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Math Help - A quick question about Increasing and decreasing Functions.

  1. #1
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    A quick question about Increasing and decreasing Functions.

    Ok so it says here

    A manufacturing company finds the profit for a production level of x motors per hour is

    p = 400 x sqr(12x - x^2) - 100 x e (0,12]
    For what range of production is the profit
    Increasing
    Decreasing


    Now the answer is (0, 6) increasing and (6, 12) decreasing.

    When I found the derivative, I got

    (-200x +2400) (12x-x^2)^-1/2

    Now the critical number for the first bracket is 12, which I found, makes sence because the function decreases at 12. But I cant find where the 6 comes from.

    Can anyone help?
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  2. #2
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    Quote Originally Posted by Sky12 View Post
    Ok so it says here

    A manufacturing company finds the profit for a production level of x motors per hour is

    p = 400 x sqr(12x - x^2) - 100 x e (0,12]
    For what range of production is the profit
    Increasing
    Decreasing
    Thus you have,
    P=400x\sqrt{12x-x^2}-100
    And I assume you mean, x\in (0,12]

    So find the derivative und make it zero.
    P'=(400x)'(\sqrt{12x-x^2})+(400x)(\sqrt{12x-x^2})'
    Thus,
    P'=400\sqrt{12x-x^2}+\frac{400x(12-2x)}{2\sqrt{12x-x^2}}=0
    Multiply through by the radical,
    400(12x-x^2)+200x(12-2x)=0
    Divide by 200,
    2(12x-x^2)+x(12-2x)=0
    24x-2x^2+12x-2x^2=0
    -4x^2+36x=0
    -4x(x-9)=0
    Thus,
    x=0,9
    But,
    x\in (0,12].
    Thus,
    x=9
    Is the only solution.

    Now take any point to the left of 9 (and more than zero) and see what the sign is. Positive increasing and negative decreasing. Same for any point to the right of 9 (and less than 12)
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  3. #3
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    Ah sorry for the confusion but i ment the x as (times) so it would be more like

    400sqr(12x - x^2) - 100
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