document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=164175>Question 164175</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a cylindrical soda pop can of radius r and height h is to be manufactured to hold exactly 500 milliliters of liquid when completely full.  A manufacturer wishes to find the dimensions of the can with the minimum surface area.   Your task is to numerically, algebraically and graphically investigate this situation.\r<BR>\n" );
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document.write( "Determine the radius x and height h that yields a can of minimal surface area.  Compute this minimal surface area. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #120977 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=120977\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  a cylindrical soda pop can of radius r and height h is to be manufactured to hold exactly 500 milliliters of liquid when completely full. A manufacturer wishes to find the dimensions of the can with the minimum surface area. Your task is to numerically, algebraically and graphically investigate this situation. <BR>\n" );
document.write( "Determine the radius x and height h that yields a can of minimal surface area. Compute this minimal surface area.<BR>\n" );
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document.write( "Tough problem.<BR>\n" );
document.write( "The volume is PI*r^2*h  (r = radius, h= height) = 500 cc<BR>\n" );
document.write( "The area is 2PI*r*h + 2PI*r^2  (the 1st term is the cylinder, the 2nd the top and bottom)<BR>\n" );
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document.write( "Eliminate the h term in the area by solving for h in terms of r.<BR>\n" );
document.write( "PI*r^2*h = 500<BR>\n" );
document.write( "h = 500/(PI*r^2)<BR>\n" );
document.write( "Sub for h in the eqn for area:<BR>\n" );
document.write( "Area = 2PI*r + 2PI*r^2<BR>\n" );
document.write( "Area = (1000/r) + 2PI*r^2<BR>\n" );
document.write( "Differentiate with respect to r, and set = to 0.<BR>\n" );
document.write( "(-1000/r^2) + 4PIr = 0<BR>\n" );
document.write( "4PIr^3 = 1000<BR>\n" );
document.write( "r^3 = 1000/4PI = 79.57747155<BR>\n" );
document.write( "r = 4.30127 cm<BR>\n" );
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document.write( "Does that make sense?  Solve for h, and for the area.<BR>\n" );
document.write( "I'm not sure what is meant by \"solve numerically\", maybe an Excel sheet?<BR>\n" );
document.write( "I can graph it tomorrow, email me with any questions.\r<BR>\n" );
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