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Please Check My Answer

  1. Dec 20, 2005 #1
    ______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?
     
  2. jcsd
  3. Dec 20, 2005 #2

    benorin

    User Avatar
    Homework Helper

    [tex]400=4y + 3x \Rightarrow y=\frac{400-3x}{4}[/tex]

    substitute that into the area formula to get

    [tex]A(x)=2xy= 200x-\frac{3}{2}x^2[/tex]

    hence

    [tex]A^{\prime}(x)=\frac{d}{dx}\left( 200x-\frac{3}{2}x^2\right) = 200-3x=0\Rightarrow x= \frac{200}{3}\approx 66.67[/tex]

    and recall that [itex]y=\frac{400-3x}{4}[/itex], so

    [tex]y=\frac{400-3\frac{200}{3}}{4}= 50[/tex]

    so the dimensions are (roughly) 66.67 by 50.
     
  4. Dec 20, 2005 #3
    Thank you very much for correcting me. I guess my error was solving for y. I don't know how I made that mistake but thanks a lot.
     
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