The minimum value of \(Z = 2x + 3y\) subject to the constraints \(x + y \ge 5\), \(x \ge 1\), \(y \ge 2\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12
The region lies to the right of \(x = 1\), above \(y = 2\) and above the line \(x + y = 5\). It is unbounded, with two corner points where \(x + y = 5\) meets the other two lines:
\(x = 1\): \(y = 4\), corner (1, 4)
\(y = 2\): \(x = 3\), corner (3, 2)
Values of Z at the corner points:
\(Z(1, 4) = 2 + 12 = 14\)
\(Z(3, 2) = 6 + 6 = 12\)
Since the region is unbounded, check the half-plane \(2x + 3y \lt 12\). Every feasible point has \(x \ge 1\), \(y \ge 2\) and \(x + y \ge 5\), so
\(2x + 3y = 2(x + y) + y \ge 10 + 2 = 12\)
No feasible point gives Z below 12, so the minimum value is 12, at (3, 2).
Why the other options are incorrect
Option (2) 8 — Incorrect. This is Z at (1, 2), but there \(x + y = 3 \lt 5\), so (1, 2) is not feasible.
Option (3) 14 — Incorrect. This is Z at the other corner (1, 4), which is larger than 12.
Option (4) 10 — Incorrect. This is Z at (5, 0), but there \(y = 0 \lt 2\), so (5, 0) is not feasible.