document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=37163>Question 37163</A>: <I><SPAN STYLE=\"word-break: break-all;\"> this question is not out of the book...\r<BR>\n" );
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document.write( "Find what the minimum surface area for a cylindrical can will be to hold <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=500+cm%5E3 ALIGN=MIDDLE  ALT=\"500+cm%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Give the radius and the height.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=surface+area-+2%28pi%29r%5E2%2B2%28pi%29rh ALIGN=MIDDLE  ALT=\"surface+area-+2%28pi%29r%5E2%2B2%28pi%29rh\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=volume-%28pi%29r%5E2h ALIGN=MIDDLE  ALT=\"volume-%28pi%29r%5E2h\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #22848 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=22848\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since we wish to minimize surface area, we will need to find r when dA/dr = 0.\r<BR>\n" );
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document.write( "(1)  A = 2(pi)r^2 + 2(pi)rh\r<BR>\n" );
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document.write( "We can find h in terms of r by considering the volume formula\r<BR>\n" );
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document.write( "V = (pi)r^2h\r<BR>\n" );
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document.write( "500 = (pi)r^2h\r<BR>\n" );
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document.write( "h = 500/(pi)r^2\r<BR>\n" );
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document.write( "Substituting into (1) above, we get\r<BR>\n" );
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document.write( "A = 2(pi)r^2 + 2(pi)r(500/(pi)r^2)\r<BR>\n" );
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document.write( "A = 2(pi)r^2 + 1000/r\r<BR>\n" );
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document.write( "Now take dA/dr and set it equal to zero.\r<BR>\n" );
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document.write( "dA/dr = 4(pi)r - 1000/r^2 = 0\r<BR>\n" );
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document.write( "And r = cube root of (250/pi)\r<BR>\n" );
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document.write( "From there you can find h and A. </SPAN>\n" );
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