(1) Set X to rent trucks A and B (10-x).

Available equations:

4x+2( 10-x)>=30

x+2( 10-x)>= 13

Solution 5

Therefore, there are three schemes.

1, 5 trucks a and 5 trucks b.

2. There are six trucks A and four trucks B.

There are 7 trucks a and 3 trucks B.

(2) Let the freight be Y yuan.

y = 2000 x+ 1300( 10-x)= 13000+700 x

Therefore, when x=5, the minimum value of y is 16500 yuan.