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Math Help - Finding the function

  1. #1
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    Finding the function

    I started but got stuck...

    A Norman Window has the shape of a rectangle surmounted by a semicircle. It has a perimeter of 40 ft.
    A) find the function that models the area of the window
    I ended up with this function, but I'm not sure it's right.
    x(-3╥ -4)
    ----------- +20x
    8

    B) find the dimensions, to two decimal places, that admits the greatest amount of light.
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  2. #2
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    Re: Finding the function

    Hello, karlyjo!

    Did you make a sketch?


    A Norman window has the shape of a rectangle surmounted by a semicircle.
    It has a perimeter of 40 ft.

    A) Find the function that models the area of the window

    Code:
                  * * *
              *           *
            *               *
           *                 *
     
          *                   *
          * - - - - * - - - - *
          |    r    :    r    |
          |                   |
        h |                   | h
          |                   |
          |                   |
          *-------------------*
          : - - - - 2r  - - - :
    The radius of the semicircle is r.
    The height of the rectangle is h.
    The width of the rectangle is 2r.

    The circumference of the circle is 2\pi r.
    The perimeter of the semicircle is \pi r.
    The perimeter of the rectangle is 2r + 2h.

    The total perimeter is 40 ft.
    . . \pi r + 2r + 2h \:=\:40 \quad\Rightarrow\quad h \:=\:\frac{40 - \pi r - 2r}{2} .[1]

    The area of the semicircle is: \tfrac{1}{2}\pi r^2
    The area of the rectangle is: (2r)(h) \:=\:2rh

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

    Substitute [1] into [2]: . A \;=\;\tfrac{1}{2}\pi r^2 + 2r\left(\frac{40 - \pi r - 2r}{2}\right)

    And we have: . A \;=\;40r - \tfrac{1}{2}(\pi + 4)r^2




    B) Find the dimensions, to two decimal places, that admits the greatest amount of light.

    We want, of course, maximum area of the window.

    We have: . A \;=\;40r - \tfrac{1}{2}(\pi + 4)r^2

    Then: . A' \:=\:40 - (\pi + 4)r \;=\;0 \quad\Rightarro\quad r \:=\:\frac{40}{\pi+4} \:=\:5.600991535

    Substitute into [1]: . h \;=\;\frac{40 - \pi\left(\frac{40}{\pi+4}\right) - 2\left(\frac{40}{\pi+4}\right)}{2} \;=\;\frac{40}{\pi + 4}

    Therefore: . \begin{Bmatrix} r &\approx & 5.60\text{ ft} \\ \\[-3mm] h &\approx & 5.60\text{ ft} \end{Bmatrix}
    Thanks from karlyjo
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