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Math Help - Optimizing a Norman Window - What am I doing wrong?

  1. #1
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    Optimizing a Norman Window - What am I doing wrong?

    A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 45 feet?

    A(h) = 1/2 \pi r^2 + 2r ((45-\pi(r)-2r)/2r)

     A'(h) = \pi(r) - \pi-2

    So solving for 0 i get (\pi +2)/\pi as the critical point

    pluging back in:

    1/2 \pi[(\pi +2)/\pi]^2 + 45-\pi[(\pi+2)/\pi)]-2[(\pi +2)/\pi)]<br />

    the hell is wrong here? i keep getting an incorrect answer. something simple i suppose
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  2. #2
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    Quote Originally Posted by pseizure2000 View Post
    A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 45 feet?

    A(h) = 1/2 pi r^2 + 2r ((45-pi(r)-2r)/2r)
    Why divide by 2r?

    Divide by 2 only.

    Perimeter, P = 45 = (1/2)(2pi*r) +2r +2y
    2y = 45 -pi(r) -2r
    y = (45 -pi(r) -2r) / 2 only.
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  3. #3
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    [size=13Hello, pseizure2000![/size]

    A Norman window has the shape of a semicircle atop a rectangle
    so that the diameter of the semicircle is equal to the width of the rectangle.
    What is the area of the largest possible Norman window with a perimeter of 45 feet?
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          * - - - - + - - - - *
          |    r         r    |
          |                   |
        h |                   | h
          |                   |
          |                   |
          * - - - - - - - - - *
                   2r

    The radius of the semicircle is r.
    Then the width of the rectangle is 2r.
    Let h = height of the rectangle.

    The perimeter is: . \pi r + 2h + 2r \:=\:45\quad\Rightarrow\quad h \:=\:\frac{45 - 2r - 2\pi}{2} . [1]

    The area of the window is: . A \;=\;\frac{1}{2}\pi r^2 + 2rh . [2]

    Substitute [1] into [2]: . A \;=\;\frac{1}{2}\pi r^2 + 2r\left(\frac{45 - 2r - \pi r}{2}\right)

    . . which simplifies to: . A \;=\;45r - \left(\frac{4+3\pi}{2}\right)r^2 . [3]


    Differentiate and equate to zero: . A' \;=\;45 - (4+3\pi)r \;=\;0

    . . Hence: . r \;=\;\frac{45}{4+3\pi}


    Substitute into [3]: . A \;=\;45\left(\frac{45}{4+3\pi}\right) - \left(\frac{4+3\pi}{2}\right)\left(\frac{45}{4+3\p  i}\right)^2<br />

    . . and we get: . A \;=\;\frac{2025}{2(4+3\pi)} \;\approx\;\boxed{7542\text{ ft}^2}

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  4. #4
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    in more detail...

    sorry if this seems a bit elementary but how did you simplify to step #3?
    where does the 3\pi in numerator come from?

    I am not sure where I am making a mistake but I keep getting:
    A \;=\;45r - \left(\frac{\pi-4}{2}\right)r^2

    thanks!



    Quote Originally Posted by Soroban View Post

    . . which simplifies to: . A \;=\;45r - \left(\frac{4+3\pi}{2}\right)r^2 . [3]
    Last edited by fitswell; March 26th 2011 at 08:53 PM.
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  5. #5
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    Quote Originally Posted by fitswell View Post
    sorry if this seems a bit elementary but how did you simplify to step #3?
    where does the 3\pi in numerator come from?

    I am not sure where I am making a mistake but I keep getting:
    A \;=\;45r - \left(\frac{\pi-4}{2}\right)r^2

    thanks!
    Yes there is a mistake in the post. An easy mistake to make.

    I get A \;=\;45r - \left(\frac{\pi+4}{2}\right)r^2

    Note the + ....
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  6. #6
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    yes, very easy to get lost in the numbers, many thanks for the rework!
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