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Math Help - Find the dimensions of the field that maximizes the total area

  1. #1
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    Find the dimensions of the field that maximizes the total area

    A rectangular field is to be subdivided in 2 equal fields. There is 8350 feet of fencing available.
    Find the dimensions of the field that maximizes the total area. (List the longer side first)
    Width =_____________ feet
    Length =________________ feet
    What is the maximum area ? Area =_________________
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  2. #2
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    Re: Find the dimensions of the field that maximizes the total area

    width = w, length = l, P = 8350 (perimeter), A = total area. Units are everywhere feet.

    P = 2w + 3l (That's 2w + 2l to bound the area, and another fence of length l down the middle to split the fields.)

    Thus l = (P-2w)/3.

    Note w>0, l>0. Thus also (P-2w)/3 >0, so w < P/2.

    A = lw = w(P-2w)/3.

    Translation: Maximize A(w) = w(P-2w)/3 where 0<w<P/2, and where P is constant.
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  3. #3
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    Re: Find the dimensions of the field that maximizes the total area

    We have P = 2w + 3l = 8350, and A = wl.

    By AM-GM inequality, \frac{2w + 3l}{2} \ge \sqrt{6wl} \Rightarrow \sqrt{6wl} \le \frac{8350}{2}

    The equality case occurs when 2w = 3l.
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  4. #4
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    Re: Find the dimensions of the field that maximizes the total area

    Hello, spyder12!

    Did you make a sketch?


    A rectangular field is to be subdivided into two equal fields.
    . .
    Actually, the two subfields do not have to be equal!
    There is 8350 feet of fencing available.

    Find the dimensions of the field that maximizes the total area.
    What is the maximum area?

    Code:
          : - - - x - - - :
          *-----*---------*
          |     |         |
         y|     |y        |y
          |     |         |
          *-----*---------*
          : - - - x - - - :

    The perimeter is: . 2x + 3y \:=\:8350 \quad\Rightarrow\quad y \:=\:\tfrac{8350-2x}{3}

    The area is: . A \;=\;xy \;=\;x\left(\tfrac{8350-2x}{3}\right)

    . . . . . . . . . . A \;=\;\tfrac{8350}{3}x - \tfrac{2}{3}x^2

    Maximize: . . A' \;=\;\tfrac{8350}{3} - \tfrac{4}{3}x \;=\;0 \quad\Rightarrow\quad \tfrac{4}{3}x \:=\:\tfrac{8350}{3}

    . . . . . . . . . . x \:=\:\frac{4175}{2} \:=\:2087\tfrac{1}{2}\text{ ft}

    . . . . . . . . . . y \:=\:\frac{4175}{3} \:=\:1391\tfrac{2}{3}\text{ ft}


    \text{Area } \;=\;xy \;=\;\left(\frac{4175}{2}\right)\left(\frac{4175}{  3}\right) \;=\;\frac{17,\!430,\!625}{6} \;=\;2,\!905,\!104\tfrac{1}{6}\text{ ft}^2
    Thanks from spyder12
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