Question 390085
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*rs = pi*r*sqrt(h^2 + r^2))}}}.  s = slant height.  The volume is given by {{{V= (pi*r^2h)/3}}}, which is constant in this problem.  We proceed to use Lagrange multipliers.
Consider 
{{{F(r, h, alpha) = pi*r*sqrt(h^2 + r^2) + alpha*( (pi*r^2h)/3 - K)}}}, where K is constant.
Setting the partial derivatives of F to 0: 
F_r = {{{pi*(sqrt(h^2+ r^2) + r^2/sqrt(h^2+r^2)) + (alpha*pi*2rh)/3 = 0}}},
F_h = {{{(pi*rh)/sqrt(h^2+r^2) + (alpha*pi*r^2)/3 = 0}}},
F_{{{alpha}}} = {{{(pi*r^2h)/3 - K = 0}}}.

The first equation gives, after simplification, {{{(h^2 + 2r^2)/sqrt(h^2 + r^2) + (2alpha*rh)/3=0}}}. <-----(A)
The second equation gives {{{h/sqrt(h^2 + r^2) + (alpha*r)/3=  0}}}, or {{{alpha = (-3h)/(r*sqrt(h^2+r^2))}}}.  Putting this into (A), we get
{{{h^2 + 2r^2 = -(2/3)*((-3h)/(r*sqrt(h^2+r^2)))*rh*sqrt(h^2 + r^2)}}}, or {{{h^2 + 2r^2  = 2h^2}}}, or {{{2r^2 = h^2}}}, which gives {{{h = sqrt(2)*r}}}.  This gives a condition for an optimum lateral surface area given a fixed volume.
Now let r = 1.  Then {{{h = sqrt(2)}}}.  The fixed volume is then {{{V = (pi*sqrt(2))/3}}}.  The corresponding LSA is {{{sqrt(3)*pi}}}.

If r = 2, then {{{(pi/3)*2^2h = (pi*sqrt(2))/3}}}, or {{{h = sqrt(2)/4}}}. 
(This comes from equating the two volume values.)

The corresponding LSA is {{{2*pi*sqrt(33/8)}}}.
But {{{2*pi*sqrt(33/8)}}} >  {{{sqrt(3)*pi}}}.
Therefore the condition {{{h = sqrt(2)*r}}} yields a minimum value for the lateral surface area.