Optimization — Question 10

Question 10

A right circular cylinder is inscribed in a right circular cone of height $H$ and base radius $R$, such that the cylinder shares the same central axis as the cone.

Find the dimensions of the cylinder that maximize its volume.

Question 10 - Solution

Let the height of the cylinder be $h$, and its base radius be $r$. We want to maximize: $$V = \pi r^2 h$$

Step 1: Relationship between $r$ and $h$

From similar triangles (cone vs cylinder section), we have: $$\frac{r}{H - h} = \frac{R}{H} \Rightarrow r = \frac{R(H - h)}{H}$$

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

Step 2: Maximize $V(h)$

Let’s simplify: $$V(h) = \frac{\pi R^2}{H^2} (H - h)^2 h$$

Let $k = \frac{\pi R^2}{H^2}$, then: $$V(h) = k(H - h)^2 h$$

Use the product rule: $$V'(h) = k \left[ 2(H - h)(-1) \cdot h + (H - h)^2 \right] = k \left[ -2h(H - h) + (H - h)^2 \right]$$

$$= k(H - h)[(H - h) - 2h] = k(H - h)(H - 3h)$$

Set $V'(h) = 0$: $$(H - h)(H - 3h) = 0 \Rightarrow h = H \text{ or } h = \frac{H}{3}$$

But $h = H$ gives 0 volume. So maximum occurs at $\boxed{h = \frac{H}{3}}$

Substitute into $r$: $$r = \frac{R(H - H/3)}{H} = \frac{2R}{3}$$

Step 3: Final Answer

$$\boxed{ \begin{aligned} &\text{Height of cylinder: } h = \frac{H}{3} \\ &\text{Radius of cylinder: } r = \frac{2R}{3} \\ &\text{Maximum Volume: } V = \pi r^2 h = \pi \cdot \left(\frac{2R}{3}\right)^2 \cdot \frac{H}{3} = \frac{4\pi R^2 H}{27} \end{aligned} }$$