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Thread: 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

    $\displaystyle y=8-x^2$ 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

    $\displaystyle 2x(8-x^2)$

    and differentiate

    $\displaystyle (2x(-2x))+2(8-x^2)$

    $\displaystyle -4x^2+16+2x^2$

    Solve for 0

    $\displaystyle -2x^2+16$

    $\displaystyle +-\sqrt{8}$

    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?

    $\displaystyle A = 2x(8 - x^2) = 16x - 2x^3$

    $\displaystyle A' = 16 - 6x^2$

    $\displaystyle 0 = 16 - 6x^2$

    $\displaystyle 16 = 6x^2$

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

    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


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

    $\displaystyle x = \sqrt {16/6}$

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

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