Calculus
posted by DJ on .
A right circular cone is inscribed in a sphere of radius r. Find the dimensions of the cone that maximize the volume of the cone.

Draw a diagram, with the top of the cone at the top of the sphere. If the cone extends past the diameter of the sphere of radius R, its base will have a radius r, and height h > R.
From the center of the sphere, draw a radius to the base of the cone. Then
r^2 + (hR)^2 = R^2
r^2 + h^2  2rR = 0
Now, the cone has a volume of V = 1/3 πr^2 h
To avoid a bunch of square roots, let's square things:
V^2 = 1/9 π^2 r^4 h^2
= π^2/9 r^4 (2Rr  r^2)^2
2VV' = 4π/9 r^3 (2Rr  r^2)^2 + 2π^2/9 r^4 (2Rrr^2)(2R2r)
Now, V' is a bunch of stuff all divided by V. V is not zero, so V' will be zero when the numerator is zero. Toss out all that π^2 stuff, and we are left with a bunch of algebra that ends up giving h = 4/3 R.
That will give you the value of r.
do a google for maximum volume cone inscribed in sphere to find more details.