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Math Help - Please Check My Answer

  1. #1
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    Please Check My Answer

    ______y__________y______
    l------------l--------------l
    l------------l--------------l
    l------------l--------------l
    l--x---------l--x-----------l x
    l------------l--------------l
    l------------l--------------l
    l___________l____________l

    A person has 400 ft. of fencing to maximize the area. What are the dimensions?

    400=4y + 3x
    A=2xy

    57.143=y+x
    y=57.143-x

    A=2(57.143-x)(x)
    A=114.286x-2x^2
    A'=114.286-4x=0
    114.286=4x
    x=28.572

    400=4y+3(28.572)
    400=4y+85.715
    315.286=4y
    y=78.571

    x=28.572 ft.
    y=78.571 ft.

    Does this look right?
    Last edited by Vigo; December 20th 2005 at 01:32 PM.
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  2. #2
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    Ah! an optimization problem, I love those:
    Given:
    400=4y+3x--------Probably x,y>0 are the other conditions
    Maximize:
    2xy

    Since
    400=4y+3x
    we have by solving for y:
    100-.75x=y
    Thus, 2xy becomes:
    2x(100-.75x) same as 200x-1.5x^2
    Now maximize that function by taking the derivative:
    200-3x=0
    Thus,
    x=200/3
    Then,
    100-(3/4)(200/3)=y
    Thus,
    100-50=y
    Thus, the solution is
    x=200/3 and y=50
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  3. #3
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    Thank you for correcting me. I made my error when solving for y. Tried to simplify too much. Thanks a lot.
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  4. #4
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    I've seen this thread on physicsforums.com. I think many of the users here are users there.
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  5. #5
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    Yeah I posted this on that forum too because sometimes I don't get an answer if I only post it in one forum.
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