1. **Problem statement:** A right triangle with hypotenuse $\sqrt{3}$ meters is revolved about one of its legs to form a right circular cone. We need to find the radius $r$, height $h$, and volume $V$ of the cone with the greatest volume. 2. **Setup and formula:** Let the legs of the right triangle be $r$ and $h$. The hypotenuse is given by the Pythagorean theorem: $$r^2 + h^2 = (\sqrt{3})^2 = 3$$ The volume of a cone is: $$V = \frac{1}{3} \pi r^2 h$$ 3. **Express volume in one variable:** From the Pythagorean relation: $$h = \sqrt{3 - r^2}$$ Substitute into volume formula: $$V(r) = \frac{1}{3} \pi r^2 \sqrt{3 - r^2}$$ 4. **Maximize volume:** We find the critical points by differentiating $V(r)$ with respect to $r$ and setting it to zero. Calculate derivative: $$V'(r) = \frac{1}{3} \pi \left(2r \sqrt{3 - r^2} + r^2 \frac{d}{dr} \sqrt{3 - r^2} \right)$$ Compute derivative inside: $$\frac{d}{dr} \sqrt{3 - r^2} = \frac{-r}{\sqrt{3 - r^2}}$$ So: $$V'(r) = \frac{1}{3} \pi \left(2r \sqrt{3 - r^2} - \frac{r^3}{\sqrt{3 - r^2}} \right) = \frac{\pi r}{3 \sqrt{3 - r^2}} \left(2(3 - r^2) - r^2 \right) = \frac{\pi r}{3 \sqrt{3 - r^2}} (6 - 3r^2)$$ Set $V'(r) = 0$: $$\frac{\pi r}{3 \sqrt{3 - r^2}} (6 - 3r^2) = 0$$ This gives: - $r = 0$ (no volume) - $6 - 3r^2 = 0 \Rightarrow r^2 = 2 \Rightarrow r = \sqrt{2}$ (valid since $r^2 \leq 3$) 5. **Find corresponding height:** $$h = \sqrt{3 - r^2} = \sqrt{3 - 2} = 1$$ 6. **Calculate maximum volume:** $$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2)(1) = \frac{2\pi}{3}$$ **Final answers:** - Radius $r = \sqrt{2}$ meters - Height $h = 1$ meter - Maximum volume $V = \frac{2\pi}{3}$ cubic meters