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" );
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document.write( "A = Area of square + Area of circle.\r<BR>\n" );
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document.write( "   = (S/4)^2 + C^2/4(pi symbol)     where C = 2(pi symbol)r\r<BR>\n" );
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document.write( " 1) S + C =____________\r<BR>\n" );
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document.write( "2) A =_______ + _______________  (Write the equation in terms of C. )\r<BR>\n" );
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document.write( "3) Find dA/dC.\r<BR>\n" );
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document.write( "4) Set dA/dC = 0 and solve for C.\r<BR>\n" );
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document.write( "5) Check if A’’ is positive. If it is positive, the total area is minimum.\r<BR>\n" );
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document.write( "6) S = ___________   and C =_______________\r<BR>\n" );
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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 #598578 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=598578\">put this solution on YOUR website!</A><BR><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.<BR>\n" );
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document.write( "s = side length of the square.<BR>\n" );
document.write( "Circumference of the circle = 16 - 4s = 2pi*r<BR>\n" );
document.write( "r = (16-4s)/2pi = (8-2s)/pi<BR>\n" );
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document.write( "Area of circle = pi*(8 - 2s)^2<BR>\n" );
document.write( "Area of square = s^2<BR>\n" );
document.write( "Total = s^2 + pi*(8 - 2s)^2<BR>\n" );
document.write( "dA/ds = 2s + pi*2*(8-2s)*(-2) = 2s - 4pi*(8-2s) = 0<BR>\n" );
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document.write( "2s - 4pi*(8-2s) = 0<BR>\n" );
document.write( "s = 16pi - 4pi*s<BR>\n" );
document.write( "s*(4pi + 1) = 16pi<BR>\n" );
document.write( "s = 16pi/(4pi + 1)<BR>\n" );
document.write( "s =~ 3.705 inches<BR>\n" );
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document.write( "dA/ds = 2s - 4pi*(8-2s)<BR>\n" );
document.write( "2nd derivative = 2 + 8pi which is positive --> a minimum<BR>\n" );
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document.write( "You can do it using dA/dC in a similar manner.<BR>\n" );
document.write( "I prefer making the side of the square the variable.<BR>\n" );
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