Solution

Let's first consider the speed skater. The centripetal acceleration of the skater is provided by the static friction force against the ice:
$$F_{\text{fr}} = \mu N$$
Since the skater does not move in the vertical direction, the normal reaction force $N$ is equal in magnitude to the gravity force $Mg$. Therefore, the maximum friction force is:
$$F_{\text{fr}} = \mu Mg$$
According to Newton's second law for circular motion, the maximum speed is determined by:
$$\frac{Mv^2}{R} = \mu Mg \implies v = \sqrt{\mu g R}$$
When making a turn, the skater covers a distance proportional to the radius, $S \sim R$ (for example, for a semicircle $S = \pi R$). The time taken to complete the turn is:
$$t = \frac{S}{v} = \frac{\pi R}{\sqrt{\mu g R}} = \pi\sqrt{\frac{R}{\mu g}}$$
The larger the radius $R$ of the circle along which the skater moves, the longer this time. Although the maximum speed increases with the turning radius ($\sim \sqrt{R}$), the distance traveled increases even more rapidly ($\sim R$). This is why the speed skater tries to take the turn as close to the inner edge as possible to minimize their time.

For the cyclist on the velodrome, the mechanics change due to the banked track. The angle of inclination increases as one moves further from the inner edge. On a steeply banked track, the horizontal component of the normal reaction force provides a massive centripetal force, allowing for much higher speeds ($v \sim \sqrt{g R \tan\alpha}$). Because the steepness ($\tan\alpha$) of the velodrome increases towards the outer edge, the maximum permissible speed increases much faster than $\sqrt{R}$, which more than compensates for the longer path distance $S$. Thus, the cyclist minimizes their time by riding higher up on the steeper, outer part of the track.