A car rental agency rents 210 cars per day at a rate of 27 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?

x=7.5

Step-by-step explanation:

Rental: R(x)=(27+x) dollar per day

Number of cars rented: N(x)

=(210-5x)

Income: I(x) =(27+x)(210-5x)

=5670 - 135x + 210x-5x^2

=5670+75x-5x^2

The maximum will be achieved when the derivative of

I(x) =zero.

I(x)=5670+75x-5x^2

dI(x)/dx=75-10x=0

75-10x=0

75=10x

x=75/10

=7.5

x=7.5

x=7.5 is the rate at which the car will be rented to produce maximum income.

The maximum income could be derived by using x=7 or x=8

27+7=$34 per day

27+8=$35 per day

When x=7

I(x)=5670+75x-5x^2

=5670+75(7)-5(7)^2

=5670+525-5(49)

=5670+525-245

=5,950

When x=8

I(x)=5670+75x-5x^2

=5670+75(8)-5(8)^2

=5670+600-5(64)

=5670+600-320

=5950

Either x=7 or 8 will yield the same income

ian gonna lie that is way to much