More Optimization — Question 9

Question 9 - Solution

Let the cylinder have: - Radius: $r$ - Height: $h$

It is inscribed in a sphere of radius $R$, so by the Pythagorean Theorem applied to the right triangle formed from the center of the sphere to the top edge of the cylinder: $$r^2 + \left( \frac{h}{2} \right)^2 = R^2$$

Volume of the cylinder: $$V = \pi r^2 h$$

From the constraint: $$r^2 = R^2 - \left( \frac{h}{2} \right)^2 = R^2 - \frac{h^2}{4}$$

Substitute into the volume: $$V(h) = \pi \left( R^2 - \frac{h^2}{4} \right) h = \pi \left( R^2 h - \frac{h^3}{4} \right)$$

Differentiate and maximize: $$V'(h) = \pi \left( R^2 - \frac{3h^2}{4} \right)$$

Set $V'(h) = 0$: $$R^2 - \frac{3h^2}{4} = 0 \Rightarrow h^2 = \frac{4R^2}{3} \Rightarrow h = \frac{2R}{\sqrt{3}}$$

Then: $$r^2 = R^2 - \left( \frac{h}{2} \right)^2 = R^2 - \frac{1}{4} \cdot \frac{4R^2}{3} = R^2 - \frac{R^2}{3} = \frac{2R^2}{3} \Rightarrow r = \sqrt{\frac{2}{3}} R$$

Answer: $$\boxed{ h = \frac{2R}{\sqrt{3}}, \quad r = \sqrt{\frac{2}{3}} R }$$

These are the dimensions of the inscribed cylinder with maximum volume.