Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is (8/27)of the volume of the sphere.

Let
R = radius sphere
r = base radius cone
R + h = height cone
V = volume cone
_______

V = (1/3)πr²(R + h)

By the Pythagorean Theorem:
r² = R² - h²

Plug into the formula for volume.
V = (1/3)π(R² - h²)(R + h) = (1/3)π(R³ + R²h - Rh² - h³)

Take the derivative and set equal to zero to find the critical points.

dV/dh = (1/3)π(R² - 2Rh - 3h²) = 0
R² - 2Rh - 3h² = 0
(R - 3h)(R + h) = 0
h = R/3, -R

But h must be positive so:
h = R/3

Calculate the second derivative to determine the nature of the critical points.

d²V/dh² = (π/3)(-2R - 6h) < 0
So this is a relative maximum which we wanted.

Solve for r².
r² = R² - h² = R² - (R/3)² = R²(1 - 1/9) = (8/9)R²

Calculate maximum volume.

V = (π/3)[(8/9)R²](R + R/3) = (8/27)πR³(4/3) = 32πR³/81

V=(8/27)(4/3)πR³

Hence proved

For R = 3 maximum volume is:

V = 32π(3³)/81 = 32π(27)/81 = 32π/3