document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=976989>Question 976989</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Take any wire of length 16 inches. Cut the wire into two pieces such that the sum of the areas of the square and circle formed by the pieces of wire is the minimum.  \r<BR>\n" );
document.write( "\n" );
document.write( "A = Area of square + Area of circle.\r<BR>\n" );
document.write( "\n" );
document.write( "   = (S/4)^2 + C^2/4(pi symbol)     where C = 2(pi symbol)r\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " 1) S + C =____________\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "2) A =_______ + _______________  (Write the equation in terms of C. )\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3) Find dA/dC.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "4) Set dA/dC = 0 and solve for C.\r<BR>\n" );
document.write( "\n" );
document.write( "5) Check if A’’ is positive. If it is positive, the total area is minimum.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "6) S = ___________   and C =_______________\r<BR>\n" );
document.write( "\n" );
document.write( "Please help me with this problem. Thank you so much!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #598924 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=598924\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let L = the length of the wire<BR>\n" );
document.write( "Let s = the perimeter of the square, then s/4 = the length of one side<BR>\n" );
document.write( "Then the circumference of the circle, C = L - s<BR>\n" );
document.write( "Area of the square = (s/4)^2<BR>\n" );
document.write( "Area of the circle = C^2/(4*pi)<BR>\n" );
document.write( "A = the sum of the areas = (s/4)^2 + C^2/(4*pi)<BR>\n" );
document.write( "In terms of C, since s = L - C, we can write<BR>\n" );
document.write( "A = (L-C)^2/16 + C^2/(4*pi)<BR>\n" );
document.write( "For the sum of the areas to be a minimum, dA/dC = 0:<BR>\n" );
document.write( "0 = 2(L-C)(-1)/16 + 2C/(4*pi)<BR>\n" );
document.write( "0 = (C-L)/8 + C/(2*pi)<BR>\n" );
document.write( "Solve for C in terms of L:<BR>\n" );
document.write( "C(1/8 + 1/(2*pi)) = L/8<BR>\n" );
document.write( "C(1 + 4/pi) = L -> C = L/(1 + 4/pi)<BR>\n" );
document.write( "Substituting L = 16 gives C = 7.04 inches<BR>\n" );
document.write( "Therefore, the perimeter of the square is s = 16 - 7.04 = 8.96 inches<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );