1. **State the problem:** A manufacturer wants to design a cylindrical can that holds a volume of 1500 cm³. We need to find the 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 cylinder: $$A = 2\pi r h + 2\pi r^2$$ 3. **Given:** $$V = 1500 = \pi r^2 h$$ We want to minimize $$A = 2\pi r h + 2\pi r^2$$ 4. **Express $h$ in terms of $r$ using volume:** $$h = \frac{1500}{\pi r^2}$$ 5. **Substitute $h$ into surface area formula:** $$A = 2\pi r \left(\frac{1500}{\pi r^2}\right) + 2\pi r^2 = \frac{3000}{r} + 2\pi r^2$$ 6. **Minimize $A$ by finding critical points:** Take derivative with respect to $r$: $$\frac{dA}{dr} = -\frac{3000}{r^2} + 4\pi r$$ 7. **Set derivative to zero to find minimum:** $$-\frac{3000}{r^2} + 4\pi r = 0$$ 8. **Solve for $r$:** $$4\pi r = \frac{3000}{r^2}$$ Multiply both sides by $r^2$: $$4\pi r^3 = 3000$$ Divide both sides by $4\pi$: $$r^3 = \frac{3000}{4\pi} = \frac{750}{\pi}$$ 9. **Calculate $r$:** $$r = \sqrt[3]{\frac{750}{\pi}}$$ 10. **Calculate $h$ using $r$:** $$h = \frac{1500}{\pi r^2}$$ 11. **Final answer:** The radius and height that minimize the surface area are: $$r = \sqrt[3]{\frac{750}{\pi}} \approx 6.203 \text{ cm}$$ $$h = \frac{1500}{\pi (6.203)^2} \approx 12.407 \text{ cm}$$ These dimensions minimize the material used for the can.