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Math Help - Find length and width

  1. #1
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    Find length and width

    of a rectangle that has the given perimeter and a maximum area.

    Perimeter : 80 meters
    Im still not sure how to set these up to prove with calculus.
    2x+2y=80?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Nightasylum View Post
    of a rectangle that has the given perimeter and a maximum area.

    Perimeter : 80 meters
    Im still not sure how to set these up to prove with calculus.
    Let a rectangle have length l and width w.

    Then by the first condition, we have P=2l+2w=80.

    Let A=lw. It follows from above that l=40-w.

    Substituting this into the area equation, we have A=(40-w)w=40w-w^2.

    Since we want to maximize area, it follows that \frac{\,dA}{\,dw}=40-2w. Now, we find the critical point:

    40-2w=0\implies w=20 (To verify its a max, we differentiate again to get A^{\prime\prime}=-2<0)

    Since w=20, it follows that l=40-(20)=20.

    Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Let a rectangle have length l and width w.

    Then by the first condition, we have P=2l+2w=80.

    Let A=lw. It follows from above that l=40-w.

    Substituting this into the area equation, we have A=(40-w)w=40w-w^2.

    Since we want to maximize area, it follows that \frac{\,dA}{\,dw}=40-2w. Now, we find the critical point:

    40-2w=0\implies w=20 (To verify its a max, we differentiate again to get A^{\prime\prime}=-2<0)

    Since w=20, it follows that l=40-(20)=20.

    Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.

    Does this make sense?
    Yes perfect sense. Thanks alot
    Last edited by Nightasylum; October 25th 2009 at 06:10 PM.
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