1. **State the problem:** We want to find the values of $r$ and $h$ that maximize the total volume of two stacked cylinders: a small cylinder with radius $r$ and height $h$, and a large cylinder with radius $2r$ and height $2h$. The total surface area of both cylinders is $400\pi$ cm². 2. **Recall formulas:** - Surface area of a closed cylinder: $S = 2\pi r^2 + 2\pi r h$ - Volume of a cylinder: $V = \pi r^2 h$ 3. **Given:** - Small cylinder surface area: $S_s = 2\pi r^2 + 2\pi r h$ - Large cylinder surface area: $S_l = 2\pi (2r)^2 + 2\pi (2r)(2h) = 8\pi r^2 + 8\pi r h$ - Total surface area: $S_s + S_l = 400\pi$ 4. **Express the surface area equation:** $$2\pi r^2 + 2\pi r h + 8\pi r^2 + 8\pi r h = 400\pi$$ Simplify: $$10\pi r^2 + 10\pi r h = 400\pi$$ Divide both sides by $10\pi$: $$r^2 + r h = 40$$ 5. **Express $h$ in terms of $r$:** $$r h = 40 - r^2 \implies h = \frac{40 - r^2}{r} = \frac{40}{r} - r$$ 6. **Express total volume $V$ in terms of $r$:** - Volume of small cylinder: $V_s = \pi r^2 h$ - Volume of large cylinder: $V_l = \pi (2r)^2 (2h) = 8\pi r^2 h$ - Total volume: $$V = V_s + V_l = \pi r^2 h + 8\pi r^2 h = 9\pi r^2 h$$ Substitute $h$: $$V = 9\pi r^2 \left(\frac{40}{r} - r\right) = 9\pi (40 r - r^3) = 360\pi r - 9\pi r^3$$ 7. **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: $$360\pi - 27\pi r^2 = 0 \implies 360 = 27 r^2 \implies r^2 = \frac{360}{27} = \frac{40}{3}$$ $$r = \sqrt{\frac{40}{3}} \approx 3.65$$ 8. **Find $h$ using $r$:** $$h = \frac{40}{3.65} - 3.65 \approx 10.96 - 3.65 = 7.31$$ 9. **Verify maximum by second derivative test:** $$\frac{d^2V}{dr^2} = -54\pi r$$ At $r=3.65$, second derivative is negative, confirming a maximum. **Final answers:** $$r \approx 3.65 \text{ cm}, \quad h \approx 7.31 \text{ cm}$$