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Math Help - Area of an inscribed Rectangle

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    Exclamation Area of an inscribed Rectangle

    A rectangle that is x feet wide is inscribed in a circle of radius 8 ft.

    a) Express th area of the rectangle as a function of x.
    b) find the domain of the function
    c) What dimmensions maximize the area of the rectangle?

    I would appreciate help.
    thank you
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  2. #2
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    Quote Originally Posted by ludiaz223 View Post
    A triangle that is x feet wide is inscribed in a circle of radius 8 ft.

    a) Express th area of the rectangle as a function of x.
    b) find the domain of the function
    c) What dimmensions maximize the area of the rectangle?

    I would appreciate help.
    thank you
    Always draw a picture

    Area of an inscribed Rectangle-circ.bmp

    The Point P is located at (x,\sqrt{64-x^2})

    The red box is one quarter of the area of the rectangle so

    A(x)=4\cdot x \cdot \sqrt{64-x^2}

    The domain is all values of the radicand(the stuff under the square root sign) is greater than or equal to zero.

    64-x^2 \ge 0 \iff (8-x)(8+x) \ge 0 when we test our values we get x \in [-8,8]

    To maximize the are we take the derivative of A(x)

    \frac{dA}{dx}=4[\sqrt{64-x^2}+x\cdot \frac{1}{2}\cdot \frac{-2x}{\sqrt{64-x^2}}]=4\left[ \frac{64-x^2-x^2}{\sqrt{64-x^2}}\right]

    setting the numerator equal to zero and solving we get

    64-2x^2=0 \iff x^2=32 \iff x = \pm \sqrt{32}

    Since we are talking about a distance we use only the postive root.

    Now when x=\sqrt{32} plugging this into \sqrt{64-x^2} to get the height of the box we get

    \sqrt{64-(\sqrt{32})^2}=\sqrt{64-32}=sqrt{32}
    So the base and the height of the rectangle are the same so it is a square.
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