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Thread: Maximum area problem

  1. #1
    Junior Member
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    Jan 2008
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    Maximum area problem

    An indoor physical fitness room consists of a rectangular region with a semicircle on each end. The perimeter of the room is to be a 200-meter running track. Find the dimensions that will make the area of the rectangular region as large as possible.

    I know that I need the area of a rectangle (bh) and am guessing also the radius of a circle (because of the semicircles on ends) but have no idea how to come up with a primary equation for this problem because I have no similar examples in my book. I'm assuming the perimeter is circular because that would give the greatest area so would my primary equation, then, relate to the area of a circle? Thank you!

  2. #2
    Elite Member
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    Re: Maximum area problem

    The perimeter is given as 200 meters. But, the perimeter can be expressed as the length around the track.

    Which is [tex]2y+2{\pi}r=200[/tex]....[1]

    The area can be expressed as [tex]A={\pi}r^{2}+2ry[/tex]........[2]

    Solve [1] for y and sub into [2]:

    [tex]y=100-{\pi}r[/tex]

    [tex]A={\pi}r^{2}+2r(100-{\pi}r)=200r-{\pi}r^{2}[/tex]

    This is what must be maximized. Find dA/dr and the rest will follow.

  3. #3
    Junior Member
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    Re: Maximum area problem

    Well, some of it followed! I took the derivative of 200r - (pi)r^2, set = 0 and obtained a radius of 31.83, which is a relative maximum. So I'm now trying to figure out how I'm to use this information to get the sides of the rectangle. If I plug the radius into (pi)r^2 + 2ry, I get 3182.9 for the area of the circle, but I'm not getting how I can solve this problem without using the formula for the area of a rectangle or, if I am, how to do this. Thanks for your patience!

  4. #4
    Elite Member
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    Re: Maximum area problem

    Yes, the radius is [tex]\frac{100}{\pi}\approx{31.83}[/tex]

    Just plug that into the equation for area: [tex]\;\ A={\pi}r^{2}+2r(100-{\pi}r)=\frac{10,000}{\pi}\approx{3183.1} \;\ m^{2}[/tex]

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