -173.25.