Consider the inequalities x > y, x + y ≤ 3 and y ≤ 1. The feasible region is :
Answer & explanation
Correct answer: option 3
The correct answer is Option 3: open half plane containing the point (2, 1)
Given inequalities
$x>y$
$x+y\le3$
$y\le1$
Check origin $(0,0)$
$0>0$ is false
So origin is not in region
Check point $(0,3)$
$y=3\le1$ is false
So not in region
Check point $(2,1)$
$2>1$ true
$2+1=3\le3$ true
$1\le1$ true
So $(2,1)$ lies in region
Since $x>y$ is strict inequality, region is open
Also region is not bounded
Correct answer: open half plane containing the point $(2,1)$