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.
