1. **State the problem:** We want to find the dimensions (radius $r$ and height $h$) of a cylinder that contains a volume of 30 cubic inches and has the smallest possible surface area. 2. **Given formulas:** - Volume: $$V = \pi r^2 h$$ - Surface Area: $$SA = 2\pi r^2 + 2\pi r h$$ 3. **Express $h$ in terms of $r$ using the volume constraint:** $$30 = \pi r^2 h \implies h = \frac{30}{\pi r^2}$$ 4. **Substitute $h$ into the surface area formula:** $$SA = 2\pi r^2 + 2\pi r \left(\frac{30}{\pi r^2}\right) = 2\pi r^2 + \frac{60}{r}$$ 5. **Minimize $SA$ by finding critical points:** Take the derivative of $SA$ with respect to $r$: $$\frac{d(SA)}{dr} = 4\pi r - \frac{60}{r^2}$$ 6. **Set derivative equal to zero to find minimum:** $$4\pi r - \frac{60}{r^2} = 0$$ 7. **Solve for $r$:** Multiply both sides by $r^2$: $$4\pi r^3 - 60 = 0$$ $$4\pi r^3 = 60$$ $$r^3 = \frac{60}{4\pi} = \frac{15}{\pi}$$ $$r = \sqrt[3]{\frac{15}{\pi}}$$ 8. **Find $h$ using $r$:** $$h = \frac{30}{\pi r^2} = \frac{30}{\pi \left(\sqrt[3]{\frac{15}{\pi}}\right)^2} = \frac{30}{\pi \left(\frac{15}{\pi}\right)^{2/3}} = 30 \pi^{-1} \times \left(\frac{\pi}{15}\right)^{2/3} = 30 \times \pi^{-1} \times \pi^{2/3} \times 15^{-2/3} = 30 \times \pi^{-1/3} \times 15^{-2/3}$$ Simplify: $$h = 30 \times \frac{1}{\pi^{1/3}} \times \frac{1}{15^{2/3}} = 30 \times \frac{1}{\pi^{1/3} \times 15^{2/3}}$$ 9. **Final answer:** - Radius: $$r = \sqrt[3]{\frac{15}{\pi}}$$ - Height: $$h = \frac{30}{\pi r^2} = 30 \times \frac{1}{\pi^{1/3} \times 15^{2/3}}$$ These dimensions minimize the surface area for a cylinder containing 30 cubic inches of volume.