document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=167653>Question 167653</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a large soup can is to be designed so that the can will hold 16&#960; in^3 (about 28oz) of soup. find what height and radius of the cylindrical can will minimize the amount of metal needed. note: you will need to use the \"minimum\" program on your calculator. -- I have no idea where to even start with this problem.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #123567 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=123567\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> From the solution above:<BR>\n" );
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document.write( "Start at S.A. = 2PIr^2 + 2V/r<BR>\n" );
document.write( "Set the 1st derivate of dSA/dr = 0 (derivative of SA with respect to r)<BR>\n" );
document.write( "4PIr - 2V/r^2 = 0<BR>\n" );
document.write( "4PIr^3 - 2V = 0<BR>\n" );
document.write( "2PIr^3 - 16PI = 0<BR>\n" );
document.write( "r^3 = 8<BR>\n" );
document.write( "r = 2<BR>\n" );
document.write( "h = 4\r<BR>\n" );
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document.write( " </SPAN>\n" );
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