Approach: Put the semicircle on coordinates and parametrise the top corner by an angle, so the area collapses to a single sine term whose maximum is obvious.
Step 1: Let the semicircle have radius $r$ with its diameter on the x-axis and centre at the origin. By symmetry the rectangle's base lies on the diameter, with bottom corners $(\pm x,0)$ and top corners $(\pm x,y)$ on the arc, so $x^2+y^2=r^2$.
Step 2: Width $=2x$, height $=y$, so Area $=2xy$. Put $x=r\cos\theta,\ y=r\sin\theta$; then Area $=2r^2\sin\theta\cos\theta=r^2\sin 2\theta$.
Step 3: $\sin 2\theta$ is greatest when $2\theta=90^\circ$, i.e. $\theta=45^\circ$, giving $x=y=\dfrac{r}{\sqrt2}$.
Step 4: Larger side (width) $=2x=r\sqrt2$; smaller side (height) $=y=\dfrac{r}{\sqrt2}$. Ratio $=\dfrac{r\sqrt2}{r/\sqrt2}=2$.
Answer: The ratio of the larger to the smaller side is $2:1$.