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Math Help - Calculus is killing me!

  1. #1
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    Calculus is killing me!

    Three sides of a rectangular playing field are to be fenced with 400 m of fencing. Find the dimensions so that the area of the field will be a maximum.

    Can any one solve this question?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    OK here you go

    Quote Originally Posted by Raina View Post
    Three sides of a rectangular playing field are to be fenced with 400 m of fencing. Find the dimensions so that the area of the field will be a maximum.

    Can any one solve this question?
    You have this P_{field}=400=x+2y only one y becuase one side is missing...and Area_{field}=xy...using substitution you have tha A=y\cdot(400-2y)...diffentiating we get A'=400-4y... y=50...now we know this is a max since A''<0,for every y since it is,A''-4...so we know that y=50 is a max...so now put that back into your permiter equation and get x
    Last edited by Jameson; April 13th 2008 at 09:59 PM. Reason: fixed Latex
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  3. #3
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    i noticed sth
    when u slove 400-4y=0
    u get 100
    NOT 50...

    any way thanks for ur time and help
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  4. #4
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    Hello,

    Let x and y be the sides (don't know how to name it ) such as the perimeter P is :

    P=x+2y=400 (one side is missing) -> x=400-2y

    Plus, we want the area be at its maximum. The area A is :

    A=xy

    -> A=(400-2y)y=-2y+400y

    If you derivate : A'=-4y+400 and if y<100, A' will be positive, if y>100, A' will be negative. So it firstly increases and then decreases.
    y=100 is a maximum, and so x=400-200=200
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