document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=247985>Question 247985</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This is a quadratic word problem. \r<BR>\n" );
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document.write( "A farmer wishes to enclose a rectangular region bordering a river using 600 ft. of fencing. He wants to divide the region into two equal parts using some of the fence material. What is the maximum area that can be enclosed with the fencing?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #180777 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=180777\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a quadratic word problem.<BR>\n" );
document.write( "A farmer wishes to enclose a rectangular region bordering a river using 600 ft. of fencing. <BR>\n" );
document.write( "He wants to divide the region into two equal parts using some of the fence material. <BR>\n" );
document.write( "What is the maximum area that can be enclosed with the fencing? <BR>\n" );
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document.write( "To divide it into two equal parts, we have the fence equation:<BR>\n" );
document.write( "2L + 3W = 600<BR>\n" );
document.write( "2L = 600 - 3W<BR>\n" );
document.write( "divide by 2<BR>\n" );
document.write( "L = 300 - 1.5W<BR>\n" );
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document.write( "Area = L*W<BR>\n" );
document.write( "Substitute for L<BR>\n" );
document.write( "A = W*(300-1.5W)<BR>\n" );
document.write( "A = 300W - 1.5W^2<BR>\n" );
document.write( "As a quadratic equation<BR>\n" );
document.write( "A = -1.5W^2 + 300W<BR>\n" );
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document.write( "max area occurs at the axis of symmetry, Formula for that: x = -b/(2a)<BR>\n" );
document.write( "in our equation: W = x; b = 300; a = -1.5<BR>\n" );
document.write( "W = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-300%29%2F%282%2A-1.5%29 ALIGN=MIDDLE  ALT=\"%28-300%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=%28-300%29%2F%28-3%29 ALIGN=MIDDLE  ALT=\"%28-300%29%2F%28-3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "W = +100 ft width for max area<BR>\n" );
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document.write( "Find the area; W=100<BR>\n" );
document.write( "A = -1.5(100^2) + 300(100)<BR>\n" );
document.write( "A = -1.5(10000) + 30000<BR>\n" );
document.write( "A = -15000 + 30000<BR>\n" );
document.write( "A = 15000 sq/ft is max area<BR>\n" );
document.write( ";<BR>\n" );
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document.write( "Check<BR>\n" );
document.write( "Find the length<BR>\n" );
document.write( "L = 300 - 1.5(100)<BR>\n" );
document.write( "L = 300 - 150<BR>\n" );
document.write( "L = 150 ft is the length<BR>\n" );
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document.write( "then: 100 * 150 = 15000 sq/ft is the area </SPAN>\n" );
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