1. There are x tons of heavy vehicles and y tons of light vehicles.

Then 2x+4y=98 3x+2y=9 1 and x=2 1 y= 14 is obtained.

Assume that the number of heavy-duty vehicle A and light-duty vehicle B on the 23rd is 2 1A+ 14b = 77. Because both a and b are integers, B = 4A = 3B = 1.

There are two kinds.

2 because 400/21>; 280/ 14 try to use large cars.

As mentioned above, on 2 1, we should choose 4 heavy vehicles, 1 light vehicles, on 22, 3 heavy vehicles, 2 small vehicles, and on 23, 3 heavy vehicles, 1 light vehicles.

3 Save money: 2 * 400+4 * 280+3 * 400+2 * 280+1* 400+4 * 280-(4 * 400+1* 280+3 * 400+2 * 280+3.