X & gt=0, y & gt=0, it is available.

x+2y & gt； =0,

Also because

x+2y & lt； =2,

therefore

0 & lt= x+2y & lt； =2

, both are available.

0 & lt= x & lt=2,

0 & lt= y & lt= 1

if

0 & lt= ax+by & lt； =2

,

Then you can get 0.

0 & lt= b & lt=2,

Because when a person

B<= by & lt=0, then

ax+by & lt； =0

With the known ax+by & gt；; =0 contradiction.

When a>0 and b<0 can get 0 < = ax & lt=2a and b<= by & lt=0, then

b & lt= ax+by & lt； =2a, and b

When a<0, b>0, 2a

2a & lt= ax+by & lt； =b，

Also at this time 2a

So only a and b are positive, that is, a >；; 0 and b>0, you can get 0 < = ax < = 2a, 0 < = by < = b, then 0.

0 & lt= ax+by & lt； =2, so 0

To sum up, A and B must have the same number and a>0, b>0,

2a+b & lt； When =2, the meaning of the problem is satisfied.

Consider the case that A and B take 0, when a=0, B! =0, yes

0<= by & lt=b, and

Also know 0

When a! = 0, when b = 0, there is 0.

0<= ax & lt=2, then 2a

When both a and b are 0, the meaning of the problem is obviously satisfied.

So in summary, you can get 0.