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Math Help - 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, P. Show the values r and h are both 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, P. Show the values r and h are both P/(pi+4) when the area is a maxiumum.

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

    and ...

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

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

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

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

    r=-b/2a
    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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    A(r) = 2r \cdot \frac{P - (2+\pi)r}{2} + \frac{\pi r^2}{2}

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

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

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

    try again
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  5. #5
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    ok...

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

    rmax=-b/2a
    =-P/-2(pi/2+2)
    =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...?

    h=P-2(P/(pi+4))+pi(P/(pi+4))

    When I simplify this, I get

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

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

    h = \frac{P}{2} - \frac{2+\pi}{2} \cdot \frac{P}{\pi+4}<br />

    common denominator ...

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

    combine fractions ...

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

    combine like terms ...

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

    simplify ...

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


    you need to work on your algebra.
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