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Area optimization problem

  1. Jun 5, 2006 #1
    Given 80 feet of fencing, what is the maximum area that you can enclose along a wall?

    Solution:
    L=lenght, W=width, A=area

    2L+2W=80 (perimeter) ==> L+W=40 ==> L=40-W

    LW=A ==> (40-W)(W)= -W^2+ 40W= A

    (dA/dW)=0=-2W+40=0 ==> W=20

    W=20, L=20--------------------------------------------------ANSWERS

    Is this right? any input would be appreciated.
     
  2. jcsd
  3. Jun 5, 2006 #2

    StatusX

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    I would think only three sides would be fence and the other would be the wall.
     
  4. Jun 5, 2006 #3
    Yea, hes right only 3 sides.
     
  5. Jun 5, 2006 #4
    Doh!!!
    here's the updated solutions:

    L+2W=80 ==> L=80-2W

    A=LW= (80-2W)(W)= 80W-2W^2

    (dA/dW)=-4W+80=0 ==> W=20, L=40 ---------------------Answers
     
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