document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=58256>Question 58256</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A farmer has 96 meters of fencing. He wants to enclose a rectangular field and build a fence across the middle, parallel to the shorter ends. \r<BR>\n" );
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document.write( "a. Express the total area of the field as a function of the width w, where w is the length of the shorter ends (hint: result is a quadratic)\r<BR>\n" );
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document.write( "b. Algebraically determine the value of x for which the area is maximum. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #39932 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=39932\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A farmer has 96 meters of fencing. He wants to enclose a rectangular field and build a fence across the middle, parallel to the shorter ends.<BR>\n" );
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document.write( "a. Express the total area of the field as a function of the width w, where w is the length of the shorter ends (hint: result is a quadratic)<BR>\n" );
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document.write( "Because of that fence down the middle, the perimeter is: <BR>\n" );
document.write( "2L + 3W = 96<BR>\n" );
document.write( "2L = 96 - 3W<BR>\n" );
document.write( "L = (48 - 1.5W); divided eq by 2. <BR>\n" );
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document.write( "Area = L*W<BR>\n" );
document.write( "Let y = area<BR>\n" );
document.write( "Substitute (48-1.5W) for L<BR>\n" );
document.write( "y = (48 - 1.5W) * W<BR>\n" );
document.write( "y = 48W - 1.5W^2<BR>\n" );
document.write( "Arrange as a Quadraic eq: y = -1.5W^2 + 48W<BR>\n" );
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document.write( "b. Algebraically determine the value of x for which the area is maximum.<BR>\n" );
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document.write( "Use the vertex equation: x = -b/(2a); in our eq, a=-1.5, b= 48<BR>\n" );
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document.write( "x = -48/(2*-1.5)<BR>\n" );
document.write( "x = -48/-3<BR>\n" );
document.write( "x = +16 meters is the width for max area<BR>\n" );
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document.write( "Find the max area, substitute 16 for x in the equation: y = -1.5x^2 + 48x<BR>\n" );
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document.write( "y = -1.5(16^2) + 48(16)<BR>\n" );
document.write( "y = -1.5(256) + 768<BR>\n" );
document.write( "y = -384 + 768<BR>\n" );
document.write( "y = +384 sq meters is max area and occurs when the width is 16 meters.<BR>\n" );
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