For the linear programming problem: Maximise \(Z = x + 4y\) subject to \(2x + y \le 4\), \(x \ge 3\), \(y \ge 0\), the maximum value of Z is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Z has no maximum value, as no point satisfies all the constraints
Since \(y \ge 0\), the first constraint gives
\(2x \le 4 - y \le 4\)
\(x \le 2\)
This contradicts \(x \ge 3\). No point satisfies all the constraints, so the feasible region is empty and the problem has no optimal value.
Why the other options are incorrect
Option (1) 16 — Incorrect. This is Z at (0, 4), which breaks \(x \ge 3\).
Option (2) 3 — Incorrect. This is Z at (3, 0), which breaks \(2x + y \le 4\) since 6 > 4.
Option (4) 2 — Incorrect. This is Z at (2, 0), which breaks \(x \ge 3\).