Minimise and Maximise
`Z=5x+10y`………………..1
Subject to `x+2yle120`………………2
`x+yge60`…………………3
`x-2yge0`…………………4
`xge0,yge0`………………….5
First we draw the graph of line `x+2y=120`
Put `(0,0)` in the inequation `x+2yle120`
`0+2xx0le120implies0le120`
Therefore half plane contains the origin,
Now, drain the graph of the line `x+y=60`
Put `(0,0)` in the inequation `x+yge60`,
`0+0ge60implies0ge60` (False)
Therefore, half plane does not contain the origin.
Now, draw the graph of the line `x-2y=0`
Put `(5,0)` in the inequation `x-2yge0,`
`5-2xx0ge0implies5ge0` (True)
which is above `X` -axis.
Since `x,yge0`
Therefore feasible region is in first quadrant.
`:.` Feasible region is ABCDA.
From the equations `x-2y=0` and `x+y=60` the point of intersection is `D(40,20)` and from the equations `x-2y=0` and `x+2y=120` the point of intersection is `C(60,30)`.
Therefore the vertices of the feasible region are `A(60,0),B(120,0),C(60,30)` and `D(40,20)`. We find the value of `Z` at these points.
The minimum value of `Z` is 300 at point `(60,0)` and maximum value of `Z` is 600 at each point of the line joining the points `(120,0)` and `(60,30)`.
