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.