1. **Problem statement:** We need to find the dimensions (radius $r$ and height $h$) of a closed right circular cylinder that holds a volume of 1 litre (1000 cubic centimeters) and uses the minimum amount of material, i.e., has minimum surface area. 2. **Formulas:** - Volume of cylinder: $$V = \pi r^2 h$$ - Surface area of closed cylinder: $$S = 2\pi r^2 + 2\pi r h$$ 3. **Given:** - Volume $V = 1000$ cm³ 4. **Express height $h$ in terms of $r$ using volume:** $$h = \frac{V}{\pi r^2} = \frac{1000}{\pi r^2}$$ 5. **Substitute $h$ into surface area formula:** $$S = 2\pi r^2 + 2\pi r \times \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r}$$ 6. **Minimize surface area $S$ by differentiating with respect to $r$ and setting derivative to zero:** $$\frac{dS}{dr} = 4\pi r - \frac{2000}{r^2} = 0$$ 7. **Solve for $r$:** $$4\pi r = \frac{2000}{r^2} \implies 4\pi r^3 = 2000 \implies r^3 = \frac{2000}{4\pi} = \frac{500}{\pi}$$ $$r = \sqrt[3]{\frac{500}{\pi}}$$ 8. **Calculate $r$ approximately:** $$r \approx \sqrt[3]{159.15} \approx 5.42 \text{ cm}$$ 9. **Calculate $h$ using $r$:** $$h = \frac{1000}{\pi (5.42)^2} \approx \frac{1000}{\pi \times 29.38} \approx \frac{1000}{92.28} \approx 10.83 \text{ cm}$$ **Final answer:** - Radius $r \approx 5.42$ cm - Height $h \approx 10.83$ cm These dimensions minimize the material used for the can while holding 1 litre volume.