In a linear programming problem, the constraints on the decision variables x and y are, \(x-3y \ge 0\), \(y \ge 0\), \(0 \le x \le 3\). The feasible region:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → is bounded in the first quadrant
Each constraint describes one part of the plane:
\(x - 3y \ge 0\) means \(y \le \frac{x}{3}\): on or below the line through (0, 0) and (3, 1)
\(y \ge 0\): on or above the x-axis
\(0 \le x \le 3\): between the y-axis and the line x = 3
Together these give the triangle with corners (0, 0), (3, 0) and (3, 1). For example, (2, 0.5) satisfies all of them, so the region exists.
All its points have \(x \ge 0\), \(y \ge 0\), so it lies in the first quadrant, and since \(0 \le x \le 3\) and \(0 \le y \le 1\) it is enclosed in a finite triangle, so it is bounded.