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Math Help - revenue maximization

  1. #1
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    revenue maximization

    A tool rental shop can rent 30 floor sanders per day at a daily rate of $12 per sander. For each additional $2 a day charged per sander, one fewer sander is rented. If the store wants to maximize its revenue, what should it charge to rent a sander for one day? (Give your answer to the nearest dollar)

    i used excel to find all the possible revenue by multiplying different daily rates and number of sanders rented to find the answer.
    however, is there an equation? i cannot seem to set one up.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    The current revenue is 30*12 = $360

    It can be said to be of the form xy = R

    where x is the number of floor sanders rented, y the price per sander and R the revenue.

    Also, we see that as y increase, x decreases, both linearly. We can say that y = -kx + c

    First case, x = 30, y = 12.

    12 = -k(30) + c
    12 = -30k + c

    Second case, x = 29, y = 14.

    14 = -k(29) + c
    14 = -29k + c

    Using elimination;

    2 = k
    k = 2

    Then, c becomes 72

    So;

    y = -2x + 72

    Replace that in the initial equation;
    x(-2x + 72) = R

    simplify;

    R = -2x^2 + 72x

    At R max, dR/dx = 0.

    R' = -4x + 72 = 0

    Solving for x gives x = 18.
    When x = 18, y = 36.
    Revenue becomes 18*36 = $648
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  3. #3
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    ok this may be a stupid question..but how do you know its linear???
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    MHF Contributor Unknown008's Avatar
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    What I did was imagine some cases. In each case, as y increases by 2, x increases by 1.

    The first coordinate was (30, 12), then, x decreases as y increases giving (29, 14). If x increases, it becomes (31, 12). There is a constant gradient throughout which can be said to have the vector \binom{-1}{2} or \binom{1}{-2}.

    Sorry if that is difficult to understand... maybe a graph will help you.

    y = -2 x +5 - Wolfram|Alpha

    Ok, I input the line y = -2x + 5

    As you can see, as y becomes larger (ie you go higher up), x decreases (ie you go towards the left). There is a point where there aren't any sales, and that's when the price is too high. There is a point too where you will have the maximum number of people buying, that when the price is zero, that is when the sanders are free.

    I hope it helped.
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  5. #5
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    ok this may be a stupid question..but how do you know its linear???
    For each additional $2 a day charged per sander, one fewer sander is rented
    This tells you that quantity is a linear function of price. (every time price goes up a fixed amount, quantity goes down a fixed amount)
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  6. #6
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    Hello, jamesk486!

    A tool rental shop can rent 30 floor sanders per day at a daily rate of $12 per sander.
    For each additional $2 a day charged per sander, one fewer sander is rented.
    Tto maximize its revenue, what should they charge to rent a sander for one day?

    We will use an obvious formula:

    . . \text{(Revenue)} \:=\:\text{(Daily rate}) \times \text{(No. of sanders)}


    Presently, we have: . \begin{Bmatrix}\text{Rate} & \$12 \\ \text{Sanders} & 30 \end{Bmatrix} \quad\text{ Revenue: }\$360


    Let x = number of $2 increases.
    . . The daily rate will be: . 12 + 2x dollars.

    Then 30-x = number of sanders rented.

    Hence, the revenue will be: . R \;=\;(12+2x)(30-x) dollars.


    We have: . R \;=\;360 + 48x - 2x^2

    This is a down-opening paraloba.
    . . Its maximum occurs at its vertex.

    The vertex is at: . x \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}48}{2(\text{-}2)} \:=\:12


    Therefore, they should have 12 two-dollar increases.

    The new rate will be: . 12 + 12(2) \:=\:\boxed{36\text{ dollars per day}}


    Only 30-12 \,=\,18 sanders will be rented,
    . . but the revenue will be: . (\$36)(18) \:=\:\$648

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