document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=869952>Question 869952</A>: <I><SPAN STYLE=\"word-break: break-all;\">  A farmer has 3000 feet of wire to enclose a rectangular field. He plans to fence the entire area and then subdivide it by running a perpendicular fence across the middle. Find the dimensions of the field that would enclose the maximum area . What is the maximum area? <BR>\n" );
document.write( "Can you please help me ? Thanks so much in advance I really appreciate it:) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #524640 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=524640\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  A farmer has 3000 feet of wire to enclose a rectangular field.<BR>\n" );
document.write( " He plans to fence the entire area and then subdivide it by running a perpendicular fence across the middle.<BR>\n" );
document.write( " Find the dimensions of the field that would enclose the maximum area .<BR>\n" );
document.write( " What is the maximum area?<BR>\n" );
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document.write( "With a fence down the middle we have 2 lengths and 3 widths, therefore:<BR>\n" );
document.write( "2L + 3W = 3000<BR>\n" );
document.write( "simplify, divide by 2<BR>\n" );
document.write( "L + 1.5W = 1500<BR>\n" );
document.write( "L = -1.5W + 1500, we can use this form for substitution<BR>\n" );
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document.write( "Area<BR>\n" );
document.write( "A = L * W<BR>\n" );
document.write( "Replace L with (-1.5W+1500)<BR>\n" );
document.write( "A = (-1.5W+1500) * W<BR>\n" );
document.write( "A = -1.5W^2 + 15W <BR>\n" );
document.write( "From this we can write a quadratic equation, y = A, x=W<BR>\n" );
document.write( "y = -1.5W^2 + 1500W<BR>\n" );
document.write( "Max area occurs on the axis of symmetry, x=-b/(2a), so we have<BR>\n" );
document.write( "x = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1500%29%2F%282%2A-1.5%29 ALIGN=MIDDLE  ALT=\"%28-1500%29%2F%282%2A-1.5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "x = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1500%29%2F%28-3%29 ALIGN=MIDDLE  ALT=\"%28-1500%29%2F%28-3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "x = +500 ft is the width for max area<BR>\n" );
document.write( "Find L, replace W with 500<BR>\n" );
document.write( "L = -1.5(500) + 1500<BR>\n" );
document.write( "L = 750 ft is the length for max area<BR>\n" );
document.write( "Find the Area<BR>\n" );
document.write( "A = 750 * 500<BR>\n" );
document.write( "A = 375,000 sq/ft<BR>\n" );
document.write( ":<BR>\n" );
document.write( ":<BR>\n" );
document.write( "You can confirm this your self, find the area using the equation, W=500<BR>\n" );
document.write( "A = -1.5W^2 + 1500W<BR>\n" );
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