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12 May, 08:23

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?

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Answers (2)
  1. 12 May, 08:31
    0
    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
  2. 12 May, 09:35
    0
    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
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