The optimal configuration is when the two ends $X$ and $Y$ of the semicircle lie on the arc of the quarter circle. Let $O$ and $P$ be the centers of the quarter circle and semicircle, respectively. Also, let $M$ and $N$ be the points where the semicircle is tangent to the radii of the quartercircle. Let $r$ be the radius of the semicircle. Since $P M=P N, P M O N$ is a square and $O P=\sqrt{2} r$. By the Pythagorean theorem on triangle $O P X, 1=2 r^{2}+r^{2}$, so $r=1 / \sqrt{3}$. The area of the semicircle is therefore $\frac{\pi}{2} \frac{1}{3}=\frac{\pi}{6}$.