200 tents, food 120.
(1) Assuming that A and B have X and Y trucks respectively, then x+y=8, that is, Y = 8-X...( 1 formula)
It is known that Class A trucks can hold up to 40 tents and 10 food, while Class B trucks can hold up to 20 tents and 20 food respectively.
40x+20y & gt； =200 (tent)
Bring (1 formula) into:
40x+20*(8-x)>=200
Solution: x & gt=2
10x+20y & gt； = 120 (food)
Bring (1 formula) into:
10x+20*(8-x)>= 120
Solution: x
So, x = 2,3,4.
y=6、5、4
That is, there are three schemes (2, 6), (3, 5) and (4, 4).
(2) Under the condition of (1), the total transportation costs of the three schemes are as follows:
The first type: 2*4000+6*3600=29600.
The second type: 3*4000+5*3600=30000.
The third type: 4*4000+4*3600=30400.
Therefore, the transportation cost of scheme 1 is the least, which is 29,600 yuan.