1. **Problem Statement:** We need to find the radius $r$ and cylindrical height $h$ of a reservoir shaped as a right circular cylinder with a hemispherical top that maximizes the volume, given a total exterior surface area constraint of 700 m². 2. **Formulas and Definitions:** - Volume $V$ of the reservoir = volume of cylinder + volume of hemisphere: $$V = \pi r^2 h + \frac{2}{3} \pi r^3$$ - Surface area $S$ (exterior) = lateral surface area of cylinder + surface area of hemisphere (no base area since it's underground): $$S = 2 \pi r h + 2 \pi r^2$$ - Constraint: $S = 700$ 3. **Express $h$ in terms of $r$ using the surface area constraint:** $$2 \pi r h + 2 \pi r^2 = 700 \implies 2 \pi r h = 700 - 2 \pi r^2 \implies h = \frac{700 - 2 \pi r^2}{2 \pi r} = \frac{700}{2 \pi r} - r = \frac{350}{\pi r} - r$$ 4. **Substitute $h$ into the volume formula:** $$V = \pi r^2 \left(\frac{350}{\pi r} - r\right) + \frac{2}{3} \pi r^3 = 350 r - \pi r^3 + \frac{2}{3} \pi r^3 = 350 r - \frac{1}{3} \pi r^3$$ 5. **Maximize $V$ by finding critical points:** Take derivative with respect to $r$: $$\frac{dV}{dr} = 350 - \pi r^2$$ Set derivative to zero: $$350 - \pi r^2 = 0 \implies r^2 = \frac{350}{\pi} \implies r = \sqrt{\frac{350}{\pi}}$$ 6. **Calculate $r$ numerically:** $$r \approx \sqrt{\frac{350}{3.1416}} \approx \sqrt{111.41} \approx 10.55 \text{ m}$$ 7. **Calculate $h$ using $r$:** $$h = \frac{350}{\pi \times 10.55} - 10.55 \approx \frac{350}{33.15} - 10.55 \approx 10.56 - 10.55 = 0.01 \text{ m}$$ 8. **Interpretation:** The optimal height $h$ is approximately 0.01 m, meaning the reservoir is almost a hemisphere with a very short cylindrical section. 9. **Final answer:** - Radius $r \approx 10.55$ m - Cylindrical height $h \approx 0.01$ m These dimensions maximize the storage volume under the surface area constraint of 700 m².