Heres a calc problem.
I have to find the maximum volume of the upside down cone inside the larger cone.
Here is a cut on a plane
Volume of "small" cone is v = 1/3 pi rē h
It is possible to express r and h function of theta, R and H
After differentiating I find an extremum for
With R=6' and H=12'
Corresponding volume is
which is about 15% of the volume of the "big" cone
This is a little different than the usual cylinder in a cone and what not.
But, like most others we can use similar triangles. Here it is in general. You can add the constants H=12 and R=6
Let H=height of big cone, R=radius of big cone, r=radius of small cone, h=height of small cone.
Plug this into the cone volume formula:
Now differentiate w.r,t h, set to 0 and solve for h. H=12, R=6 are constants
You should get and go from there.