Question

I need help with this question: A car rental agency rents 180 cars per day at a rate of 28 dollars per day. For each 1 dollar increase in the daily rate, 5 fewer cars are rented. At what rate should the cars be rented to produce the maximum income, and what is the maximum income? My teacher provided notes on this question, but I am still having trouble. First he said that you should set up the equation: A = CP (where A is the amount of money they make, C is the number of cars and P is the price. He also set up another equation where: C=(-5(P-28)+180)P, to help us. But, I still need help!

Answer 1

let rate change on top of $28 = d. This will lead to a loss of 5*d cars a day
i.e. At rate = 28+d, Number of cars rented = 180 - 5*d
revenue =R= (28+d) * ( 180 - 5*d)
now maximize R. I believe you'll get d=4 i.e. $32 is the ideal daily rate

Answer 2

This may be a stupid question, but how do you get d from maximizing the revenue? Would you need a value for the derivative of R in respect of d?

Answer 3

A few different ways -

1. Graph y = (28+d)*(180-5d) and see that d=4 the curve is at its peak

2. differentiate wrt d and solve for d

3. Its a quadratic in d. So complete the square on the right side of R and you'll see that it is max at d=4

Answer 4

Okay, thanks!