Answer:
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,
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
Answer:
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
