1. **Problem Statement:** We have a cylinder inscribed in a sphere of radius $R=3$. We want to find the height $h$ of the cylinder that maximizes its volume. 2. **Formula for Volume of Cylinder:** The volume $V$ of a cylinder is given by: $$V = \pi r^2 h$$ where $r$ is the radius of the cylinder's base and $h$ is its height. 3. **Relating Cylinder Dimensions to Sphere:** Since the cylinder is inscribed in the sphere, the diagonal from the center of the cylinder's base to the top edge lies on the sphere's radius. Using the Pythagorean theorem: $$r^2 + \left(\frac{h}{2}\right)^2 = R^2$$ 4. **Express $r$ in terms of $h$ and $R$:** $$r = \sqrt{R^2 - \left(\frac{h}{2}\right)^2} = \sqrt{9 - \frac{h^2}{4}}$$ 5. **Volume as a function of $h$:** $$V(h) = \pi \left(9 - \frac{h^2}{4}\right) h = \pi \left(9h - \frac{h^3}{4}\right)$$ 6. **Maximize $V(h)$ by finding critical points:** Take derivative: $$V'(h) = \pi \left(9 - \frac{3h^2}{4}\right)$$ Set derivative to zero: $$9 - \frac{3h^2}{4} = 0$$ 7. **Solve for $h$:** $$9 = \frac{3h^2}{4}$$ Multiply both sides by 4: $$36 = 3h^2$$ Divide both sides by 3: $$12 = h^2$$ Take square root: $$h = \sqrt{12} = 2\sqrt{3}$$ 8. **Check the options:** The height that maximizes volume is $h = 2\sqrt{3}$, which corresponds to option (d). **Final answer:** $h = 2\sqrt{3}$