For the linear programming problem: Maximise \(Z = 2x + y\) subject to \(x + y \le 8\), \(x \le 5\), \(y \ge 2\), \(x \ge 0\), the optimal feasible solution \((x, y)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((5, 3)\)
The region lies above \(y = 2\), to the left of \(x = 5\) and below \(x + y = 8\). It is bounded, with corner points (0, 2), (5, 2), (5, 3) and (0, 8), since \(x = 5\) meets \(x + y = 8\) at \(y = 3\).
\(Z(0, 2) = 2\)
\(Z(5, 2) = 12\)
\(Z(5, 3) = 13\)
\(Z(0, 8) = 8\)
The largest value 13 occurs only at (5, 3), so (5, 3) is the optimal feasible solution.
Why the other options are incorrect
Option (1) \((5, 2)\) — Incorrect. It is feasible, but Z = 12 there, less than 13.
Option (2) \((8, 0)\) — Incorrect. It gives Z = 16, but \(x = 8 \gt 5\) and \(y = 0 \lt 2\), so it is an infeasible solution.
Option (4) \((0, 8)\) — Incorrect. It is feasible, but Z = 8 there.