Area optimization problem

  • Thread starter ksle82
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
StatusX
Homework Helper
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I would think only three sides would be fence and the other would be the wall.
 
  • #3
87
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Yea, hes right only 3 sides.
 
  • #4
30
0
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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