1. **State the problem:** We want to find the values of $r$ and $h$ that maximize the total volume of the combined cylindrical cans, given the surface area constraint. 2. **Recall given information:** - Small cylinder radius = $r$, height = $h$ - Large cylinder radius = $2r$, height = $2h$ - Total surface area of both cylinders = $400\pi$ 3. **Surface area formulas:** - Surface area of a cylinder = $2\pi r^2 + 2\pi r h$ - For small cylinder: $S_1 = 2\pi r^2 + 2\pi r h$ - For large cylinder: $S_2 = 2\pi (2r)^2 + 2\pi (2r)(2h) = 8\pi r^2 + 8\pi r h$ 4. **Total surface area constraint:** $$S_1 + S_2 = 2\pi r^2 + 2\pi r h + 8\pi r^2 + 8\pi r h = 10\pi r^2 + 10\pi r h = 400\pi$$ Divide both sides by $\pi$: $$10 r^2 + 10 r h = 400$$ Simplify: $$r^2 + r h = 40$$ 5. **Express $h$ in terms of $r$:** $$r h = 40 - r^2 \implies h = \frac{40 - r^2}{r} = \frac{40}{r} - r$$ 6. **Volume formulas:** - Volume of small cylinder: $V_1 = \pi r^2 h$ - Volume of large cylinder: $V_2 = \pi (2r)^2 (2h) = 8 \pi r^2 h$ - Total volume: $$V = V_1 + V_2 = \pi r^2 h + 8 \pi r^2 h = 9 \pi r^2 h$$ 7. **Substitute $h$ into $V$:** $$V = 9 \pi r^2 \left(\frac{40}{r} - r\right) = 9 \pi r^2 \left(\frac{40}{r}\right) - 9 \pi r^2 (r) = 9 \pi (40 r - r^3) = 360 \pi r - 9 \pi r^3$$ 8. **Maximize $V$ by finding critical points:** Take derivative with respect to $r$: $$\frac{dV}{dr} = 360 \pi - 27 \pi r^2$$ Set derivative to zero: $$360 \pi - 27 \pi r^2 = 0 \implies 360 = 27 r^2 \implies r^2 = \frac{360}{27} = \frac{40}{3}$$ 9. **Calculate $r$:** $$r = \sqrt{\frac{40}{3}} \approx 3.65$$ 10. **Calculate $h$ using $h = \frac{40}{r} - r$:** $$h = \frac{40}{3.65} - 3.65 \approx 10.96 - 3.65 = 7.31$$ 11. **Verify maximum by second derivative test:** $$\frac{d^2V}{dr^2} = -54 \pi r < 0$$ for $r > 0$, so $V$ is maximized at $r \approx 3.65$. **Final answers:** $$r \approx 3.65 \text{ cm}, \quad h \approx 7.31 \text{ cm}$$