11. (10 points) On the weekend, Xiaohu goes boating in the park, departing from the boat rental place at 9:00 AM, and plans to return to the boat rental place by no later than 11:00 AM. It is known that the boat rental place is in the middle of the river, the river channel is straight, and the river current speed is 1.5 kilometers/hour. Every half hour of rowing, Xiaohu needs to rest for ten minutes, allowing the boat to drift with the current. The boat's speed in still water is 3 kilometers/hour; Question: How far can Xiaohu's boat be from the boat rental place at the farthest?
【Analysis】Xiaohu's total time for rowing is 120 minutes. He rows for 30 minutes and rests for 10 minutes, with a cycle of 40 minutes. Therefore, it can be divided into 3 periods of 30 minutes of rowing and 3 periods of 10 minutes of rest. When rowing, the ratio of the boat's speed downstream to upstream is $4.5: 1.5=3: 1$. Since Xiaohu needs to row as far away from the boat rental point as possible, he can only change the direction of rowing once, and it should be at the farthest point. There are two scenarios to consider: starting by rowing downstream or starting by rowing upstream, and then further solving accordingly.

【Solution】According to the problem, we can divide it into two scenarios: starting by rowing downstream or starting by rowing upstream;
(1) Assume he starts by rowing downstream. If he rows for 30 minutes, he will row downstream $(3+1.5) \times 0.5 + 1.5 \times \frac{1}{6} = 2.5$ (kilometers);

When returning, he rows upstream for half an hour, and the boat drifts downstream for 10 minutes in one cycle, he can row upstream $(3-1.5) \times 0.5 - 1.5 \times \frac{1}{6} = 0.5$ (kilometers);

In the remaining 40 minutes, he cannot return to the boat rental point.
Assume he starts by rowing downstream for $x (x < 30)$ minutes,
then in the first 40 minutes, he can row downstream $(3+1.5) \times \frac{x}{60} - (3-1.5) \times \frac{30-x}{60} + 1.5 \times \frac{1}{60} = 0.1x - 0.25$ (kilometers)

To ensure he can return to the boat rental point, it is required that $0.1x - 0.25 \leq 0.5 + 0.75$, which means $x \leq 15$;
So, he can row at most $(3+1.5) \times \frac{1}{4} = 1.125$ (kilometers);
(2) Start by rowing upstream, from (1) we know that rowing upstream for half an hour, and the boat drifting downstream for 10 minutes in one cycle ($T=40$ minutes), he can row upstream $(3-1.5) \times 0.5 - 1.5 \times \frac{1}{6} = 0.5$ (kilometers);
Assume he rows upstream for $2 \times 40 + x (x \leq 30)$ minutes, then he can row upstream $2 \times 0.5 + (3-1.5) \times \frac{x}{60} = 1 + \frac{x}{40}$; the remaining time he can row downstream $(3+1.5) \times \frac{30-x}{60} + 1.5 \times \frac{1}{6} = 2.5 - \frac{3}{40}x$; to ensure he can return to the boat rental point, it is required that $1 + \frac{x}{40} \leq 2.5 - \frac{3}{40}x$, solving for $x \leq 15$; so, he can leave the boat rental point at most $1 + \frac{15}{40} = 1.375$ (kilometers); comparing the two scenarios, he can leave the boat rental point at most 1.375 kilometers.

Answer: Xiaohu's boat can be at most 1.375 kilometers away from the boat rental point.