The solution set of the linear constraints \(x - 2y \ge 0, 2x - y \le -4, x \ge 0\) and \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\phi\)
Writing the first two constraints in terms of y:
\( \displaystyle x - 2y \ge 0 \Rightarrow y \le \frac{x}{2} \)
\(2x - y \le -4 \Rightarrow y \ge 2x + 4\)
A point satisfying both needs
\( \displaystyle 2x + 4 \le \frac{x}{2} \)
\(4x + 8 \le x\)
\(3x \le -8\)
\( \displaystyle x \le -\frac{8}{3} \)
This contradicts \(x \ge 0\). So no point satisfies all the constraints, and the solution set is the empty set \(\phi\).
Why the other options are incorrect
Option (2) \(\{(-1, 0), (0, 2)\}\) — Incorrect. (−1, 0) has \(x \lt 0\), and (0, 2) gives \(x - 2y = -4 \lt 0\).
Option (3) \(\{(-1, 0), (0, 2), (0, 0)\}\) — Incorrect. (−1, 0) has \(x \lt 0\), (0, 2) gives \(x - 2y = -4 \lt 0\), and (0, 0) gives \(2x - y = 0\), which is not \(\le -4\).
Option (4) unbounded region in first quadrant — Incorrect. No point at all satisfies the constraints, so there is no region.