A car rental agency rents 220 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?

Jul 9, 2015

Assuming an even dollar rental is required;
The cars should be rented at $36 per day for a maximum income of$6300 per day.

Explanation:

If the daily rental is increased by $$x$then Rental: $R \left(x\right) = \left(27 + x\right)$dollars per car-day Number of cars rented: $N \left(x\right) = \left(220 - 5 x\right)$and Income: $I \left(x\right) = \left(27 + x\right) \left(220 - 5 x\right) = 5840 + 85 x - 5 {x}^{2}$dollars/day The maximum will be achieved when the derivative of $I \left(x\right)$is zero. $\frac{d I \left(x\right)}{\mathrm{dx}} = 85 - 10 x = 0$$\Rightarrow x = 8.5$For an even dollar rental amount, and increase of$8/day or $9/day will generate the same income. So $27+$8 =$35/day
or
$27+$9 = $36/day would both be valid answers. However,$36/day involves renting fewer cars and thus reduced expenses.

Using basic substitution and arithmetic
$\textcolor{w h i t e}{\text{XXXX}}$$I \left(9\right) = 6300$