SOLUTION: Find the conditions for a right circular cone to have optimum lateral surface area given it has a fixed volume.

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Question 390085: Find the conditions for a right circular cone to have optimum lateral surface area given it has a fixed volume.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
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.
Consider
, where K is constant.
Setting the partial derivatives of F to 0:
F_r = ,
F_h = %28pi%2Arh%29%2Fsqrt%28h%5E2%2Br%5E2%29+%2B+%28alpha%2Api%2Ar%5E2%29%2F3+=+0,
F_alpha = %28pi%2Ar%5E2h%29%2F3+-+K+=+0.
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)
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
, 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.
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.
If r = 2, then %28pi%2F3%29%2A2%5E2h+=+%28pi%2Asqrt%282%29%29%2F3, or h+=+sqrt%282%29%2F4.
(This comes from equating the two volume values.)
The corresponding LSA is 2%2Api%2Asqrt%2833%2F8%29.
But 2%2Api%2Asqrt%2833%2F8%29 > sqrt%283%29%2Api.
Therefore the condition h+=+sqrt%282%29%2Ar yields a minimum value for the lateral surface area.