![(3,0)](/tpl/images/0934/0333/a5006.png)
Step-by-step explanation:
The integer points which fall in the common region bounded by the lines are
![(0,2),(2,3),(3,0)](/tpl/images/0934/0333/ca089.png)
Since, one of the points is
we round it to ![(2,3)](/tpl/images/0934/0333/be74f.png)
If we round up then the point will be
which does not fall in the bounded region.
Also
and
no negative values are considered.
Applying in the values in the LP problem ![Z=5x_1+x_2](/tpl/images/0934/0333/22abf.png)
![Z=5\times 0+2\\\Rightarrow Z=2](/tpl/images/0934/0333/ec172.png)
![Z=5\times 2+3\\\Rightarrow Z=13](/tpl/images/0934/0333/7efa3.png)
![Z=5\times 3+0\\\Rightarrow Z=15](/tpl/images/0934/0333/f80c5.png)
So, the LP problem is maximum at point
.
![Consider the following IP problem:
Maximum Z= 5x1+x2
subject to
-x1+2x2 <=4
x1-x2 <=4
4x1+ x2](/tpl/images/0934/0333/98ca2.jpg)