Question

Prove that the height of the cylinder of maximum volume inscribed in a sphere of radius \( R \) is \( \frac{2R}{\sqrt{3}} \). 

Hint:

When maximizing the volume of an object inscribed in a sphere, use geometric relationships such as the Pythagorean theorem and apply calculus to find the maximum.

Answer 1

Step 1: Let the radius of the sphere be \( R \). The volume of the cylinder inscribed in the sphere depends on its height and radius. Let the height of the cylinder be \( h \) and its radius be \( r \). The equation for the volume \( V \) of the cylinder is: \[ V = \pi r^2 h. \] Step 2: Use the Pythagorean theorem. For the cylinder inscribed in the sphere, the radius of the cylinder \( r \) and half of the height \( \frac{h}{2} \) form a right triangle with the radius of the sphere: \[ r^2 + \left( \frac{h}{2} \right)^2 = R^2. \] This simplifies to: \[ r^2 = R^2 - \frac{h^2}{4}. \] Step 3: Substitute the expression for \( r^2 \) into the volume formula: \[ V = \pi \left( R^2 - \frac{h^2}{4} \right) h. \] Step 4: Maximize the volume by differentiating \( V \) with respect to \( h \) and setting the derivative equal to zero: \[ \frac{dV}{dh} = 0. \] Solving this gives the value of \( h \) that maximizes the volume: \[ h = \frac{2R}{\sqrt{3}}. \] Thus, the height of the cylinder of maximum volume inscribed in the sphere is \( \frac{2R}{\sqrt{3}} \).