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.