Problem 2. The lighthouse keeper of Finisterre has received communication that there will be a power outage and he must operate the lighthouse using a diesel generator. This generator consumes 6 liters of diesel per hour and an additional half liter each time it needs to be started (initially, it is off). During the 10 hours of the night, the lighthouse cannot be out of service for more than 10 minutes at a time. And when it is operating, it must do so for at least 15 minutes at a time. How many liters of diesel does he need, at a minimum, to comply with the lighthouse operating regulations?
## Solution:

The best way to save fuel is to make 10-minute stops (longer stops are not allowed). With each stop, we save 1 liter of diesel, but we spend half a liter to restart the generator. It is obvious that, in principle, the best savings strategy involves alternating 15-minute operating periods (shorter periods are not allowed) with 10-minute stops. In each "start-15 minutes operating-10 minutes stop" process, 0.5 + 0.25 * 6 = 2 liters of diesel are consumed. To complete the 10 hours of night, we need 600 / 25 = 24 of these processes, so it seems that the minimum amount of fuel required is 48 liters. However, there is a way to save an additional half liter. When the 10 hours of night begin, we can go 10 minutes without turning on the lighthouse. If we then chain 23 "start-15 minutes operating-10 minutes stop" processes, we complete 10 hours minus 15 minutes. We would have to restart the generator and run it for another 15 minutes, consuming 48 liters again. However, if, for example, we make one of the operating periods 30 minutes and "save" the 10-minute intermediate stop for the last 10 minutes of the night, the operating time is the same but we save one start, i.e., half a liter of diesel. Only 47.5 liters are needed.