Consider the top of the cone touching the inner sphere and consider the height of the cone h
then [h-R],[ R=radius of sphere] will equal one leg of a right angle triangle, the other two sides of the triangle would be r the radius of the cone that touches the sphere and the radius of the sphere, R.
So we can say,[h-R^2]+r^2=R^2, thus r^2=R^2-[h-R^2]
Expanding the bracket and rearranging, r^2=[2Rh-h^2]......[1]
The volume of a cone is pi/3[r^2h], substituting for r in ....[1]
V=pi/3[2Rh-h^2]h....=pi/3[2Rh^2-h^3] ........[2]
Differentiating [2] with respect to h, keeping R constant,
DV/dh=[4piRh]/3-[3pih^2]/3=0 for max/min.........[3]
evaluating [3] yields h=4/3R, and this value of h will maximise the volume of the cone. Taking the second derivative and substituting for h=4/3r will give a negative result confirming this value of h will ensure the maximum volume of the cone.
Substituting for h=4/3R, in ...[2] volume of cone will equal,
pi/3[2piR[[4R]/3]^2-[[4r]/3]^3]. Evaluated this will give the above answer. I hope this was helpful , and I will request someone to check the answer.
