1. **Problem statement:** A closed cylindrical oil can must hold a volume of 1 litre (1000 cm³). We want to find the dimensions (radius $r$ and height $h$) that minimize the surface area (material used). 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:** $$V = 1000 = \pi r^2 h$$ 4. **Express $h$ in terms of $r$ using volume:** $$h = \frac{1000}{\pi r^2}$$ 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 $S$ 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 $r$:** $$4\pi r = \frac{2000}{r^2}$$ $$4\pi r^3 = 2000$$ $$r^3 = \frac{2000}{4\pi} = \frac{500}{\pi}$$ $$r = \sqrt[3]{\frac{500}{\pi}}$$ 8. **Calculate $h$ using $r$:** $$h = \frac{1000}{\pi r^2} = \frac{1000}{\pi \left(\sqrt[3]{\frac{500}{\pi}}\right)^2} = \frac{1000}{\pi \left(\frac{500}{\pi}\right)^{2/3}}$$ 9. **Final dimensions:** $$r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \text{ cm}$$ $$h = \frac{1000}{\pi (5.42)^2} \approx 10.84 \text{ cm}$$ These dimensions minimize the material used for the can.