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?

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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?

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Answer 1 · 2015-07-09T13:23:48

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 $ $x$ $x$
then
Rental: $R (x) = (27 + x)$ $R(x) =(27+x)$ dollars per car-day
Number of cars rented: $N (x) = (220 - 5 x)$ $N(x) =(220-5x)$
and
Income: $I (x) = (27 + x) (220 - 5 x) = 5840 + 85 x - 5 x^{2}$ $I(x) =(27+x)(220-5x) = 5840+85x-5x^2$ dollars/day

The maximum will be achieved when the derivative of $I (x)$ $I(x)$ is zero.

$\frac{d I (x)}{d x} = 85 - 10 x = 0$ $(d I(x))/(dx) = 85-10x = 0$

$\Rightarrow x = 8.5$ $rArr x = 8.5$

For an even dollar rental amount, and increase of $8/day or $9/day will generate the same income.
So $$ 27 + $ 8 = $ 35$ $$27+$8 = $35$ /day
or
$$ 27 + $ 9 = $ 36$ $$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
$XXXX$ $color(white)("XXXX")$ $I (9) = 6300$ $I(9) = 6300$

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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?

1 Answer

Explanation:

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