document.write( "Question 196947: A retangular field is to be enclosed by 500m of fence. What dimensions will give the maximum area? What is the maximum area? Be sure to use \"let statements\" \n" ); document.write( "
Algebra.Com's Answer #147655 by solver91311(24713)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Let's solve this one in general, that is for any given length of fence.\r
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\n" ); document.write( "\n" ); document.write( "Let F represent the length of fence available.\r
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\n" ); document.write( "\n" ); document.write( "Let w represent the width of the field.\r
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\n" ); document.write( "\n" ); document.write( "Let l represent the length of the field.\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( "So, in order to enclose the field,\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 available fencing. 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( "The area function is continuous and twice differentiable across 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.\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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