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Math Help - Problem solving

  1. #1
    Member Mr Rayon's Avatar
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    Smile Problem solving

    A farmer wishes to make a rectangular pen using an existing section of straight fence and 36 m of relocatable fencing materials. What is the largest possible area of the pen?
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  2. #2
    Senior Member topher0805's Avatar
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    Is the existing section however long you need it to be?

    Any rectangle must have a length L and a width W. For the perimeter to be 32+x, 2L + 2W must be equal to 32+x.

    We know that by the definition of a rectangle, the section of pre-existing fence will be equal in length to either the length or the width. So, we set x equal to L.

    This gives us 2L + 2W = 36+L, or L + 2W = 36.

    Also, we know that the maximum area of any rectangle is actually a rectangle with four equal sides. This is also known as a square.

    With that knowledge, when looking for the maximum area we can say that L = W.

    If L = W, and L + 2W = 36, then W + 2W = 36.

    After that, it's just basic algebra.

    3W = 36
    W = 36/3

    We know the width, and since all the sides are equal, we can square it to find the area.

    (36/3)^2 = 144 m^2
    Last edited by topher0805; February 7th 2008 at 12:37 PM.
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Mr Rayon View Post
    A farmer wishes to make a rectangular pen using an existing section of straight fence and 36 m of relocatable fencing materials. What is the largest possible area of the pen?
    Perimeter must equal 36m: 36=2(w+l)

    We want the largest area: A=wl

    Solve the first equal for either length or width and plug it into the second equation. Calculus will give you the value for the largest area, but without knowledge of that math, you will need to use some intuition/trial and error...

    18=w+l

    A=(18-w)w
    A=18w-w^2

    w=9; l=9
    A=81
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  4. #4
    Senior Member topher0805's Avatar
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    Quote Originally Posted by colby2152 View Post
    Perimeter must equal 36m: 36=2(w+l)

    We want the largest area: A=wl

    Solve the first equal for either length or width and plug it into the second equation. Calculus will give you the value for the largest area, but without knowledge of that math, you will need to use some intuition/trial and error...

    18=w+l

    A=(18-w)w
    A=18w-w^2

    w=9; l=9
    A=81
    I am sorry but with all due respect, that is just not right. Your response ignores the fact that there is another section of straight fence.
    Last edited by topher0805; February 11th 2008 at 12:01 AM.
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