5x+3y≥30。

x+2y≥ 13

In order to solve the above inequality, X≥3, and vehicle A has at most 4 vehicles, so vehicle A has 3 or 4 vehicles, and the corresponding vehicle B is y≥5 or y≥4.

The minimum calculation cost is 3X4000+5X3500=29500.

Comparing the cost of 29,500 with the above three methods, we can know that A uses 3 vehicles and B uses 5 vehicles, with the lowest cost.