SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 312 feet, what dimension would yield the maximum area? What is the maximum area?

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Question 334094: A carpenter is building a rectangular room with a fixed perimeter of 312 feet, what dimension would yield the maximum area? What is the maximum area?
Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
As a general rule, the maximum area of a rectangle will be the special case when it is a square.
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Let L=length
Let W=width
Let P=perimeter of rectangle
Let A=area of rectangle
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we know that P=2*L+2*W
and that A=L*W
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Given: P=312, find L and W such that A is maximum
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Area: A=L*W
Perimeter: 312=2*L+2*W
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312=2*L+2*W
156=L+W (divide both sides by 2 to simplify)
L=156-W (solve for L)
A=L*W=(156-W)*W (substitute for L)
A=156%2AW-W%5E2 (distribute W)
A=-1%2A%28W%5E2-156%2AW%29+ (rearrange and factor out -1)
A=-%28w%5E2-156%2AW%2B6084-6084%29 (complete the square by adding and subtracting %28156%2F2%29%5E2=78%5E2=6084)
A=-%28W-78%29%5E2-6084
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This is a parabola with vertex at (78,-6084)
Since the leading coefficient of the square term is negative, the curve turns downward, thus this is a maximum.
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Therefore W=78 is the maximum. And since L=156-W=156-78=78
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Answer: the maximum area is when L=78 and W=78 which is a square.