1. **State the problem:** We want to find the radius $r$ and height $h$ of an open-top cylindrical can that holds volume $500$ cm³ and minimizes the surface area (material used). 2. **Given:** - Volume constraint: $$V = \pi r^2 h = 500$$ - Surface area to minimize (open top): $$A = \pi r^2 + 2 \pi r h$$ 3. **Express height $h$ in terms of $r$ using the volume:** $$h = \frac{500}{\pi r^2}$$ 4. **Rewrite surface area $A$ as a function of $r$ only:** $$A = \pi r^2 + 2 \pi r \left( \frac{500}{\pi r^2} \right) = \pi r^2 + \frac{1000}{r}$$ 5. **Minimize $A(r)$ by taking the derivative and setting it to zero:** $$\frac{dA}{dr} = 2 \pi r - \frac{1000}{r^2} = 0$$ 6. **Solve for $r$: ** $$2 \pi r = \frac{1000}{r^2}$$ $$2 \pi r^3 = 1000$$ $$r^3 = \frac{1000}{2 \pi} = \frac{500}{\pi}$$ $$r = \left( \frac{500}{\pi} \right)^{\frac{1}{3}}$$ 7. **Find $h$ using $r$: ** $$h = \frac{500}{\pi r^2} = \frac{500}{\pi \left( \frac{500}{\pi} \right)^{\frac{2}{3}}} = 500 \pi^{-1} \left( \frac{\pi}{500} \right)^{\frac{2}{3}} = 500 \pi^{-1} \pi^{\frac{2}{3}} 500^{-\frac{2}{3}} = 500^{1 - \frac{2}{3}} \pi^{-1 + \frac{2}{3}} = 500^{\frac{1}{3}} \pi^{-\frac{1}{3}} = \left( \frac{500}{\pi} \right)^{\frac{1}{3}}$$ 8. **Final answers:** $$\boxed{r = \left( \frac{500}{\pi} \right)^{\frac{1}{3}}}$$ $$\boxed{h = \left( \frac{500}{\pi} \right)^{\frac{1}{3}}}$$ The height equals the radius in the optimal design.