Task 3. A landscaping team worked on a large and a small football field, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 6 more workers than in the part that worked on the small field. When the landscaping of the large field was completed, the part of the team that was on the small field was still working. What is the maximum number of workers that could have been in the team
Answer: 16.

Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+6$, and the total number of people in the team is $2n+6$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of each part of the team are $an$ and $a(n+6)$. If the area of the smaller field is $S$, then the area of the larger field is $2S$. The time spent on completing the entire work by each team is $\frac{S}{an}$ and $\frac{2S}{a(n+6)}$, respectively. According to the problem, $\frac{S}{an} > \frac{2S}{a(n+6)}$. Due to the positivity of all variables, this inequality is equivalent to the inequality $n+6 > 2n$, or $n < 6$. Therefore, $n \leqslant 5$, and $2n+6 \leqslant 16$. The equality situation is clearly possible: just take any positive $S$ and $a$.