Maximize `Z=-x+2y`……………..1
Subject to `xge3`……………….2
`x+yge5`………………3
`x+2yge6`………………4 ltbr. `yge0`…………………5 ltbr. First we draw the graph of the line `x+y=5`
Put `(0,0)` in the inequation `x+yge5`,
`0+0ge6implies0ge5` (False)
Therefore, half plane does not contain the origin.
Now we draw the graph of the line `x+2y=6`.
PUt `(0,0)` in the inequation `x+2yge6`
`0+2xx0ge6`
`implies0ge6` (False)
Therefore, the half plane does not contain the origin.
Since `xge3, yge0`
Therefore the feasible region in first quadrant in right of `x=3`.
The point of intersection of the lines `x=3` and `x+y=5` is `C(3,2)` and the point of intersection of the lines `x+2y=6` and `x+y=5` is `B(4,1)`. Clearly the feasible region is unbounded.
Thus, the vertices of the feasible region are `A(6,0),B(4,1)` and `C(3,2)` and we find the value of `Z` at these vertices.
Since the feasible region is unbounded.
Therefore `Z` has no maximum value.
