The minimum value of \(Z = 3x + 2y\) subject to the constraints \(x + y \ge 5\), \(x \ge 1\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11
The region lies to the right of \(x = 1\) and above \(x + y = 5\). It is unbounded, with corner points:
(1, 4), where \(x = 1\) meets \(x + y = 5\)
(5, 0), where \(x + y = 5\) meets \(y = 0\)
Values of Z:
\(Z(1, 4) = 3 + 8 = 11\)
\(Z(5, 0) = 15\)
Since the region is unbounded, 11 is the minimum only if the half-plane \(3x + 2y \lt 11\) has no point in common with the region. Every feasible point satisfies
\(3x + 2y = 2(x + y) + x \ge 2 \times 5 + 1 = 11\)
So the minimum value is 11, at (1, 4).
Why the other options are incorrect
Option (1) 10 — Incorrect. 10 is Z at (0, 5), which breaks \(x \ge 1\).
Option (3) 15 — Incorrect. 15 is Z at (5, 0), which is larger than 11.
Option (4) Z has no minimum value — Incorrect. An unbounded region can still have a minimum: \(3x + 2y \ge 11\) at every feasible point, with equality at (1, 4).