document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=746002>Question 746002</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Minimizing area. A 46-in piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the wire be cut so that the sum of the areas is minimum? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #454108 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=454108\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let L = the length of the string = 46 in<BR>\n" );
document.write( "Let x = the length of one piece<BR>\n" );
document.write( "Then L-x = the length of the remaining piece<BR>\n" );
document.write( "If the circle is formed from the 1st piece, the radius of the circle will be x/(2*pi)<BR>\n" );
document.write( "The area of the circle will be Ac = pi*(x/(2*pi))^2 = x^2/(4*pi)<BR>\n" );
document.write( "The perimeter of the square = L-x, so each side s = (L-x)/4<BR>\n" );
document.write( "The area of the square will be s^2 = (L-x)^2/16<BR>\n" );
document.write( "To minimize the sum of the areas, we take the derivative of the sum and set = 0<BR>\n" );
document.write( "If you perform these operations and solve for x in terms of L, you should get x = pi/(4+pi)*L<BR>\n" );
document.write( "So the areas are minimized if the cut is made at about 0.44*46 = 20.24 in. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );