document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1012461>Question 1012461</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A piece of wire 6 metres long is cut into 2 parts. One part is used to form a square and the other to make a rectangle whose length is 3 times its breadth. Find the lengths of the 2 parts if the sum of the two areas is a minimum. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #628414 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=628414\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  A piece of wire 6 metres long is cut into 2 parts.<BR>\n" );
document.write( " One part is used to form a square and the other to make a rectangle whose length is 3 times its breadth.<BR>\n" );
document.write( " Find the lengths of the 2 parts if the sum of the two areas is a minimum.<BR>\n" );
document.write( ":<BR>\n" );
document.write( "let s = the side of the square<BR>\n" );
document.write( "let L = the length of the rectangle<BR>\n" );
document.write( "let W = the width<BR>\n" );
document.write( "therefore the two perimeters = 6<BR>\n" );
document.write( "4s + 2L + 2W = 6<BR>\n" );
document.write( "Simplify, divide equation by 2<BR>\n" );
document.write( "2s + L + W = 3<BR>\n" );
document.write( "Given that<BR>\n" );
document.write( "L = 3W<BR>\n" );
document.write( "therefore<BR>\n" );
document.write( "2s + 3W + W = 3<BR>\n" );
document.write( "2s + 4W = 3:<BR>\n" );
document.write( "Simplify, divide by 2<BR>\n" );
document.write( "s + 2W = 1.5<BR>\n" );
document.write( "s = -2W+1.5<BR>\n" );
document.write( ": <BR>\n" );
document.write( "the total area<BR>\n" );
document.write( "A = s^2 + LW<BR>\n" );
document.write( "Replace s with (-2W+1.5), replace L with 3W<BR>\n" );
document.write( "A = (-2W+1.5)^2 + 3W(W)<BR>\n" );
document.write( "FOIL<BR>\n" );
document.write( "A = 4W^2 - 3W - 3W + 2.25 + 3W^2<BR>\n" );
document.write( "Combine like terms<BR>\n" );
document.write( "A = 7W^2 - 6W + 2.25<BR>\n" );
document.write( "Find the minimum by finding the axis of symmetry, x=-b/(2a)<BR>\n" );
document.write( "w = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-%28-6%29%29%2F%282%2A7%29 ALIGN=MIDDLE  ALT=\"%28-%28-6%29%29%2F%282%2A7%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=6%2F14 ALIGN=MIDDLE  ALT=\"6%2F14\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "W = .42857 is the width for minimum area<BR>\n" );
document.write( "Find L<BR>\n" );
document.write( "3(.42857) = 1.2857 is the length<BR>\n" );
document.write( "Find the length of the portion made into a rectangle (perimeter)<BR>\n" );
document.write( "P = 2(1.2857 + 2(.42857)<BR>\n" );
document.write( "P = 3.42856m is the portion made into a rectangle<BR>\n" );
document.write( "then<BR>\n" );
document.write( " 6 - 3.42856 = 2.57m is the portion made into a square<BR>\n" );
document.write( ":<BR>\n" );
document.write( ":<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );