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 :
Answer & explanation
Correct answer: option 2
constraints.
$x-3 y ≥ 0$
$y ≥ 0$
$0 ≤ x ≤ 3$
plotting
x - 3y = 0
| x | 0 | 3 |
| y | 0 | 1 |
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