7.5. Andrey and Seva are going to visit Borya. Andrey is at point $A$, and Borya is at point $B$, 30 km away from point $A$ along a straight highway. Seva is at point $C$, exactly halfway between $A$ and $B$. The friends decided to leave simultaneously: Andrey on a bicycle, and Seva on foot, but Andrey will leave the bicycle at an agreed location so that Seva can use it (Andrey will finish the journey on foot). The boys move at a speed of 20 km/hour on the bicycle and 5 km/hour on foot. Where should the bicycle be left so that the friends can arrive at Borya's place together as soon as possible?
Answer: 5 km before point $B$. Solution. Let $a=15$ (km), $u=5$ (km/h), $v=20$ (km/h). Let $x$ (km) be the distance from point $B$ to the place where the bicycle is left. Then Andrei's travel time is $t_{A}=\frac{2 a-x}{v}+\frac{x}{u}$, and Seva's travel time is $t_{C}=\frac{a-x}{u}+\frac{x}{v}$. We need to find such a value of $x$ for which the maximum of the two times $t_{A}$ and $t_{C}$ is as small as possible (if, for example, $t_{A}t_{C}$ and in each of these cases it is easy to show how to improve the time by leaving the bicycle a little closer or, respectively, a little further from the given place). Without proving this fact, the solution is rated less than half of the maximum score.