Answer
height $h=2$, radius $r=\sqrt 2$
Work Step by Step
Step 1. We illustrate the problem with the shown figure, where $R=\sqrt 3$ is the radius of the sphere and $r, h$ are the radius and height of the cylinder.
Step 2. Using Pythagorean Theorem, we have $R^2=r^2+(\frac{h}{2})^2$, which gives $r^2=3-(\frac{h}{2})^2$
Step 3. The volume of the cylinder is given by $V=\pi r^2 h=\pi h(3-(\frac{h}{2})^2)=\pi(3h-\frac{1}{4}h^3)$
Step 4. To maximize $V$, let $V'=0$, we have $3-\frac{3}{4}h^2=0$, which gives $h=2$ (positive answer only). Thus $r^2=3-(\frac{h}{2})^2=2$ and $r=\sqrt 2$
Step 5. Check $V''=-(\frac{3}{2})h\lt0$; we confirm the above dimensions give a maximum volume.
