A car rental agency rents 180 cars per day at a rate of 30 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?

Rate = $33/day

Maximum Income = $5,445

Explanation:

Rental: (Rx) = (30+x) dollars per day

Number of car rented: N (x) = (180 - 5x)

Income

I (x) = (30+x) * (180 - 5x)

Open bracket by (30 + x)

30 (180-5x) + x (180-5x)

5400-150x+180x-5x²

= 5400+30x-5x²

Therefore the maximum that will be achieved when the derivative of I (x) is zero

dI (x) / dx

30-10x=0

=30/10x

x = 3

For an even dollar rental amount, an increase of $3/days will generate the same income

$30+$3 = $33/day

Hence: The rate in which the cars should be rented in order to produce maximum income is $33/day

Maximum Income

180 - (5*3)

=180-15

=165

165*33=$5,445

Maximum Income = $5,445

The maximum income is $5,445

Rate = $33

Maximum Income = $5445

Explanation:

Let x be the amount of increase in rental to achieve maximum profit.

So, Rate = (30+x)

When rate increase by x, the quantity decreases by (180 - 5x).

Income = (30+x) * (180 - 5x)

Income = 5400 - 150x + 180x - 5x²

Income = - 5x² + 30x + 5400

The income will be maximized when derivative of Income is zero.

Taking derivative,

dI/dx = 2 * - 5x + 1 * 30x° + 0 - 10x + 30 = 0 - 10x = - 30 x = - 30 / - 10 x = 3

The rate at which cars should be rented to earn maximum income is 30 + 3 = $33 per day per car.

Maximum Income will be,

Rate = 33

Quantity = 180 - 5 (3) = 165 cars

Max Income = 33 * 165 = $5445