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Math Help - optimization

  1. #1
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    optimization

    I cant seem to get the right answer.

    A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 20 ft, find the value of x so that the greatest possible amount of light is admitted.

    for my critical number i get 80/(4+pi) for my critical number of x which isnt working.
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  2. #2
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    Quote Originally Posted by Evan.Kimia View Post
    I cant seem to get the right answer.

    A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 20 ft, find the value of x so that the greatest possible amount of light is admitted.

    for my critical number i get 80/(4+pi) for my critical number of x which isnt working.
    What does x stand for?
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  3. #3
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    Hello, Evan.Kimia!

    Your answer is "close", so you're doing most of it Right.
    You probably made a very simpler error.


    A Norman window has the shape of a rectangle surmounted by a semicircle.
    If the perimeter of the window is 20ft, find the value of x
    so that the greatest possible amount of light is admitted.

    For my critical number, i get 80/(pi + 4) . It shoud be: . {\color{blue}\frac{20}{\pi + 4}}
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          * - - - - * - - - - *
          |    r         r    |
          |                   |
         y|                   |y
          |                   |
          |                   |
          * - - - - - - - - - *
                   2r

    The radius of the semicircle is r.
    Hence, the width of the rectangle is 2r.
    The height of the rectangle is y.

    The perimeter of the semicircle is: . \pi r
    The perimeter of the rectangle is: . 2r + 2y

    The total perimeter is 20:
    . . \pi r + 2r + 2y \:=\:20 \quad\Rightarrow\quad y \:=\:10 - \left(\frac{\pi+2}{2}\right)r .[1]


    The area of the semicircle is: . \tfrac{1}{2}\pi r^2
    The area of the rectangle is: . (2r)(y)

    The total area is: . A \;=\;\tfrac{1}{2}\pi r^2 + 2ry .[2]


    Substitute [1] into [2]: . A \;=\;\tfrac{1}{2}\pi r^2 + 2r\bigg[10 - \frac{\pi+2}{2}\,r\bigg]

    . . And we have: . A \;=\;\tfrac{1}{2}\pi r^2 + 20r - (\pi+2)r^2


    Differentiate and equate to zero: . A' \;=\;\pi r + 20 - 2(\pi+2)r \;=\;0

    . . . (\pi+4)r \:=\:20 \quad\Rightarrow\quad\boxed{ r \:=\:\frac{20}{\pi+4}}


    Substitute into [1]: . y \:=\:10 - \left(\frac{\pi+2}{2}\right)\left(\frac{20}{\pi+4}  \right) \quad\Rightarrow\quad\boxed{y \;=\;\frac{20}{\pi+4}}

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