12.18 A pedestrian set out for a walk from point $A$ at a speed of $v$ km/h. After he had walked 6 km from $A$, a cyclist set out after him from $A$ at a speed 9 km/h greater than the pedestrian's speed. When the cyclist caught up with the pedestrian, they turned back and returned to $A$ together at a speed of 4 km/h. For what value of $v$ will the pedestrian's walking time be the shortest?
12.18 The time it takes for the cyclist to catch up with the pedestrian is $\frac{6}{9}=\frac{2}{3}$ (hours). Before the meeting, the pedestrian had been walking for $\frac{6}{v}+\frac{2}{3}$ (hours) and covered $v\left(\frac{6}{v}+\frac{2}{3}\right)$ (km). They covered this same distance on the way back at a constant speed of 4 km/h and spent $\frac{v\left(\frac{6}{v}+\frac{2}{3}\right)}{4}$ (hours). Therefore, the total time spent by the pedestrian on the walk can be expressed by the function

$$
t(v)=\frac{6}{v}+\frac{2}{3}+\frac{v\left(\frac{6}{v}+\frac{2}{3}\right)}{4}=\frac{6}{v}+\frac{2}{3}+\frac{3}{2}+\frac{v}{6}=\frac{v^{2}+13 v+36}{6 v}
$$

where $v>0$. We find

$$
t^{\prime}(v)=\frac{(2 v+13) v-\left(v^{2}+13 v+36\right)}{6 v^{2}}=\frac{v^{2}-36}{6 v^{2}}
$$

from which $t^{\prime}(v)=0$ when $v=6$. It is easy to verify that $v=6$ is a minimum point of the function $t(v)$. Therefore, the pedestrian will spend the least amount of time on the walk if they initially walk at a speed of 6 km/h.

Answer: 6 km/h