document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=852390>Question 852390</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, could you please help me solve this question. A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its side. The 500m of fence used to enclose a maximum area. What are the dimensions of the fence? (leave answers in fraction form) Thank you!\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #513535 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=513535\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its side.<BR>\n" );
document.write( " The 500m of fence used to enclose a maximum area.<BR>\n" );
document.write( " What are the dimensions of the fence? (leave answers in fraction form)<BR>\n" );
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document.write( "Because of the divided area, the perimeter will be:<BR>\n" );
document.write( "2L + 3W = 500<BR>\n" );
document.write( "therefore<BR>\n" );
document.write( "2L = -3W + 500<BR>\n" );
document.write( "L = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3%2F2 ALIGN=MIDDLE  ALT=\"-3%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >W + <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=500%2F2 ALIGN=MIDDLE  ALT=\"500%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "L = -1.5W + 250<BR>\n" );
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document.write( "Area = L*W<BR>\n" );
document.write( "Replace L with (-1.5W+250)<BR>\n" );
document.write( "A = (-1.5W+250)*W<BR>\n" );
document.write( "A = -1.5W^2 + 250W<BR>\n" );
document.write( "A quadratic equation, the max area occurs on the axis of symmetry x=-b/(2a)<BR>\n" );
document.write( "In this equation, x=w, a=-1.5, b=250<BR>\n" );
document.write( "w = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-250%29%2F%282%2A-1.5%29 ALIGN=MIDDLE  ALT=\"%28-250%29%2F%282%2A-1.5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "W = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-250%29%2F%28-3%29 ALIGN=MIDDLE  ALT=\"%28-250%29%2F%28-3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "W = +83<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F3 ALIGN=MIDDLE  ALT=\"1%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > meters is with width that gives max area<BR>\n" );
document.write( "Find the Length<BR>\n" );
document.write( "L = -1.5w+250<BR>\n" );
document.write( "L = -1.5(83.33) + 250<BR>\n" );
document.write( "L = -125 + 250<BR>\n" );
document.write( "L = 125 meters is length for max area<BR>\n" );
document.write( "Then<BR>\n" );
document.write( "125 * 83<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F3 ALIGN=MIDDLE  ALT=\"1%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 10416<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2F3 ALIGN=MIDDLE  ALT=\"2%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > sq meters is the max area </SPAN>\n" );
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