document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=736368>Question 736368</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A cylinder with a top and bottom has a volume of 30 cubic inches. Find the minimum amount of material needed to create the can. (Surface area=2pir^2 + 2pirh & volume = pir^2h) <BR>\n" );
document.write( "I tried to create two separate equations, <BR>\n" );
document.write( "30 = pir^2h (and then solve for r^2)<BR>\n" );
document.write( "and<BR>\n" );
document.write( "2pir^2 <BR>\n" );
document.write( "and then I was going to substitute the r value into the other equation. This did not work. <BR>\n" );
document.write( "Thanks in advance!<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #449825 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=449825\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> we have two unknown variables, the radius (r) and the height (h).  <BR>\n" );
document.write( " Volume (V) = pir^2h<BR>\n" );
document.write( "  <BR>\n" );
document.write( " <BR>\n" );
document.write( "30 = pir^2h<BR>\n" );
document.write( "h =  30/pir2 <BR>\n" );
document.write( "  Substitute this into the surface area equation for h, we get  <BR>\n" );
document.write( "A = (2pir( 30 )/pir^2) + 2pir2) <BR>\n" );
document.write( "  <BR>\n" );
document.write( "A =  (60/r)  + 2pir2 <BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Take the derivative  for r<BR>\n" );
document.write( "  <BR>\n" );
document.write( " A = 60r-1  + 2pir2<BR>\n" );
document.write( "  <BR>\n" );
document.write( " A' = -60r-2  + 4pir<BR>\n" );
document.write( "  <BR>\n" );
document.write( "    <BR>\n" );
document.write( "when slope is zero  <BR>\n" );
document.write( " <BR>\n" );
document.write( " 0 = -60r-2  + 4pir<BR>\n" );
document.write( "  <BR>\n" );
document.write( " <BR>\n" );
document.write( " 60r-2  = 4pir<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Multiply both sides by r^2  <BR>\n" );
document.write( " <BR>\n" );
document.write( " 60  = 4pir3<BR>\n" );
document.write( "  <BR>\n" );
document.write( " <BR>\n" );
document.write( "4.775 = r3<BR>\n" );
document.write( "  <BR>\n" );
document.write( " 1.68 = r <BR>\n" );
document.write( "h =  30/pir^2 <BR>\n" );
document.write( "h=30/pi*(1.68)^2\r<BR>\n" );
document.write( "\n" );
document.write( "h = 3.38<BR>\n" );
document.write( "  <BR>\n" );
document.write( "    <BR>\n" );
document.write( "A  = 2pirh + 2pir^2 <BR>\n" );
document.write( "A  = 2*pi*(1.68)(3.38) + 2pi*(1.68)^2 <BR>\n" );
document.write( "A  = 35.68 + 17.73 <BR>\n" );
document.write( "A = 53.41 square inches  \r<BR>\n" );
document.write( "\n" );
document.write( "CHECK<BR>\n" );
document.write( "  <BR>\n" );
document.write( "V=    pi*1.68^2*3.38= 29.96\r<BR>\n" );
document.write( "\n" );
document.write( "you can check for maxima or minima by taking the second derivate for more detailing<BR>\n" );
document.write( "  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );