8.5. When fully fueled, a motorboat can travel exactly 40 km upstream or exactly 60 km downstream. What is the greatest distance the motorboat can travel along the river if the fuel must be enough for the round trip, back to the starting point? Justify your answer.
# Solution:

Method 1. Let's launch a raft along with the boat, which will move with the current, and observe the boat from this raft. Then, no matter which direction the boat goes, upstream or downstream, it will be at the same distance from the raft by the time the fuel runs out. This means that the raft will be at the midpoint of the segment between the extreme possible positions of the boat, i.e., (60 - 40) : 2 = 10 km downstream from the starting point. The boat will be 50 km away from it. This means the boat's speed is 50 : 10 = 5 times greater than the speed of the current. This means that downstream, the boat travels at a speed 6 times greater than the current, and upstream at a speed 4 times greater. Thus, downstream the boat travels 1.5 times faster than upstream. To cover the same distance there and back, we need to travel upstream 1.5 times longer than downstream, i.e., 3/5 of the total travel time. During this time, the boat will be 30 km away from the raft, and the raft will be carried 6 km downstream by the river. This means the boat can travel 24 km in one direction.

Method 2. Let the speed of the boat be \( x \) km/h, the current speed be \( -y \) km/h, and the fuel reserve in the boat is enough for \( t \) hours of travel. The problem conditions can be written as the system

\[
\left\{\begin{array}{l}
(x+y) \cdot t=60 \\
(x-y) \cdot t=40
\end{array}\right.
\]

By adding the equations of the system term by term, we get \( x t = 50 \), and by subtracting them term by term, we get \( y t = 10 \). Suppose we now set off upstream and travel for such a time \( T \) that in the remaining time \( t - T \) we return to the starting point. This means that the condition \((x-y) T = (x+y)(t-T)\) must be satisfied. Taking into account the first equation of the system, we get from it that \( x T = 30 \). Since \( x t = 50 \), then \( T = 0.6 t \). Then, during time \( T \), the boat will be at a distance from the starting point equal to \((x-y) \cdot 0.6 t = 0.6 \cdot 40 = 24\) km.

Answer: 24 km.

Grading Recommendations:

| present in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| There is an arithmetic error in the correct solution, possibly leading to an incorrect answer | 6 points |
| The system of equations correctly describing the conditions of the problem is obtained (but not solved) | 1 point |
| Incorrect solution or its absence | 0 points |