Optimization — Question 8

Question 8 - Solution

Let the height of the cylinder be $h$ and its radius be $r$. From the diagram, the diagonal of the right triangle gives the constraint:

$$r^2 + \left(\frac{h}{2}\right)^2 = R^2 \Rightarrow h = 2\sqrt{R^2 - r^2}$$

Step 1: Volume Function

$$V = \pi r^2 h = \pi r^2 \cdot 2\sqrt{R^2 - r^2} = 2\pi r^2 \sqrt{R^2 - r^2}$$

Step 2: Maximize $V(r)$

Let: $$V(r) = 2\pi r^2 (R^2 - r^2)^{1/2}$$

Differentiate using the product and chain rule:

$$V'(r) = 2\pi \left[ 2r(R^2 - r^2)^{1/2} - \frac{r^3}{(R^2 - r^2)^{1/2}} \right] = \frac{2\pi}{(R^2 - r^2)^{1/2}} \left( 2r(R^2 - r^2) - r^3 \right)$$

Set $V'(r) = 0$: $$2r(R^2 - r^2) - r^3 = 0 \Rightarrow r(2R^2 - 3r^2) = 0 \Rightarrow r = 0 \text{ or } r^2 = \frac{2R^2}{3}$$

Step 3: Dimensions

$$r = \sqrt{\frac{2}{3}} R, \quad h = 2\sqrt{R^2 - r^2} = 2\sqrt{R^2 - \frac{2R^2}{3}} = 2\sqrt{\frac{R^2}{3}} = \frac{2R}{\sqrt{3}}$$

Step 4: Max Volume

$$V = \pi r^2 h = \pi \cdot \left(\frac{2R^2}{3}\right) \cdot \frac{2R}{\sqrt{3}} = \frac{4\pi R^3}{3\sqrt{3}}$$

$$\boxed{ \begin{aligned} &\text{Radius: } r = \sqrt{\frac{2}{3}}R \\ &\text{Height: } h = \frac{2R}{\sqrt{3}} \\ &\text{Max Volume: } V = \frac{4\pi R^3}{3\sqrt{3}} \end{aligned} }$$