Show that the altitude of the right circular cone of maximum volume that can be described in a sphere of radius r is 4r/3. Also show that the maximum volume of the cone is 8/27 of the volume of the sphere.

Let R and h be the radius and height of the cone. r be the radius of the sphere. To show h = 4r/3 and Maximum volume of sphere = 8/27 Volume of sphere
In ΔABC, AC = h - r
Therefore, using Pythagoras theorem,
(h - r)² + R² = r² ⇒ R² = r² - (h - r)²
Volume of cone: V = (1/3)πR²h
or V = (1/3)π(r² - (h - r)²)h
V = (1/3)π[r² - h² - r² + 2hr]h
V = (1/3)π[2h²r - h³]
For maxima or minima, dV/dh = 0
Now, dV/dh = (1/3)π[4hr - 3h²]
Putting, dV/dh = 0 we get
4hr = 3h²
h = 4r/3
d²V/dh² = (1/3)π[4r - 6h]
Putting h = 4r/3
d²V/dh² = (1/3)π[4r - 6(4r/3)] ⇒ (1/3)π[-4r]
Which is less than zero, therefore h = 4r/3 is a maxima and the volume of the cone at h = 4r/3 will be maximum,
V = (1/3)πR²h = (1/3)π[r² - (h - r)²]h = (1/3)π[r² - (4r/3 - r)²][4r/3] = (1/3)π[8r²/9][4r/3] = 8/27(4πr³/3) = 8/27 x volume of sphere