document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=668609>Question 668609</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The area of a rectangle is 384 square feet. If the perimeter is 80 feet, find the length and width of the rectangle.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #415697 by </B><A HREF=/tutors/aboutme.mpl?userid=swincher4391><B>swincher4391(1107)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=swincher4391><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=swincher4391><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=415697\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A = lw<BR>\n" );
document.write( "P = 2l + 2w\r<BR>\n" );
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document.write( "384 = lw<BR>\n" );
document.write( "80 = 2l + 2w\r<BR>\n" );
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document.write( "384/l = w\r<BR>\n" );
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document.write( "80 = 2l + 2*(384/l)\r<BR>\n" );
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document.write( "80l/l = (2l^2+ 2*384)/l\r<BR>\n" );
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document.write( "80l = 2l^2 + 768\r<BR>\n" );
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document.write( "2l^2-80l+768 = 0\r<BR>\n" );
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document.write( "2(l^2-40l + 384) = 0\r<BR>\n" );
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document.write( "l = 16, l =24\r<BR>\n" );
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document.write( "Case 1:  l=16\r<BR>\n" );
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document.write( "If l =16 then w = 384/16 = 24\r<BR>\n" );
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document.write( "If l =24 then w = 384/24 = 16\r<BR>\n" );
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document.write( "So, we could have 16x24 or 24x16.\r<BR>\n" );
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