document.write( "Question 1033164: You have been asked to design a can shaped like a right circular cylinder with height h and radius r. Given that the can must hold exactly 410 cm3, what values of h and r will minimise the total surface area (including the top and bottom faces)? Give your answers correct to 2 decimal places as a list [in brackets] of the form: [ h, r ]
\n" ); document.write( "for constants h (height), r (radius), in that order.
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Algebra.Com's Answer #647830 by KMST(5328)\"\" \"About 
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the volume of the cylinder with height \"h\" and radius \"r\" (both in cm) is
\n" ); document.write( "\"pi%2Ar%5E2%2Ah=410\" \"cm%5E3\" .
\n" ); document.write( "\"pi%2Ar%5E2%2Ah=410\" ---> \"h=410%2F%28pi%2Ar%5E2%29\" .
\n" ); document.write( "The total surface area (including the top and bottom faces)of a right circular cylinder with height \"h\" and radius \"r\" is
\n" ); document.write( "\"A=2pi%2Ar%2Ah%2B2pi%2Ar%5E2\" .
\n" ); document.write( "(We would measure \"h\" and radius \"r\" in cm, and \"A\" would be in \"cm%5E2\", of course).
\n" ); document.write( "Substituting the expression found for \"h\" ,
\n" ); document.write( "\"A=2pi%2Ar%2A%28410%2F%28pi%2Ar%5E2%29%29%2B2pi%2Ar%5E2\" <--> \"A=2%2A410%2Fr%2B2pi%2Ar%5E2\" <--> \"A=2%2A%28410%2Bpi%2Ar%5E3%29%2Fr\" .
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\n" ); document.write( "We need to find the value of \"r\" that yields the minimum for \"A\" .
\n" ); document.write( "There may be another way to find that value.
\n" ); document.write( "Maybe you are expected to do it using a graphing calculator,
\n" ); document.write( "or maybe you are expected to use calculus,
\n" ); document.write( "and specifically derivatives.
\n" ); document.write( "The result should be the same.
\n" ); document.write( "Using calculus:
\n" ); document.write( "A local minimum of \"A\" happens only for a value of \"r\" that makes the derivative zero.
\n" ); document.write( "The derivative of \"A=2%2A410%2Fr%2B2pi%2Ar%5E2\" is
\n" ); document.write( "\"dA%2Fdr=-2%2A410%2Fr%5E2%2B4pi%2Ar=%282%2Fr%5E2%29%28-410%2B2pi%2Ar%5E3%29\" .
\n" ); document.write( "\"%282%2Fr%5E2%29%28-410%2B2pi%2Ar%5E3%29=0\"<--->\"-410%2B2pi%2Ar%5E3=0\"<--->\"r%5E3=410%2F%282%2Api%29=205%2Fpi\" ---> \"r=root%283%2C%28205%2Fpi%29%29=highlight%284.03%29\" (correct to 2 decimal places).
\n" ); document.write( "Then,
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