document.write( "Question 390085: Find the conditions for a right circular cone to have optimum lateral surface area given it has a fixed volume. \n" ); document.write( "

Algebra.Com's Answer #276575 by robertb(5830) $\"\"$ $\"About$

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Let h = height of right circular cone, and r = radius of same cone. Then the lateral surface area of the cone is given by $\"SA+=+pi%2Ars+=+pi%2Ar%2Asqrt%28h%5E2+%2B+r%5E2%29%29\"$ . s = slant height. The volume is given by $\"V=+%28pi%2Ar%5E2h%29%2F3\"$ , which is constant in this problem. We proceed to use Lagrange multipliers.
\n" ); document.write( "Consider
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, where K is constant.
\n" ); document.write( "Setting the partial derivatives of F to 0:
\n" ); document.write( "F_r =

,
\n" ); document.write( "F_h = $\"%28pi%2Arh%29%2Fsqrt%28h%5E2%2Br%5E2%29+%2B+%28alpha%2Api%2Ar%5E2%29%2F3+=+0\"$ ,
\n" ); document.write( "F_ $\"alpha\"$ = $\"%28pi%2Ar%5E2h%29%2F3+-+K+=+0\"$ .\r
\n" ); document.write( "\n" ); document.write( "The first equation gives, after simplification, $\"%28h%5E2+%2B+2r%5E2%29%2Fsqrt%28h%5E2+%2B+r%5E2%29+%2B+%282alpha%2Arh%29%2F3=0\"$ . <-----(A)
\n" ); document.write( "The second equation gives $\"h%2Fsqrt%28h%5E2+%2B+r%5E2%29+%2B+%28alpha%2Ar%29%2F3=++0\"$ , or $\"alpha+=+%28-3h%29%2F%28r%2Asqrt%28h%5E2%2Br%5E2%29%29\"$ . Putting this into (A), we get
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, or $\"h%5E2+%2B+2r%5E2++=+2h%5E2\"$ , or $\"2r%5E2+=+h%5E2\"$ , which gives $\"h+=+sqrt%282%29%2Ar\"$ . This gives a condition for an optimum lateral surface area given a fixed volume.
\n" ); document.write( "Now let r = 1. Then $\"h+=+sqrt%282%29\"$ . The fixed volume is then $\"V+=+%28pi%2Asqrt%282%29%29%2F3\"$ . The corresponding LSA is $\"sqrt%283%29%2Api\"$ .\r
\n" ); document.write( "\n" ); document.write( "If r = 2, then $\"%28pi%2F3%29%2A2%5E2h+=+%28pi%2Asqrt%282%29%29%2F3\"$ , or $\"h+=+sqrt%282%29%2F4\"$ .
\n" ); document.write( "(This comes from equating the two volume values.)\r
\n" ); document.write( "\n" ); document.write( "The corresponding LSA is $\"2%2Api%2Asqrt%2833%2F8%29\"$ .
\n" ); document.write( "But $\"2%2Api%2Asqrt%2833%2F8%29\"$ > $\"sqrt%283%29%2Api\"$ .
\n" ); document.write( "Therefore the condition $\"h+=+sqrt%282%29%2Ar\"$ yields a minimum value for the lateral surface area. \n" ); document.write( "

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