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Math Help - Showing that a square has largest area

  1. #1
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    Showing that a square has largest area

    This is the question:

    "Let S be the set of all rectangles with perimeter P. Show that the square is the element of S with the largest area"

    I'm having trouble proving this. Any assistance would be greatly appreciated.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hint:

    Try to find a function which is describes an area of rectangles with perimeter P.

    remember that:

    P=2x+2y (when a & y are sides or some rectangle)
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  3. #3
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    I did that, but I don't know what to do with it.

    I wrote down A = x^2 (for a square) and A = xy (for any rectangle). But I don't know how to prove the square is bigger.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    let A=xy=x(p/2 -x)=f(x)

    f'(x)=p/2-2x=0

    p=4x

    ==> 2x=2y ==>x=y


    or something like that...
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  5. #5
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    Put y in terms of x then find A'=0
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  6. #6
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    I don't think I'm supposed to use calculus. This is pre-calculus material.
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  7. #7
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    Quote Originally Posted by Glitch View Post
    I don't think I'm supposed to use calculus. This is pre-calculus material.
    Well it should be posted in pre-calculus sub-forum not in the university level sub-forum.

    If you need to solve it w/out calculus take a number to fix P i.e 10 then with trial and error for x and y you will find a maximum area when x=y
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  8. #8
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    Yeah, I don't know where to post things any more. My last question was from the same problem set, and I was told it should be put here.
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Glitch View Post
    I don't think I'm supposed to use calculus. This is pre-calculus material.
    ok....

    f(x)=x(p/2 -x)=-x^2+xp/2 this is sad parabola, when she reach the maximum?
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