1. **Problem Statement:** You need to design a right circular cylindrical can with volume 1 liter (1000 cm³) that uses the least material, i.e., minimizes the surface area. 2. **Formulas and Known Values:** - Volume of cylinder: $$V = \pi r^2 h = 1000$$ - Surface area of cylinder: $$A = 2 \pi r^2 + 2 \pi r h$$ 3. **Express Surface Area as a Function of One Variable:** Solve for height $$h$$ from the volume formula: $$h = \frac{1000}{\pi r^2}$$ Substitute into surface area: $$A = 2 \pi r^2 + 2 \pi r \left(\frac{1000}{\pi r^2}\right) = 2 \pi r^2 + \frac{2000}{r}$$ 4. **Find Critical Points by Differentiation:** Differentiate $$A$$ with respect to $$r$$: $$\frac{dA}{dr} = 4 \pi r - \frac{2000}{r^2}$$ Set derivative to zero to find minima: $$0 = 4 \pi r - \frac{2000}{r^2}$$ Multiply both sides by $$r^2$$: $$4 \pi r^3 = 2000$$ Solve for $$r$$: $$r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42$$ cm 5. **Confirm Minimum Using Second Derivative:** Second derivative: $$\frac{d^2A}{dr^2} = 4 \pi + \frac{4000}{r^3} > 0$$ Since this is positive for all $$r > 0$$, the critical point is a minimum. 6. **Find Corresponding Height $$h$$:** Substitute $$r$$ back into height formula: $$h = \frac{1000}{\pi r^2} = 2 \sqrt[3]{\frac{500}{\pi}} = 2r \approx 10.84$$ cm 7. **Conclusion:** The can with minimum surface area has radius $$r \approx 5.42$$ cm and height $$h \approx 10.84$$ cm, meaning the height equals the diameter.