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Math Help - Tricky Applied Extrema Problem

  1. #1
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    Question Tricky Applied Extrema Problem

    A company needs to erect a fence around a rectangular storage yard next to a 250 m long side of a warehouse. They have 1000 m of available fencing. They will not fence in the side along the warehouse but the fence can extend out from that side. What is the largest area that can be enclosed?

    I know that I must find the derivative of the Area with respect to either x or y. How would I approach this problem?
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  2. #2
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    Quote Originally Posted by castle View Post
    A company needs to erect a fence around a rectangular storage yard next to a 250 m long side of a warehouse. They have 1000 m of available fencing. They will not fence in the side along the warehouse but the fence can extend out from that side. What is the largest area that can be enclosed?

    I know that I must find the derivative of the Area with respect to either x or y. How would I approach this problem?
    Let x and y be the sides of the fence, with the side y parallel to the warehouse if you like.

    Then xy=area of yard

    \frac{d}{dx}\left(xy\right)=0

    2x+y=1000

    y=1000-2x

    xy=x(1000-2x)=1000x-2x^2

    Differentiate this and equate to zero to find x corresponding to max area.
    Use this x to find y

    If the fence has one side equal to the warehouse side,
    then it's 3 sides sum to 1000, while one side is 250.
    Hence the other 2 sides sum to 750, so they are 375.
    That is the only rectangular shape possible if one side is 250.

    The above solution, using differentiation gives an area of 125,000 square units,
    which is a greater area than if the fence encloses a square by touching the sides of the warehouse,
    however the fence does not form a closed shape!
    Last edited by Archie Meade; July 14th 2010 at 03:41 AM.
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  3. #3
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    Thanks, have tried that and end up with x = 250 and y = 500. But that is not the correct answer; the dimensions are supposed to be 312.5 x 312.5.

    How did they get this solution?
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  4. #4
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    The maximum area in cases like this is always a square.
    A drawing would have been helpful.
    When it states that the fence can extend out from the side,
    it apparently means that the fence can be longer than the warehouse against the side of the warehouse itself,
    in other words, the fence is square-shaped with sides longer than the warehouse side,
    with the fencing starting at the warehouse edges.

    In this case then, the side of the fence against the warehouse has lengths sticking out equal to x.

    Hence the side of the fence parallel to the warehouse has length 250+2x
    and the other side is y.

    250+4x+2y=1000 since no fence is against the warehouse side.

    4x+2y=750

    2x+y=375

    y=375-2x

    Area=(375-2x)(250+2x)=(375)(250)+750x-500x-4x^2

    Differentiate this with respect to x and equate to zero to get x corresponding to max area

    250-8x=0

    8x=250

    x=31.25

    y=375-62.5=312.5

    250+2x=312.5
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  5. #5
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    OK, thanks. When I tried this approach I just forgot that x is really 250 + x, that caused me errors. Thanks.
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