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Math Help - Rectangle Inscribed in an Area

  1. #1
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    Rectangle Inscribed in an Area

    I have a rectangle inscribed in the parabola defined by

    y=8-x^2 with one edge on the x-axis ?

    Would I assume that edge on the x-axis is x or 2x?
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  2. #2
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    Your width is 2x, your height is y.
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  3. #3
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    So area = b*h

    2x(8-x^2)

    and differentiate

    (2x(-2x))+2(8-x^2)

    -4x^2+16+2x^2

    Solve for 0

    -2x^2+16

    +-\sqrt{8}

    which one I use? positive or negative
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  4. #4
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    So i'm guessing you're trying to maximize your area?

    A = 2x(8 - x^2) = 16x - 2x^3

    A' = 16 - 6x^2

    0 = 16 - 6x^2

    16 = 6x^2

    x = \sqrt \frac{16}{6}

    Seeing as we can't have negative dimensions, we use +.
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  5. #5
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    Thanks man it just seems I keep getting all sorts of things confused


    I get the dimsenions are


    \sqrt{\frac{16}{6}}*\frac{16}{3}
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  6. #6
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    A = 2xy

    x = \sqrt {16/6}

    y = 8 - \sqrt {16/6}^2 = 8 - 8/3 = 16/3

    Looks good to me. Just remember, technically, our base is 2x, not x.
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