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Math Help - Area of the Pins...

  1. #1
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    Area of the Pins...

    #8.
    Attached Thumbnails Attached Thumbnails Area of the Pins...-8.jpg  
    Last edited by qbkr21; April 28th 2007 at 09:44 AM.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    I'm sorry to say, but that does not seem correct.

    If there are three sections then we have a total of 4 x's.

    Code:
       _______________________________________________
      |               |               |               |
      |               |               |               |
      |               |               |               | x
      |               |               |               |
      |               |               |               | 
                             y
    So:
    P = 100 = 4x + 2y --> y = 100/2 - 4x/2 = 50 - 2x
    A = x*y = x(50 - 2x) = 50x - 2x^2

    A' = 50 - 4x
    A' = 0 = 50 - 4x --> x = 50/4 = 12.5

    P = 100 = 4(12.5) + 2y --> y = (100 - 50)/2 = 25
    A = (12.5)(25) = 312.5
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  3. #3
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    Re:

    Thanks EcMathgeek for catching the error. I was right, I just messed up on the wording when I was entering it into the computer ( I was up to late). Here is what I meant to say...
    Attached Thumbnails Attached Thumbnails Area of the Pins...-rectangle.jpg  
    Last edited by qbkr21; April 28th 2007 at 11:32 AM.
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