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Math Help - Optimization problem

  1. #1
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    Optimization problem

    This problem is giving me some trouble. It's irritating because it seems so easy.

    The question is

    "The figures are made of rectangles and semicircles.
    a.) Find a formula for the area
    b.) Find a formula for the perimeter
    c.) Find the dimensions x and y that maximize the area given that the perimeter is 100"

    The figure is a rectangle with sides x and y with a half-circle connected to one of the ends (diameter of the circle is equal to y).

    I got A=xy + 1/2*(1/2y)^2 for the area and
    P=2x + y + (1/2)(y)(pi) for the perimeter.

    The book gives the answer for c.) as x=100/(4+pi) and y=200/(4+pi)

    No matter what I try I can't seem to get their answer. I know I have find the value of x (or y) in terms of the other variable using 100 but I keep getting y=63.661977 and x=31.83098869. I'm stumped.
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  2. #2
    Junior Member
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    Re: Optimization problem

    The area of a (full) circle is \pi r^2 (with r being circle radius). I think there's a factor \pi missing in your formula for the area.
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  3. #3
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    Re: Optimization problem

    Hello, Kjcube!


    Code:
                  * *  *  *  *  *  *  *
              *     |                 *
            *       |                 *
           *        | y/2             *
                    |                 *
          *         |                 *
          *         *                 * y
          *         |                 *
                    |                 *
           *        | y/2             *
            *       |                 *
              *     |                 *
                  * *  *  *  *  *  *  *
                    : - - -  x  - - - :
    (a) Find a formula for the area.

    (b) Find a formula for the perimeter.

    (c) Find the dimensions x and y that maximize the area
    . . given that the perimeter is 100 inches.

    The book gives the answers for (c) as: . x\:=\:\frac{100}{4+\pi}\,\text{ and }\,y\:=\:\frac{200}{4+\pi}
    The book's answer are correct.


    (a) Area = (area of rectangle) + (area of semicircle)

    . . A \;=\;xy + \tfrac{1}{2}\pi\left(\tfrac{y}{2}\right)^2 \;=\; xy + \tfrac{\pi}{8}y^2 .[1]


    (b) Perimeter = (perimeter of rectangle) + (perimeter of semicircle)

    . . P \;=\;2xy + \tfrac{1}{2}(\pi y)


    (c) The perimeter is 100 inches: . 2xy + \tfrac{\pi}{2}y \:=\:100

    . . Solve for x\!:\;\;x \:=\:\frac{200 - (2+\pi)y}{4} .[2]


    Substitute into [1]: . A \;=\;\left(\frac{200-(2+\pi)y}{4}\right)y + \frac{\pi}{8}y^2

    . . which simplifies to: . A \;=\;50y - \frac{4+\pi}{8}y^2

    Then: . A' \;=\;50 - \frac{4+\pi}{4}y \;=\;0 \quad\Rightarrow\quad \boxed{y \;=\;\frac{200}{4+\pi}}


    Substitute into [2]: . x \;=\;\frac{1}{4}\left[200 - (2+\pi)\!\cdot\!\frac{200}{4+\pi}\right]

    . . which simplifies to: . \boxed{x \;=\;\frac{100}{4+\pi}}

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