A+b=50, 80a+50b≤3490, 40a+90b≤2950 (a, b∈ natural number n).

Substituting b=50-a into two inequalities to get 80a+50(50-a)≤3490, that is, 30a≤990, then a ≤ 33; 40a+90(50-a)≤2950, that is, -50a≤- 1550, then a≥3 1.

Combining two inequalities, we get 3 1≤a≤33.

Because a∈N, then a 1=3 1, a2=32, a3=33.

B=50-a, then b 1= 19, b2= 18, b3= 17.

Therefore, there are three schemes * * *, namely type A 3 1 set and type B1set; 32 sets of A-shape and 8 sets of B-shape/KLOC-0; A-shaped 33 sets, B-shaped 17 sets.

(2) Because the cost of type A is lower than that of type B, in order to reduce the cost, type A should be used as much as possible.

Therefore, in (1), there are 33 groups in A-shape and 17 groups in B-shape, with the lowest cost.

The lowest cost is 33 * 800+17 * 960 = 42,720 yuan.