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Thread: Proof regarding a norman window max area...

  1. #1
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    Proof regarding a norman window max area...

    A norman window where the semicircle's radius is r (thus the base is 2r) and the height of the rectangle is h. Perimeter is constant, $\displaystyle P$. Show the values r and h are both $\displaystyle P/(pi+4)$ when the area is a maxiumum.

    Please use a method not involving derivatives.
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  2. #2
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    Quote Originally Posted by jkrowling View Post
    A norman window where the semicircle's radius is r (thus the base is 2r) and the height of the rectangle is h. Perimeter is constant, $\displaystyle P$. Show the values r and h are both $\displaystyle P/(pi+4)$ when the area is a maxiumum.

    Please use a method not involving derivatives.
    $\displaystyle P = 2h + 2r + \pi r$

    and ...

    $\displaystyle \displaystyle A = 2rh + \frac{\pi r^2}{2}$

    solve for $\displaystyle h$ in the perimeter equation and substitute the result for $\displaystyle h$ in the area equation ... you should get area as a quadratic function of $\displaystyle r$. locate the vertex (max) using $\displaystyle \displaystyle r = \frac{-b}{2a}$.

    let's see what you can do.
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  3. #3
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    OK...
    $\displaystyle h=(P-2r-pi*r)/2$

    $\displaystyle A=(2r*(P-2r-pi*r)/2)+(pi*r^2)/2$
    $\displaystyle A(r)=r(P-2r-pi*r)+(pi*r^2)/2$

    $\displaystyle r=-b/2a$
    $\displaystyle r=-(P-2r-pi*r)/(2(1/2)*pi)$

    I get to here, but I have trouble further simplifying.
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  4. #4
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    $\displaystyle A(r) = 2r \cdot \frac{P - (2+\pi)r}{2} + \frac{\pi r^2}{2}$

    $\displaystyle A(r) = Pr - (2+\pi)r^2 + \frac{\pi r^2}{2}$

    $\displaystyle A(r) = Pr + \left(\frac{\pi}{2} - 2 - \pi\right)r^2$

    $\displaystyle A(r) = Pr - \left(\frac{\pi}{2} + 2\right)r^2$

    try again
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  5. #5
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    ok...

    $\displaystyle A(r)=Pr-(pi/2+2)r^2$

    $\displaystyle rmax=-b/2a$
    $\displaystyle =-P/-2(pi/2+2)$
    $\displaystyle =P/(pi+4)$

    So then, to prove that this is also the max value for h, I assume I should plug the result into the original equation for h...?

    $\displaystyle h=P-2(P/(pi+4))+pi(P/(pi+4))$

    When I simplify this, I get

    $\displaystyle h=(P*pi+4P-2P+P*pi)/(pi+4) * (1/2)$
    $\displaystyle =(P*pi-p)/(pi+4)$

    not the $\displaystyle P/(pi+4)$ that I was hoping for...??
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  6. #6
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    $\displaystyle h = \frac{P - (2+\pi)r}{2}$

    $\displaystyle h = \frac{P}{2} - \frac{2+\pi}{2} \cdot \frac{P}{\pi+4}
    $

    common denominator ...

    $\displaystyle h = \frac{P(\pi+4)}{2(\pi+4)} - \frac{P(2+\pi)}{2(\pi+4)}$

    combine fractions ...

    $\displaystyle h = \frac{P\pi + 4P - 2P - P\pi}{2(\pi+4)}$

    combine like terms ...

    $\displaystyle h = \frac{2P}{2(\pi+4)}$

    simplify ...

    $\displaystyle h = \frac{P}{\pi+4}$


    you need to work on your algebra.
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