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Math Help - Calc - Extreme Values - Word Problem

  1. #1
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    Calc - Extreme Values - Word Problem

    The profit, P, in dollars for selling x hamburgers is modelled by  P(x)= 2.44x - \frac {x^2}{20 000} - 5 000, where 0 \leq x \leq 35 000. For what quantities of hamburgers is the profit increasing and decreasing?


    P'(x) = 2.44 - 40000x

    0 = 2.44 - 40 000x

    6.1^{-0.5} = x

    P(6.1^{-0.5}) = -4999

    P(0) = -5000

    P(35 000) = 19 150


    Therefore, decreasing at (0, -5000) and increasing at  (35 000, 19 150)



    Textbook Answer:
    increasing - 0 \leq x \leq 24 400
    decreasing - 24 400 \leq x \leq 35 000
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Macleef View Post
    The profit, P, in dollars for selling x hamburgers is modelled by  P(x)= 2.44x - \frac {x^2}{20 000} - 5 000, where 0 \leq x \leq 35 000. For what quantities of hamburgers is the profit increasing and decreasing?


    P'(x) = 2.44 - {\color{red}40000x} << SiMoon says : it should be \frac{x}{10000} !!

    0 = 2.44 - 40 000x

    6.1^{-0.5} = x << SiMoon says : if you were right, it would have been 6.1 \cdot {\color{red}10^5} !

    P(6.1^{-0.5}) = -4999

    P(0) = -5000

    P(35 000) = 19 150


    Therefore, decreasing at (0, -5000) and increasing at  (35 000, 19 150)



    Textbook Answer:
    increasing - 0 \leq x \leq 24 400
    decreasing - 24 400 \leq x \leq 35 000
    Therefore, there are problems all along :/
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  3. #3
    Eater of Worlds
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    Check your derivative.

    P'(x)=\frac{61}{25}-\frac{x}{10,000}

    \frac{61}{25}-\frac{x}{10,000}=0

    x=24,400

    Now check the slope on either side.
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