For the linear programming problem: Maximise \(Z = 3x + 2y\) subject to \(2x + 3y \le 6\), \(x \ge 4\), \(y \ge 0\), which of the following is true?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The problem has no feasible solution.
Any point with \(x \ge 4\) and \(y \ge 0\) gives
\(2x + 3y \ge 2 \times 4 + 0 = 8\)
But the first constraint needs \(2x + 3y \le 6\). Since \(8 \gt 6\), no point satisfies all the constraints together.
The feasible region is empty, so the problem has no feasible solution and Z has no maximum value.
Why the other options are incorrect
Option (1) The maximum value of Z is 12. — Incorrect. 12 is Z at (4, 0), which breaks \(2x + 3y \le 6\) because \(2 \times 4 = 8\).
Option (2) The maximum value of Z is 9. — Incorrect. 9 is Z at (3, 0), which breaks \(x \ge 4\).
Option (3) The maximum value of Z is 6. — Incorrect. No feasible point exists, so Z takes no value at all.