The maximum value of \(Z = 3x + 2y\) subject to the constraints \(x + y \ge 2\), \(x \le 4\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Z has no maximum value
The constraints limit x to \(0 \le x \le 4\), but put no upper limit on y. The feasible region is unbounded, with corner points (2, 0), (4, 0) and (0, 2).
\(Z(2, 0) = 6\)
\(Z(4, 0) = 12\)
\(Z(0, 2) = 4\)
The largest corner value 12 is the maximum only if no feasible point gives Z above 12. But every point (0, t) with \(t \ge 2\) is feasible, and there
\(Z = 2t\)
For t = 10 this already gives Z = 20 > 12, and Z grows without limit as t grows. So Z has no maximum value.
Why the other options are incorrect
Option (1) 12 — Incorrect. This is the largest corner value, but the feasible point (0, 10) gives Z = 20.
Option (2) 6 — Incorrect. This is Z at the corner (2, 0).
Option (3) 4 — Incorrect. This is Z at the corner (0, 2); it is the minimum of Z, not the maximum.