The maximum value of \(Z = -3x - 2y\) subject to the constraints \(x + y \ge 4\), \(x + 3y \ge 6\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-8\)
The lines \(x + y = 4\) and \(x + 3y = 6\) meet where
\(2y = 2\)
\(y = 1, \; x = 3\)
The region is unbounded, with corner points (0, 4), (3, 1) and (6, 0). Values of Z:
\(Z(0, 4) = -8\)
\(Z(3, 1) = -9 - 2 = -11\)
\(Z(6, 0) = -18\)
The largest corner value is \(-8\). It is the maximum because, at every feasible point,
\(3x + 2y \ge 2x + 2y = 2(x + y) \ge 8\)
\(Z = -(3x + 2y) \le -8\)
So an unbounded region can still have a maximum value: here it is \(-8\), at (0, 4).
Why the other options are incorrect
Option (2) \(-11\) — Incorrect. This is Z at (3, 1), which is less than \(-8\).
Option (3) \(-18\) — Incorrect. This is the smallest corner value, at (6, 0).
Option (4) Z has no maximum value — Incorrect. \(Z \le -8\) at every feasible point, and \(Z = -8\) at (0, 4).