document.write( "Question 173895: A rectangle has a perimeter of 140 Square feet. Maximize the area of the rectangle \n" ); document.write( "
Algebra.Com's Answer #128777 by solver91311(24713)\"\" \"About 
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Perimeter: \"P=2L+%2B+2W\". Since the perimeter is 140 feet, \"2L+%2B+2W+=+140\"\r
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\n" ); document.write( "\n" ); document.write( "Solving for L: \"2L+=+140+-+2W\", \"L+=+70+-+W\"\r
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\n" ); document.write( "\n" ); document.write( "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\"\r
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\n" ); document.write( "\n" ); document.write( "Since A(W) is continuous and differentiable the first derivitive set to zero gives the value of the independent variable at a local extrema.\r
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\n" ); document.write( "\n" ); document.write( "\"%28dA%28W%29%29%2FdW=70-2W\", \"70-2W=0\", \"W=35\"\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "\"%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.\r
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\n" ); document.write( "\n" ); document.write( "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.
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