1. The problem asks to find the maximum volume of the combined cylindrical tank formed by a small cylinder with radius $r$ and height $h$, and a larger cylinder with radius $2r$ and height $2h$, given the total surface area constraint. 2. From previous parts, we know the surface area of both cylinders combined is $400\pi$ cm², and the height $h$ is expressed in terms of $r$ as: $$h = \frac{400 - 5\pi r^2}{10\pi r}$$ 3. The total volume $V$ of the combined cylinders is: $$V = \pi r^2 h + \pi (2r)^2 (2h) = \pi r^2 h + 8\pi r^2 h = 9\pi r^2 h$$ 4. Substitute $h$ from step 2 into the volume formula: $$V = 9\pi r^2 \times \frac{400 - 5\pi r^2}{10\pi r} = \frac{9 r (400 - 5\pi r^2)}{10}$$ 5. Simplify the volume expression: $$V = \frac{9 r (400 - 5\pi r^2)}{10} = \frac{3600 r - 45 \pi r^3}{10} = 360 r - 4.5 \pi r^3$$ 6. To find the maximum volume, differentiate $V$ with respect to $r$ and set the derivative equal to zero: $$\frac{dV}{dr} = 360 - 13.5 \pi r^2 = 0$$ 7. Solve for $r^2$: $$13.5 \pi r^2 = 360 \implies r^2 = \frac{360}{13.5 \pi} = \frac{80}{3 \pi}$$ 8. Calculate $r$: $$r = \sqrt{\frac{80}{3 \pi}} \approx 2.92 \text{ cm}$$ 9. Substitute $r$ back into the expression for $h$: $$h = \frac{400 - 5\pi (2.92)^2}{10 \pi (2.92)}$$ Calculate $5\pi (2.92)^2 \approx 5 \times 3.1416 \times 8.53 = 134.0$, so $$h = \frac{400 - 134.0}{10 \times 3.1416 \times 2.92} = \frac{266}{91.7} \approx 2.90 \text{ cm}$$ 10. Finally, calculate the maximum volume $V$: $$V = 9 \pi r^2 h = 9 \times 3.1416 \times (2.92)^2 \times 2.90$$ Calculate $(2.92)^2 = 8.53$, so $$V = 9 \times 3.1416 \times 8.53 \times 2.90 \approx 9 \times 3.1416 \times 24.74 = 9 \times 77.7 = 699.3 \text{ cm}^3$$ Answer: The maximum volume of the combined cylindrical tank is approximately $699.3$ cubic centimeters.