In a linear programming problem, the constraints on the decision variables x and y are, x - 3y ≥ 0, y ≥ 0, 0 ≤ x ≤ 3. The feasible region :

is bounded in the first quadrant

constraints.

$x-3 y ≥ 0$

$y ≥ 0$

$0 ≤ x ≤ 3$

plotting

x - 3y = 0

Checking for inequality 

x - 3y ≥ 0

for point (1, 1)

1 - 3 ≥ 0 → false

⇒  hence solution lies to side of x - 3y = 0 not containing (1, 1)

option 2 → is bounded in the first quadrant