The feasible region of a linear programming problem is the triangle with vertices \((0, 0)\), \((6, 0)\) and \((0, 3)\), including its boundary. Besides \(x \ge 0\) and \(y \ge 0\), the constraint that gives this region is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x + 2y \le 6\)
The third side of the triangle is the line through (6, 0) and (0, 3). In intercept form:
\( \displaystyle \frac{x}{6} + \frac{y}{3} = 1 \)
\(x + 2y = 6\)
The region contains the origin, and at (0, 0) we get \(0 \le 6\). So the region is on the origin side of the line:
\(x + 2y \le 6\)
Why the other options are incorrect
Option (2) \(2x + y \le 6\) — Incorrect. Its line meets the axes at (3, 0) and (0, 6), so the triangle would be (0, 0), (3, 0), (0, 6).
Option (3) \(x + 2y \ge 6\) — Incorrect. This is the side away from the origin, an unbounded region that does not contain (0, 0).
Option (4) \(x + y \le 6\) — Incorrect. Its line meets the y-axis at (0, 6), not at (0, 3).