2. Assuming that X peacocks and Y pigeons are used, the condition of 0.8*x+0.4*y (daily expenses) shall not exceed (90/30 = 3 yuan/day).
Adopt 0 peacocks, 0.8*0+0.4*y, no more than 3, and only 7 pigeons at most.
Adopt 1 peacock, that is, 0.8* 1+0.4*y, no more than 3. You can only adopt five pigeons at most.
Adopt two peacocks, with the ratio of 0.8*2+0.4*y not exceeding 3, and only three pigeons can be adopted at most.
Adopt 3 peacocks, with the ratio of 0.8*3+0.4*y not exceeding 3, and only 1 pigeon can be adopted at most.
Therefore, when it is essential to adopt two kinds of animals, the most adopted animals are 1 peacock and 5 pigeons.
3. Considering the last answer, it is easy to know that the plan is as follows, provided that the funds for activities in 90 yuan are not exceeded:
If you adopt 0 peacocks, you can adopt 7, 6, 5, 4, 3, 2, 1 pigeon;
Adopt 1 peacock, and adopt 5, 4, 3, 2, 1 pigeon;
Adopt 2 peacocks, and you can adopt 3, 2, 1 pigeon;
Adopt 3 peacocks, you can adopt 1 pigeon;
If you adopt 0 pigeons, you can adopt 3, 2, 1 peacock.
Want to be as close to the activity funds as possible in 90 yuan, and adopt the pigeons and peacocks with the largest number in various schemes:)