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Math Help - Word Problem

  1. #1
    Junior Member
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    Word Problem

    Hi,
    Can someone lend a hand with this? Thanks...

    I want to plant a garden 10 yards long and 10 yards wide. So I bought 40 yards (10 + 10 + 10 + 10 ) of fencing for the perimeter of the my garden. My wife said I need to go more along one direction and smaller on the other (Long and narrower – increase the length and reduce the width, rather than both being 10), still keeping the perimeter same to use all the wire I had brought. For example, she said go 15 yards on the length and only 5 yards wide (Total perimeter is still 15 + 5 + 15 +5 i.e. 40 yard of wire).
    However, with these measurements my garden will be only 15 * 5 or 75 square yards, 25 square yards smaller. My wife doesn’t believe me.

    I want to use algebra to prove it to her that the best shape (the largest garden) I can get, still using all the wire I have (i.e. keeping the perimeter same), is the square garden, 10 * 10 .
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Chester
    Hi,
    Can someone lend a hand with this? Thanks...

    I want to plant a garden 10 yards long and 10 yards wide. So I bought 40 yards (10 + 10 + 10 + 10 ) of fencing for the perimeter of the my garden. My wife said I need to go more along one direction and smaller on the other (Long and narrower – increase the length and reduce the width, rather than both being 10), still keeping the perimeter same to use all the wire I had brought. For example, she said go 15 yards on the length and only 5 yards wide (Total perimeter is still 15 + 5 + 15 +5 i.e. 40 yard of wire).
    However, with these measurements my garden will be only 15 * 5 or 75 square yards, 25 square yards smaller. My wife doesn’t believe me.

    I want to use algebra to prove it to her that the best shape (the largest garden) I can get, still using all the wire I have (i.e. keeping the perimeter same), is the square garden, 10 * 10 .

    Alright, we need to find the greatest amount of area possible with 40 yards of fencing.

    First things first, write out what you know:

    l\times w=A

    2w=40-2l \quad\rightarrow\quad w=\frac{40-2l}{2}=20-l

    combine the equations:
    l\left(20-l\right)=A

    multiply: 20l-l^2=A

    change things around and call the area the function of l: f(l)=-l^2+20l

    now we have a quadratic equation that opens down, so we find the vertex:

    \text{vertex}=\frac{-b}{2a}

    note: -1l^2+20l=al^2+bl

    substitute: \frac{-20}{2(-1)}

    multiply: \frac{-20}{-2}

    divide: \boxed{10}

    therefore the area is largest when l=10

    I would like to know if this is a real life problem (are you actually putting up a garden or is this in a book)
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  3. #3
    Super Member Rebesques's Avatar
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    Well actually, with the given wire, the largest garden you can fence is circular. This is called the Isoperimetric Problem, but I don't think you can convince your wife with this. Just promise to get her something and everything's ok...
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  4. #4
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    Thanks

    You are quick.

    Is there another way to express this? Can you dumb it down a little for me?

    Thanks again
    Chester
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  5. #5
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Chester
    You are quick.

    Is there another way to express this? Can you dumb it down a little for me?

    Thanks again
    Chester
    I can't dumb it down for you, but I can give you a better explanation...

    the area of the garden is of course length times width. l\times w=A

    In rectangles, the perimeter can be found with this equation:

    2l+ 2w=P

    therefore: 2w=P-2l

    then divide both sides by 2: \frac{2w}{2}=\frac{P-2l}{2}

    simplify: \frac{\not{2}w}{\not{2}}=\frac{P}{2}-\frac{2l}{2}

    remember, P=40 in this case, so subtitute: w=\frac{40}{2}-\frac{\not{2}l}{\not{2}}

    simplify: w=20-l

    now we know what the width equals, so we put that into our equation for Area:

    l(w)=A\quad\Longrightarrow\quad l(20-l)=A \quad\Longrightarrow\quad 20l-l^2=A

    now think of it like this, l=x, and, A=y

    y=20x-x^2

    See, it's a graph! now look at that graph (below) and you can see that y is highest when x=10, that means the Area is greatest when the length is 10 yards.
    Attached Thumbnails Attached Thumbnails Word Problem-areaperimeter.jpg  
    Last edited by Quick; July 22nd 2006 at 02:15 PM.
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