Solve the following linear programming problem graphically:
minimize : `z=x+2y`
Subject to: `2x+yge3`
`x+2yge6`
`xge0`
`yge0`
Show that `z` is minimum at two points.

Explanation

Minimise `Z=x+2y`…………..1
Subject to `2x+yge3`………………..2
`x+2yge6`……………………..3
`xge0,yge0`…………….4
First we draw the graph of the line `2xy=3`


Put `(0,0)` in the inequation `2x+yge3`
`2xx0+0ge3`
`implies0ge3` (False)
Therefore, half plane does not contain origin.
Since `x,yge0`
Therefore, feasible region lines in Ist quadrant.
Now, we draw the graph of the line `x+2y=16`

Put `(0,0)` in the inequation `x+2yge6`,
`0+2xx0ge6implies2ge6` (False)
Therefore, half plane does not contain origin.
From the equations `x+2y=6` and `2x+y=3`, the point of intersection is `B(0,3)`.
Thus, vertices of feasible region are `A(6,0)` and `B(0,3)`. We find the value of `Z` at these vertices.

Here, the value of `Z` at points A and B are same. If we put `(2,2)` in the line `x+2y=6`, we get `Z=6`. So the minimum value of `Z` is obtained at two or more than two points.
Therefore at each point on the line `x+2y=6, Z` has minimum value.