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Math Help - qUEstion

  1. #1
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    qUEstion

    A rectangle is inscribed in semicircle of radius 4m. Find the max possible area for such a rectangle
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let AB be the diameter, O the center of the circle, MNPQ the rectangle, with M,Q\in[AB].
    Let PQ=x,OQ=y. Then the area is A=2xy.
    We have x^2+y^2=16.
    If the area is maximum, then the squared area is also maximum.
    A^2=4x^2y^2=4x^2(16-x^2)
    Let x^2=t and f(t)=4t(16-t)=-4t^2+64t
    The maximum value of f is \displaystyle f_{max}=-\frac{\Delta}{4a}, where \Delta=b^2-4ac, \ a=-4, \ b=64, \ c=0.
    So, f_{max}=256, for t_{max}=-\frac{b}{2a}=8.
    Then x=y=2\sqrt{2}
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  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by red_dog View Post
    Let AB be the diameter, O the center of the circle, MNPQ the rectangle, with \color{red}{M,Q\in[AB]}.
    How can you claim that the statement in red is necessary?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Isomorphism View Post
    How can you claim that the statement in red is necessary?
    Because opposite vertices of an inscribed rectangle must lie on a diameter.

    RonL
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  5. #5
    Eater of Worlds
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    Here is what I was doing until I realized I used the wrong radius.

    Sorry if it seems redundant since RedDogs fine answer, but I worked it out earlier and do not want my attempt to go in vain.

    Let x and y be the dimensions of the rectangle. Then area is A=xy

    But (\frac{x}{2})^{2}+y^{2}=16

    This means y=\sqrt{16-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{64-x^{2}}

    Sub into A and we have:

    \frac{x}{2}\sqrt{64-x^{2}}

    \frac{dA}{dx}=\frac{32-x^{2}}{\sqrt{64-x^{2}}}

    32-x^{2}=0, \;\ x=4\sqrt{2}, \;\ y=2\sqrt{2}

    Max area is then (4\sqrt{2})(2\sqrt{2})=16

    Now, try the same problem with an inscribed trapezoid
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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