$\textbf{Passing the Bridge}$
It is a dark and stormy night. Four people must evacuate from an island to the mainland. The only link is a narrow bridge which allows passage of two people at a time. Moreover, the bridge must be illuminated, and the four people have only one lantern among them. After each passage to the mainland, if there are still people on the island, someone must bring the lantern back. When they cross the bridge individually, the four people take $2$, $4$, $8$ and $16$ minutes, respectively. Crossing the bridge in pairs, the slower speed is used. What is the minimum time for the entire evacuation?
$\textbf{Answer}$
The minimum time is $30$ minutes.
$\textbf{Analysis}$
It is clear that a total of $5$ passages are needed, among which three trips are going to the mainland and two are back to the island. In order to minimize the total time, it is wise to pair people with similar speed. Because if a very fast walker is paired with a very slow walker, the advantage of the former is wasted. Meanwhile, if someone has to travel solo (i.e. return to the mainland with the lamp), it will be the best to let the fastest person available go. This means, it is better to let the fast people go to the mainland first, so that they can take the lamp back as necessary.
With these guidelines in mind, we design the following arrangement:
- The fastest two people go to the mainland first. This will cost $4$ minutes.
- The fastest person comes back to the island with the lantern. This will cost $2$ minutes.
- The lowest two people go to the mainland. This will cost $16$ minutes.
- The faster person on the mainland returns to the island with the lantern. This will cost $4$ minutes.
- The remaining two people go to mainland. This will cost $4$ minutes.
Hence, the total will be $4+16+2+4+4=\boxed{30}$ minutes.
To show this is the minimum, we note that two solo travels cost at least $2+4=6$ minutes because they cannot be done by the the same person. Three two-people passages cost at least $16+4+4=24$ minutes. There is no way to get rid of the $16$-minute travel, but we can avoid a $8$-minute trip by pair this $8$-minute person with the slowest person. Meanwhile, it is impossible to have a $2$-minute paired trip because whoever travels with the fastest person will slow him down. That means the minimal time should be at least $6+24=30$ minutes.
$\textbf{Note}$
A common approach to solve a question asking for the minimum (or maximum) value is to first seek the theoretical lower bound (or upper bound) and then present a solution which archives this value.