document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=213554>Question 213554</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A farmer wants to make a rectangular enclosure along the side of a barn and then divide the enclosure into two pens with a fence constructed at a right angle to the barn. If 300 yards of fencing are available, what are the dimensions of the largest section that can be enclosed?\r<BR>\n" );
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document.write( "Firstly, the wording of this particular problem has thrown both myself, family, and friends for a loop. I have drawn, erased, and redrawn several rough sketches of this barn-and-fence problem and still feel as though it is wrong. I am at a loss of going about how to solve it. Any assistance would be greatly appreciated!<BR>\n" );
document.write( "Thank you in advance<BR>\n" );
document.write( "-B </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #161342 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=161342\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A farmer wants to make a rectangular enclosure along the side of a barn and then divide the enclosure into two pens with a fence constructed at a right angle to the barn. If 300 yards of fencing are available, what are the dimensions of the largest section that can be enclosed? <BR>\n" );
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document.write( "_________________Barn<BR>\n" );
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document.write( "I think it looks something like this:<BR>\n" );
document.write( "Then we can write an equation for the 300 yds of fencing<BR>\n" );
document.write( "L + 3W = 300<BR>\n" );
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document.write( "L = (300-3W)<BR>\n" );
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document.write( "Area = L*W<BR>\n" );
document.write( "Replace L with (300-3W)<BR>\n" );
document.write( "A = (300-3W) * W<BR>\n" );
document.write( "Arrange as a quadratic equation<BR>\n" );
document.write( "A = -3W^2 + 300W<BR>\n" );
document.write( "Find the axis of symmetry for Width that gives max area x = -b/(2a)<BR>\n" );
document.write( "in this equation a=-3; b=300 <BR>\n" );
document.write( "W = -300/(2*-3)<BR>\n" );
document.write( "W = -300/-6<BR>\n" );
document.write( "W = +50 ft is the width for max area<BR>\n" );
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document.write( "Find the length<BR>\n" );
document.write( "L = 300 - 3(50)<BR>\n" );
document.write( "L = 150 ft is the length<BR>\n" );
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document.write( "Max area would be 150 * 50 = 7500 sq/ft </SPAN>\n" );
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