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Math Help - Maximum Area of a Rectangle

  1. #1
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    Maximum Area of a Rectangle

    Please help me solve this problem:

    Question: One-hundred (100) feet of fence is used to fence in a rectangular yard. Find the length and width that give maximum area.

    Thanks in advance for your assistance.
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  2. #2
    Super Member wingless's Avatar
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    Code:
      _______x_______
     |               |
     |               |
    y|               |y
     |               |
     |_______________|
             x
    2x + 2y = 100
    x + y = 50

    Formula for the area is, A = xy.
    As x +y = 50, y = 50 -x
    So our formula becomes, A = x (50-x) = 50x - x^2.

    To find the extremum point, we will differentiate it and A' = 0 will give us the extremum point.

    A = 50x - x^2
    A' = 50 - 2x
    A' =  50 - 2x = 0
    x = 25

    There's an extremum point at x = 25, but we don't know whether it's a minima or a maxima. The second derivative test will help us to determine what it is

    If A''(25) > 0 , then it's a minima. If A''(25) < 0, it's a maxima.

    A = 50 x - x^2
    A' = 50 - 2x
    A'' = -2
    A''(25) = -2, which is less than 0 and which makes our point a maxima

    So x = 25 will give us the maximum area.
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  3. #3
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    x+y=50
    so
    50=x+y>=2*sqrt(x*y) (when x=y then x+y=sqrt(x*y))
    so we have x*y<=25^2=625 (when x=y=25)


    Calculus is unnecessary
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by invite_moon View Post
    x+y=50
    so
    50=x+y>=2*sqrt(x*y) (when x=y then x+y=sqrt(x*y))
    so we have x*y<=25^2=625 (when x=y=25)
    Your solution is OK, but can you apply it to more complicated functions?

    Quote Originally Posted by invite_moon View Post
    Calculus is unnecessary
    This question is posted under the Calculus forum, so I treated it like any optimization question.
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