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Thread: 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 $\displaystyle r.$
    The height of the rectangle is $\displaystyle h.$
    The width of the rectangle is $\displaystyle 2r.$

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

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

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

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

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

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




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

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

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

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