Find Max Volume of Right Circular Cylinder inscribed in A Sphere?

Question

My book says that there is a sphere with radius R. We are to find the maximum Volume of the Right Circular Cylinder inscribed inside of the sphere. This is all it says, not numbers, nothing else.

Answer 1

Let
R = radius of sphere
r = radius of cylinder
h = height of cylinder
V = volume cylinder

We have:

V = πr²h
r² = V/(πh)

By the Pythagorean Theorem

R² = r² + (h/2)² = r² + h²/4
r² = R² - h²/4
r = √(R² - h²/4)

Set the equations for r² equal.

V/(πh) = R² - h²/4
V = πR²h - πh³/4

Take the derivative with respect to h and set it equal to zero to find the critical point(s).

dV/dh = πR² - 3πh²/4 = 0
πR² = 3πh²/4
4R²/3 = h²
h = 2R/√3

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

d²V/dh² = -2*3πh/4 = -3πh/2 < 0 implies relative maximum

Solve for r.

r = √(R² - h²/4)
r = √{R² - (2R/√3)²/4} = √{R² - R²/3}
r = √{(2/3)R²} = R√(2/3)

Compute the volume of the cylinder.

V = πr²h = π{R√(2/3)}²{2R/√3} = π{(2/3)R²}{2R/√3}
V = 4πR³/(3√3) = (4√3)πR³/9

Answer 2

So the cylinder has for its axis one of the diameters of the sphere. Then if 2h<2R is its height, its radius r will be so that r^2+h^2=R^2 and you want to maximize hr^2 which is h(R^2-h^2). Just differentiate in h and you are done.

r is the radius of the cylinder. You get r^2+h^2=R^2 by projecting on a plane which contains the axis of the cylinder and using Pythagoras.