The maximum area that can be enclosed is 800 square feet.

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    Area Optimization
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Homework Help Overview

The problem involves maximizing the area that can be enclosed with a given length of fencing, specifically 80 feet, along a wall. The subject area relates to optimization in geometry.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the problem, particularly the interpretation of the fencing arrangement with respect to the wall. There are attempts to derive the area formula based on different configurations of length and width.

Discussion Status

Some participants have provided revised calculations based on the clarification that only three sides require fencing. There is ongoing exploration of the implications of this setup on the area calculations, but no consensus has been reached on the final answers.

Contextual Notes

Participants are working under the assumption that one side is against a wall, which alters the typical perimeter considerations. The original poster's initial interpretation led to a misunderstanding of the fencing requirements.

ksle82
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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.
 
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I would think only three sides would be fence and the other would be the wall.
 
Yea, he's right only 3 sides.
 
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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