document.write( "Question 454480: I really appreciate the help i'm getting from this website! Question: A rectangle has perimeter P. Find the maximum possible area of the rectangle. Thank you so so so so so so much for your time and effort! \n" ); document.write( "
Algebra.Com's Answer #312069 by solver91311(24713)\"\" \"About 
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\n" ); document.write( "Let P represent the fixed perimeter.\r
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\n" ); document.write( "\n" ); document.write( "Let w represent the width of the rectangle.\r
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\n" ); document.write( "\n" ); document.write( "Let l represent the length of the rectangle.\r
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\n" ); document.write( "\n" ); document.write( "The perimeter of a rectangle is:\r
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\n" ); document.write( "\n" ); document.write( "The area of a rectangle is the length times the width so a function for the area in terms of the width is:\r
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\n" ); document.write( "\n" ); document.write( "Algebra Solution:\r
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\n" ); document.write( "\n" ); document.write( "The area function is a parabola, opening downward, with vertex at:\r
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\n" ); document.write( "\n" ); document.write( "Since the parabola opens downward, the vertex represents a maximum value of the area function. The value of the width that gives this maximum value is one-fourth of the given perimeter. Therefore, the shape must be a square, and the area is the width squared.\r
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\n" ); document.write( "\n" ); document.write( "Calculus Solution:\r
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\n" ); document.write( "\n" ); document.write( "The area function is continuous and twice differentiable over its domain, therefore there will be a local extrema wherever the first derivative is equal to zero and that extreme point will be a maximum if the second derivative is negative at that point.\r
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\n" ); document.write( "\n" ); document.write( "Therefore the maximum area is obtained when\r
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\n" ); document.write( "\n" ); document.write( "And that maximum area is:\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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