Question

12. Maximum Revenue) A car-rental agency can rent 100 cars per day at a rate of $40 per day For each price increase of $1 per day, 2 less cars are rented. What rate should be charged to maximize the revenue of the agency?

Answer 1

Let X days rate = 40 + x total cars rented = 100 - 2x Income = (40 + x) (100 - 2x) f(x) = (40 + x) (100 - 2x) f(x) = 4000 - 80x + 100x - 2x^2 f(x) = 4000 + 20x - 2x^2_(1) f'(x) = 0 + 20(1) - 2(2x) {differentiation} f'(x) = 20 - 4x implies 20 = 4x implies 4x = 20 implies x = 20/4 = 5 x = 5 so for the rate = 40 + x = 40 + 5 = $45 (1) implies Maximum income = 4000 + 20(5) - 2(5)^2 = 4000 + 100 - 50 = 4000 + 50 = $ 4050