Problem 3. A team of lumberjacks was cutting trees on a large and a small plot, with the area of the small plot being 3 times less than that of the large plot. In the part of the team that worked on the large plot, there were 8 more lumberjacks than in the part that worked on the small plot. When the tree harvesting on the large plot was completed, the part of the team that was on the small plot was still working. What is the maximum number of lumberjacks that could have been in the team?
Answer: 14.

Solution. Let the number of workers on the smaller plot be denoted as $n$, then the number of workers on the larger plot is $n+8$, and the total number of workers in the team is $2n+8$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity of each part of the team is $an$ and $a(n+8)$. If the area of the smaller plot is $S$, then the area of the larger plot is $3S$. The time spent on completing the entire work by each part of the team is $\frac{S}{an}$ and $\frac{3S}{a(n+8)}$, respectively. According to the problem, $\frac{S}{an} > \frac{3S}{a(n+8)}$. Given the positivity of all variables, this inequality is equivalent to $n+8 > 3n$, or $n < 4$. Therefore, $n \leqslant 3$, and $2n+8 \leqslant 14$. The equality situation is clearly possible: just take any positive $S$ and $a$.