1. **Problem statement:** We want to find the relative dimensions (radius $r$ and height $h$) of a right circular cylinder with a closed top that maximize the volume, given a fixed total surface area. 2. **Formulas and definitions:** - Volume of cylinder: $$V = \pi r^2 h$$ - Surface area of closed cylinder: $$S = 2\pi r^2 + 2\pi r h$$ where $2\pi r^2$ is the area of the two circular ends (top and bottom) and $2\pi r h$ is the lateral surface area. 3. **Given:** Total surface area $S$ is fixed. 4. **Goal:** Maximize $V$ subject to fixed $S$. 5. **Express $h$ in terms of $r$ and $S$:** $$S = 2\pi r^2 + 2\pi r h \implies 2\pi r h = S - 2\pi r^2 \implies h = \frac{S - 2\pi r^2}{2\pi r}$$ 6. **Substitute $h$ into volume formula:** $$V = \pi r^2 \times \frac{S - 2\pi r^2}{2\pi r} = \frac{r}{2}(S - 2\pi r^2) = \frac{S r}{2} - \pi r^3$$ 7. **Maximize $V$ with respect to $r$:** Take derivative and set to zero: $$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2 = 0$$ 8. **Solve for $r$:** $$3\pi r^2 = \frac{S}{2} \implies r^2 = \frac{S}{6\pi} \implies r = \sqrt{\frac{S}{6\pi}}$$ 9. **Find $h$ using $r$:** $$h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S - 2\pi \frac{S}{6\pi}}{2\pi \sqrt{\frac{S}{6\pi}}} = \frac{S - \frac{S}{3}}{2\pi \sqrt{\frac{S}{6\pi}}} = \frac{\frac{2S}{3}}{2\pi \sqrt{\frac{S}{6\pi}}} = \frac{S}{3\pi \sqrt{\frac{S}{6\pi}}}$$ Simplify $h$: $$h = \frac{S}{3\pi} \times \sqrt{\frac{6\pi}{S}} = \frac{S}{3\pi} \times \frac{\sqrt{6\pi}}{\sqrt{S}} = \frac{\sqrt{6\pi} \sqrt{S}}{3\pi} = \sqrt{\frac{2S}{3\pi}}$$ 10. **Ratio of height to radius:** $$\frac{h}{r} = \frac{\sqrt{\frac{2S}{3\pi}}}{\sqrt{\frac{S}{6\pi}}} = \sqrt{\frac{2S}{3\pi} \times \frac{6\pi}{S}} = \sqrt{4} = 2$$ **Answer:** The volume is maximized when the height is twice the radius, i.e., $h = 2r$. This means the relative dimensions that maximize volume for fixed surface area are $h:r = 2:1$.