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Math Help - help in building fence problem

  1. #1
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    help in building fence problem

    There is a wall in your backyard. It is so long that you can’t see its endpoints. You want to build a fence of length L such that the area enclosed between the wall and the fence is maximized. The fence can be of arbitrary shape, but only its two endpoints may touch the wall.
    given : the length of the fence.
    the wanted solution : the largest area. Your answer should be rounded to 2 digits after the decimal point.

    eg :
    1 -> 0.16

    hints : take pi as 3.14159265358.

    i want help to find the formula to answer this
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  2. #2
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    The largest area you can enclose by any length of fencing is a square.

    So if you have a length \displaystyle L\,\textrm{units} of fencing, each side of the square will have length \displaystyle \frac{L}{4}\,\textrm{units}, so the area of the square is \displaystyle \frac{L^2}{16}\,\textrm{units}^2.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    The largest area you can enclose by any length of fencing is a square.

    So if you have a length \displaystyle L\,\textrm{units} of fencing, each side of the square will have length \displaystyle \frac{L}{4}\,\textrm{units}, so the area of the square is \displaystyle \frac{L^2}{16}\,\textrm{units}^2.
    it gave me wrong aswer by this way
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  4. #4
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    Hello, mido22!

    Pay no attention to the man behind the curtain . . .


    There is a wall in your backyard. It is so long that you canít see its endpoints.
    You want to build a fence of length L such that the area enclosed between
    the wall and the fence is maximized. The fence can be of arbitrary shape,
    but only its two endpoints may touch the wall.

    A semicircular region has maximum area.

    Code:
               r         r
      ----*---------*---------*----
          *                   *
    
           *                 *
            *               *
              *           *
                  * * *
                    L

    We have: . \pi r \,=\,L \quad\Rightarrow\quad r \:=\:\frac{L}{\pi}

    The area is: . A \:=\:\frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(\frac{L}{\pi}\right)^2 \:=\: \dfrac{L^2}{2\pi}

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, mido22!

    Pay no attention to the man behind the curtain . . .



    A semicircular region has maximum area.

    Code:
               r         r
      ----*---------*---------*----
          *                   *
    
           *                 *
            *               *
              *           *
                  * * *
                    L

    We have: . \pi r \,=\,L \quad\Rightarrow\quad r \:=\:\frac{L}{\pi}

    The area is: . A \:=\:\frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(\frac{L}{\pi}\right)^2 \:=\: \dfrac{L^2}{2\pi}

    hmmm..... thx very much u r right
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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, mido22!

    Pay no attention to the man behind the curtain . . .



    A semicircular region has maximum area.

    Code:
               r         r
      ----*---------*---------*----
          *                   *
    
           *                 *
            *               *
              *           *
                  * * *
                    L

    We have: . \pi r \,=\,L \quad\Rightarrow\quad r \:=\:\frac{L}{\pi}

    The area is: . A \:=\:\frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(\frac{L}{\pi}\right)^2 \:=\: \dfrac{L^2}{2\pi}

    I didn't read that you need to use an existing wall...
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