document.write( "Question 1207300: A company that manufactures dog food wishes to pack in closed cylindrical tins.
\n" ); document.write( "What should be the dimensions of each tin if it is to have a volume of 128πcm³
\n" ); document.write( "and the minimum possible surface area?
\n" ); document.write( "

Algebra.Com's Answer #845090 by ikleyn(52781)\"\" \"About 
You can put this solution on YOUR website!
.
\n" ); document.write( "A company that manufactures dog food wishes to pack in closed cylindrical tin's as,
\n" ); document.write( "what should be the dimensions of each tin if it is to have a volume of 128π cm³
\n" ); document.write( "and the minimum possible surface area.
\n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\r\n" );
document.write( "As you know, the volume of a cylinder is \r\n" );
document.write( "\r\n" );
document.write( "V = \"pi%2Ar%5E2%2Ah\", \r\n" );
document.write( "\r\n" );
document.write( "where pi = 3.14, r is the radius and h is the height.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "In your case the volume is fixed:\r\n" );
document.write( "\r\n" );
document.write( "\"pi%2Ar%5E2%2Ah\" = \"128%2Api\"  cm^3.       (1)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The surface area of a cylinder is \r\n" );
document.write( "\r\n" );
document.write( "S = \"2pi%2Ar%2Ah+%2B+2pi%2Ar%5E2\",    (2)\r\n" );
document.write( "\r\n" );
document.write( "and they ask you to find minimum of (2) under the restriction (1).\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Using (1), I can rewrite (2) in the form\r\n" );
document.write( "\r\n" );
document.write( "S(r) = \"%282pi%2Ar%5E2%2Ah%29%2Fr\" + \"2pi%2Ar%5E2\" = \"2%2A%28128%2Api%29%2Fr%29\" + \"2pi%2Ar%5E2\" = \"%28256%2Api%29%2Fr\" + \"2pi%2Ar%5E2\".   (3)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The plot below shows the function S(r) = \"256%2Api%2Fr\" + \"2pi%2Ar%5E2\", and you can clearly see that it has the minimum.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "    \r\n" );
document.write( "\r\n" );
document.write( "        Plot y = \"%28256%2Api%29%2Fr+%2B+2%2Api%2Ar%5E2\"\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "To find the minimum, use Calculus: differentiate the function to get\r\n" );
document.write( "\r\n" );
document.write( "S'(r) = \"-%28256%2Api%29%2Fr%5E2\" + \"4pi%2Ar\" = \"%28-256%2Api+%2B+4pi%2Ar%5E3%29%2Fr%5E2\"\r\n" );
document.write( "\r\n" );
document.write( "and equate it to zero.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "S'(r) = 0   leads you to equation  \"4%2Ar%5E3\" = \"256\",   which gives \r\n" );
document.write( "\r\n" );
document.write( "r = \"root%283%2C64%29\" = 4 cm.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Answer.  r = 4 cm, h = \"%28128%2Api%29%2F%28pi%2A4%5E2%29\" = 8 cm gives the minimum of the surface area.\r\n" );
document.write( "
\r
\n" ); document.write( "\n" ); document.write( "Solved.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );