Show 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 h be the height of the inscribed cylinder and x be the radius of the base of the cylinder.
Then, the relationship between h and x is given by the Pythagorean theorem:
h²/4 + x² = R²
Solving for x²:
x² = R² - h²/4
The volume V of the cylinder is given by:
V = πx²h
Substitute the expression for x²:
V = π(R² - h²/4)h = πR²h - πh³/4
To find the maximum volume, we take the derivative of V with respect to h and set it to zero:
dV/dh = πR² - (3πh²/4) = 0
Solving for h:
πR² = 3πh²/4
4R² = 3h²
h² = (4/3)R²
h = 2R/√3 = (2R√3)/3
To confirm that this is a maximum, we take the second derivative:
d²V/dh² = -6πh/4 = -3πh/2
At h = 2R/√3, d²V/dh² = -3π(2R/√3)/2 = -πR√3 < 0, indicating a maximum.
Therefore, the height of the cylinder of maximum volume is h = (2R√3)/3.
To find the maximum volume, substitute h into the volume equation:
V_max = π(R² - (1/4)((4/3)R²))(2R/√3) = π(R² - (1/3)R²)(2R/√3) = π(2/3)R²(2R/√3) = (4πR³/3√3) = (4πR³√3)/9
Alternatively:
V = πx²h
From (1),x²=4R²−h²
V=π(4R²−h²)h/4=π(4R²h−h³)/4
Differentiating with respect to h,
dV/dh=π(4R²−3h²)/4
Setting dV/dh=0 gives 4R²−3h²=0
h=2R/√3=2R√3/3
Also,d²V/dh²=−6πh/4=−(3πh/2)
At h=2R/√3
d²V/dh²=−3π(2R√3)/6=−πR√3<0.Thus V is maximum at h=2R/√3
Maximum volume at h=2R/√3=π(4R²−(4/3)R²)(2R/√3)/4=(4πR³/3√3) cubic units