In an LPP, with the constraints are x - 3y ≥ 0, y ≥ 0, 0 ≤ x ≤ 3. The feasible region is :
Answer & explanation
Correct answer: option 2
Given constraints
$x-3y \ge 0$
$y \ge 0$
$0 \le x \le 3$
From $x-3y \ge 0$
$x \ge 3y$
Since $y \ge 0$ and $x \ge 0$, the feasible region lies in the first quadrant.
Also $0 \le x \le 3$ restricts the region between $x=0$ and $x=3$.
Thus the feasible region is limited (finite) and lies entirely in the first quadrant.
The feasible region is bounded and lies in the first quadrant.