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Math Help - revenue/profit maximization

  1. #1
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    revenue/profit maximization

    The selling price for q units of a product is given by the equation p = 250-25q. The cost of producing q unites of the product is:
    C = 100 + 5q
    a) what is the max revenue that can be generated?
    b) what is the max profit that can be generated?
    Last edited by jamesk486; July 13th 2010 at 05:36 AM. Reason: solved
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  2. #2
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    Hello, jamesk486!

    You say you solved it.
    But did you notice a small snag at the end?


    The selling price for x units of a product is: . P(x) \:=\: 250-25x

    The cost of producing x units of the product is: . C(x) \:=\: 100 + 5x

    a) What is the max revenue that can be generated?

    R \;=\;x\cdot P(x) \:=\:x(250-25x) \;=\;250x - 25x^2

    Then: . R' \,=\,0 \quad\Rightarrow\quad 250-50x\:=\:0 \quad\Rightarrow\quad x \:=\:5

    Max Revenue: . R(5) \:=\:5(250-25\cdot5) \;=\;\$625




    b) What is the max profit that can be generated?

    \text{Profit}\;=\;\text{Revenue} - \text{Cost} \;=\;x\cdot P(x) - C(x)

    . . . P \;=\;x(250-25x) - (100 + 5x)

    . . . P \;=\;-25x^2 + 245x - 100


    Then: . P' \:=\:0 \quad\Rightarrow\quad -50x + 245 \:=\:0 \quad\Rightarrow\quad x \:=\:4.9



    Assuming we cannot manufacture nor sell a fraction of a unit,

    . . we have: . \begin{array}{ccccc}P(4) &=& -25(4^2) + 245(4) - 100 &=& \$480 \\<br />
P(5) &=& -25(5^2) + 245(5) - 100 &=& \$500 \end{array}


    If 5 units are made and sold, the maximum profit is $500.

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