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  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 $\displaystyle AB$ be the diameter, $\displaystyle O$ the center of the circle, $\displaystyle MNPQ$ the rectangle, with $\displaystyle M,Q\in[AB]$.
    Let $\displaystyle PQ=x,OQ=y$. Then the area is $\displaystyle A=2xy$.
    We have $\displaystyle x^2+y^2=16$.
    If the area is maximum, then the squared area is also maximum.
    $\displaystyle A^2=4x^2y^2=4x^2(16-x^2)$
    Let $\displaystyle x^2=t$ and $\displaystyle f(t)=4t(16-t)=-4t^2+64t$
    The maximum value of f is $\displaystyle \displaystyle f_{max}=-\frac{\Delta}{4a}$, where $\displaystyle \Delta=b^2-4ac, \ a=-4, \ b=64, \ c=0$.
    So, $\displaystyle f_{max}=256$, for $\displaystyle t_{max}=-\frac{b}{2a}=8$.
    Then $\displaystyle x=y=2\sqrt{2}$
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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 $\displaystyle AB$ be the diameter, $\displaystyle O$ the center of the circle, $\displaystyle MNPQ$ the rectangle, with $\displaystyle \color{red}{M,Q\in[AB]}$.
    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 $\displaystyle (\frac{x}{2})^{2}+y^{2}=16$

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

    Sub into A and we have:

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

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

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

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

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