Prove that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R√3. Also find the maximum volume.

Given: Radius of the sphere R.
Let x be the diameter of the base of the inscribed cylinder .Then, h² + x² = (2R)²
h² + x² = 4R² —(1)
Volume of the cylinder = πr²h
V = π(x/2)²h
V = (1/4)πx²h
Substituting the value of x² we get
V = (1/4)πh(4R² - h²)
From (1), x² = 4R² - h²
V = πR²h - (πh³)/4
Differentiating with respect to h,
dV/dh = πR² - (3πh²)/4
For Max volume, dV/dh = 0
π[R² - (3h²)/4] = 0
=> R² = 3h²/4
=> h = 2R√3
Also, d²V/dh² = - (6πh)/4 = - (3πh)/2
At h = 2R√3
d²V/dh² = - (3π(2R√3))/2 < 0
V is maximum at h = 2R√3
Maximum volume at h = 2R√3 => Vmax = (1/4)π[2R√3][4R² - (4R²)/3]
=> Vmax = πR²(√3)[8R²/3]
=> Vmax = (4πR³√3)/3 sq.units.