document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1163322>Question 1163322</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A two-pen corral is to be built. The outline of the corral forms two<BR>\n" );
document.write( "identical adjoining rectangles. If there is 120 m of fencing available and<BR>\n" );
document.write( "the fence width cannot be less than 6 m, what dimensions of the corral will<BR>\n" );
document.write( "maximize<BR>\n" );
document.write( "the enclosed area? [round all answers to one decimal places)<BR>\n" );
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document.write( "In the rectangular corral Horizontal line is the length and vertical line is the width.\r<BR>\n" );
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document.write( "Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #787389 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=787389\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> this turned my head around a few times, but i think i have it.<BR>\n" );
document.write( "the corral forms two rectangles that are joined together on one side.<BR>\n" );
document.write( "the total fencing available is 120 meters.<BR>\n" );
document.write( "the total fencing will be 2 * the length of the enclosure plus 3 times the width of the enclosure.<BR>\n" );
document.write( "the area of the enclosure is the length * width.<BR>\n" );
document.write( "let:<BR>\n" );
document.write( "L = length<BR>\n" );
document.write( "W = width<BR>\n" );
document.write( "A = area<BR>\n" );
document.write( "F = fence<BR>\n" );
document.write( "your equations are:<BR>\n" );
document.write( "F = 2L + 3W<BR>\n" );
document.write( "A = L * W<BR>\n" );
document.write( "since F = 120, you get:<BR>\n" );
document.write( "120 = 2L + 3W<BR>\n" );
document.write( "A = L * W<BR>\n" );
document.write( "from 120 = 2L + 3W, solve for L to get:<BR>\n" );
document.write( "L = (120 - 3W) / 2<BR>\n" );
document.write( "from A = L * W, replace L with (120 - 3W) / 2 to get:<BR>\n" );
document.write( "A = (120 - 3W) / 2 * W<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( "A = 60 * W - 1.5 * W^2<BR>\n" );
document.write( "rearrange the terms on the right side of thise equation by descendng order of degree to get:<BR>\n" );
document.write( "A = -1.5 * W^2 + 60 * W<BR>\n" );
document.write( "this is a quadratic equation in standard form where:<BR>\n" );
document.write( "a = -1.5<BR>\n" );
document.write( "b = 60<BR>\n" );
document.write( "maximum value will be when W = -b/2a = 3/60 = 20<BR>\n" );
document.write( "when W = 20, maximum area will be -1.5 * 20^2 + 60 * 20 = 600 square meters.<BR>\n" );
document.write( "since L * W = A, then L = A / W = 600 / 20 = 30<BR>\n" );
document.write( "you get:<BR>\n" );
document.write( "L = 30<BR>\n" );
document.write( "W = 20<BR>\n" );
document.write( "F = 2L + 3W becomes F = 2*30 + 3*20 = 60 + 60 = 120, so this part checks out.<BR>\n" );
document.write( "your solution is that the maximum area will be 600 square meters.<BR>\n" );
document.write( "note that the perimeter is equal to 2L + 2W, making the perimeter equal to 60 + 40 = 100 meters.<BR>\n" );
document.write( "you can graph the equation of area = -1.5 * W^2 + 60 * W by making area equal to y and W equal to x.<BR>\n" );
document.write( "the equation becomes y = -1.5 * x^2 + 60 * x<BR>\n" );
document.write( "that looks like this on the graph.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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