1. **State the problem:** We need to find the dimensions (radius $r$ and height $h$) of a cylindrical can that holds a volume of 1500 cm³ and minimizes the surface area (material used). 2. **Given formulas:** - Surface area: $$A = 2\pi rh + 2\pi r^2$$ - Volume: $$V = \pi r^2 h$$ 3. **Express $h$ in terms of $r$ using the volume constraint:** $$V = \pi r^2 h = 1500 \implies h = \frac{1500}{\pi r^2}$$ 4. **Substitute $h$ into the surface area formula:** $$A = 2\pi r \left(\frac{1500}{\pi r^2}\right) + 2\pi r^2 = \frac{3000}{r} + 2\pi r^2$$ 5. **Minimize $A$ by finding critical points:** Take derivative with respect to $r$: $$\frac{dA}{dr} = -\frac{3000}{r^2} + 4\pi r$$ 6. **Set derivative to zero to find minimum:** $$-\frac{3000}{r^2} + 4\pi r = 0$$ 7. **Solve for $r$:** $$4\pi r = \frac{3000}{r^2}$$ $$4\pi r^3 = 3000$$ $$r^3 = \frac{3000}{4\pi} = \frac{750}{\pi}$$ $$r = \sqrt[3]{\frac{750}{\pi}}$$ 8. **Calculate $r$ numerically:** $$r \approx \sqrt[3]{238.732} \approx 6.2 \text{ cm}$$ 9. **Find $h$ using $r$:** $$h = \frac{1500}{\pi (6.2)^2} \approx \frac{1500}{\pi \times 38.44} \approx \frac{1500}{120.78} \approx 12.42 \text{ cm}$$ 10. **Conclusion:** The can dimensions that minimize material are approximately: - Radius $r \approx 6.2$ cm - Height $h \approx 12.42$ cm These dimensions minimize the surface area for the given volume.