document.write( "Question 328142: A carpenter is building a rectangular room with a fixed perimeter of 332 ft. What dimensions would yield the maximum area? What is the maximum area? The length that would yield the maximum area is how many feet? \n" ); document.write( "
Algebra.Com's Answer #235019 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let's solve this one in general, that is for any given perimeter.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let w represent the width of the field.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let l represent the length of the field.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The perimeter of a rectangle is:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\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
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Algebra Solution:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The area function is a parabola, opening downward, with vertex at:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\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
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Calculus Solution:\r
\n" ); document.write( "
\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
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Therefore the maximum area is obtained when\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "And the shape is therefore a square.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "And that maximum area is:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "John
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" );