$7 \cdot 71$ A certain truck can only carry $L$ liters of gasoline, with which it can travel $a$ kilometers. Now it needs to travel $d=\frac{4}{3} a$ kilometers to a certain place, with no refueling stations along the way, but it can transport gasoline to any location along the road and store it for later use. Assuming there is only this one truck, how should it travel to reach the destination and use the least amount of gasoline? If the distance to the destination is $d=\frac{23}{15} a$ kilometers, how should it proceed?
[Solution] We apply the following general principle: If $P$ is any point on the route, the total amount of gasoline transported to point $P$ by the car, once or multiple times, must never be less than the total fuel consumption for the car's travel after point $P$.

As shown in the figure, let $O$ be the starting point, $X$ the destination, and the length of $O X$ be $\frac{4}{3} a = a + \frac{1}{3} a$ kilometers.

Consider the point $M$ on the route, which is $a$ kilometers away from $X$. The car must travel at least once between $M$ and $X$, thus consuming at least $L$ liters of fuel. According to the above principle, at least $L$ liters of fuel must be transported to point $M$.

To transport $L$ liters of fuel to point $M$, taking fuel from point $O$ once is insufficient, as some fuel will be consumed along the way. Therefore, fuel must be taken at least twice, meaning the car must travel back and forth between $O$ and $M$ at least three times.

Since $O M = \frac{1}{3} a$, traveling back and forth three times between $O$ and $M$ is exactly $a$ kilometers, consuming $L$ liters of fuel. Thus, the car can depart from $O$ with $L$ liters of fuel, store $\frac{1}{3} L$ liters of fuel at point $M$, and return to point $O$. Then, it can take another $L$ liters of fuel from $O$, consume $\frac{1}{3} L$ liters on the way to $M$, leaving $\frac{2}{3} L$ liters, and add the $\frac{1}{3} L$ liters stored at $M$, totaling $L$ liters, which is enough to reach point $X$.
Therefore, a total of $2 L$ liters of fuel is required, which is clearly the minimum fuel consumption.
If $O X = \frac{23}{15} a$ kilometers, since
$$
\frac{23}{15} a = a + \frac{1}{3} a + \frac{1}{5} a.
$$

A point $M_{1}$ can be chosen on $O X$ such that $M_{1} X = \frac{4}{3} a$ kilometers. According to the above discussion, traveling from $M_{1}$ to $X$ requires at least $2 L$ liters of fuel. Therefore, at least $2 L$ liters of fuel must be transported to $M_{1}$. Clearly, taking fuel from $O$ twice is insufficient, and the car must travel back and forth between $O$ and $M_{1}$ five times, as $O M_{1} = \frac{1}{5} a$ kilometers, and traveling back and forth five times consumes $L$ liters of fuel. Therefore, traveling from $O$ to $X$ requires at least $3 L$ liters of fuel.

For this purpose, a first fuel storage station can be set up at $M_{1}$, and a second fuel storage station at $M$, which is $a$ kilometers away from $X$. Three trips from $O$ can be made, taking a total of $3 L$ liters of fuel, and traveling back and forth five times between $O$ and $M_{1}$ consumes $L$ liters of fuel, leaving $2 L$ liters of fuel to be stored at $M_{1}$, which is exactly enough for the trip from $M_{1}$ to $X$.