First, find out the number of big trucks and small trucks, and set them as x and y respectively, which are obtained from the conditions in the problem.

X+y = 20, 15x+ 10y = 240, and the solution is X = 8 and Y = 12.

If the bus to A is A, there are cars 10-a, where 0 < = A < = 8;

At the same time, there are 8 A cars going to B, 12-( 10-A) = A+2 cars.

At least 1 15 tons will be shipped to a, so there is15a+10 (10-a) > =115, and A >；; =3, namely 3 = < a < = 8;

The total freight is t = 630a+420 (10-a)+750 (8-a)+550 (a+2) =10a+1300.

Because 3 =

The departure plan is:

A big 3, small 7, freight 1 15 tons.

B big 5, small 5, freight 125 tons.

Lowest freight 1 1330.

It's hard to finish. The landlord will give more.