SOLUTION: A rectangle has a perimeter of 140 Square feet. Maximize the area of the rectangle

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Question 173895: A rectangle has a perimeter of 140 Square feet. Maximize the area of the rectangle
Answer by solver91311(24713) About Me  (Show Source):
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Perimeter: P=2L+%2B+2W. Since the perimeter is 140 feet, 2L+%2B+2W+=+140

Solving for L: 2L+=+140+-+2W, L+=+70+-+W

Area: A%28LW%29+=+LW, but we know from the perimeter formula that L=70-W so A%28W%29=%2870-W%29W=70W-W%5E2

Since A(W) is continuous and differentiable the first derivitive set to zero gives the value of the independent variable at a local extrema.

%28dA%28W%29%29%2FdW=70-2W, 70-2W=0, W=35

Since the first derivitive is also differentiable, the sign on the second derivitive at the extreme point will characterize the extreme as a maximum or minimum.

%28d%5E2A%28W%29%29%2FdW%5E2=-2 for all W in the domain of A%28W%29, so the extreme point is a maximum.

Therefore, a rectangle with perimeter 140 has a maximum area when the width is 35, which means that the length is also 35 and the rectangle is actually a square.