The minimum value of \(Z = 2x + 3y\) subject to the constraints \(x + y \ge 4\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8
The region lies on the far side of the line x + y = 4 from the origin, so it is unbounded. Its corner points are the intercepts (4, 0) and (0, 4).
\(Z(4, 0) = 8\)
\(Z(0, 4) = 12\)
The smaller corner value is 8. Since the region is unbounded, it is the minimum only if no feasible point gives Z below 8. For every feasible point,
\(2x + 3y \ge 2x + 2y\)
\(= 2(x + y) \ge 8\)
So the minimum value is 8, at (4, 0).
Why the other options are incorrect
Option (2) 12 — Incorrect. This is Z at the other corner (0, 4), which is larger than 8.
Option (3) 0 — Incorrect. This is Z at (0, 0), which is not feasible: there x + y = 0 < 4.
Option (4) Z has no minimum value — Incorrect. An unbounded region can still have a minimum; here Z never falls below 8.