13.55 Now we need to transport gasoline from oil production site $A$ to site $B$. It is known that the maximum oil capacity of the oil tanker is exactly equal to the oil consumption required for a round trip between $A$ and $B$ (the oil tanker must return to $A$). Therefore, the oil cannot be directly transported to $B$. If a transfer station $C$ is set up between $A$ and $B$, then
(1) When $C$ is located at $\frac{1}{3} AB$ from $A$, how can we transport the gasoline from $A$ to $B$ in the most economical way, so that $B$ receives the maximum amount of gasoline? What is the oil transportation efficiency $\left(\frac{\text{amount of gasoline received at } B}{\text{amount of gasoline transported from } A}\right)$ in this case?
(2) Where should the transfer station be set up to maximize the oil transportation efficiency? What is the oil transportation efficiency in this case?
[Solution] (1) $A C=\frac{1}{3} A B$, then $C B=\frac{2}{3} A B$.
Let the maximum oil capacity of the tanker be $a$.
Since the oil consumption for a round trip from $A$ to $B$ is $a$, the oil consumption for a one-way trip from $A$ to $B$ is $\frac{a}{2}$.
The oil consumption for a one-way trip from $A$ to $C$ is $\frac{1}{3} \cdot \frac{a}{2}=\frac{a}{6}$.
The oil consumption for a one-way trip from $C$ to $B$ is $\frac{2}{3} \cdot \frac{a}{2}=\frac{a}{3}$.
The oil consumption for a round trip from $C$ to $B$ is $2 \cdot \frac{a}{3}=\frac{2 a}{3}$.
Therefore, if the tanker is fully loaded with $a$ at $C$ and travels to $B$, the maximum amount of oil that can be unloaded is $\frac{a}{3}$. Thus, the most economical method is as follows:
(1) The first trip: The tanker loads $a$ from $A$ to $C$, leaving $\frac{5}{6} a$, unloads $\frac{2}{3} a$, and has $\frac{1}{6} a$ remaining, which is just enough for the return trip to $A$. (At this point, $C$ has $\frac{2}{3} a$ of oil stored).
(2) The second trip: The tanker loads $a$ from $A$ to $C$, leaving $\frac{5}{6} a$, then loads $\frac{1}{6} a$ and continues to $B$, using $\frac{a}{3}$, leaving $\frac{2 a}{3}$, unloads $\frac{a}{3}$, and uses the remaining $\frac{a}{3}$ to return to $C$, which is just enough for the return trip to $A$ after loading $\frac{a}{6}$. (At this point, $C$ has $\frac{a}{3}$ of oil stored, and $B$ has $\frac{a}{3}$ of oil stored).
(3) The third trip is the same as the second, so the oil at $C$ is used up, and $B$ has $\frac{2}{3} a$ of oil stored.
At this point, the oil transportation efficiency is $\frac{\frac{2 a}{3}}{3 a}=\frac{2}{9} \approx 22.2 \%$.
(2)(1) If $A D=\frac{2}{3} A B$, then by (1), the best plan is also to transport the oil three times, with $B$ storing $\frac{2}{3} a$ of oil, and the oil transportation efficiency is also $22.2 \%$.
(2) By symmetry, consider setting up a storage station at the midpoint $E$ of $A B$, then the best plan is: transport the oil twice, with $B$ storing $\frac{a}{2}$ of oil, at which point the oil transportation efficiency is
$$
\frac{\frac{a}{2}}{2 a}=\frac{1}{4}=25 \% .
$$

Therefore, setting up the storage station at the midpoint $E$ results in the highest oil transportation efficiency of $25 \%$.