For the linear programming problem: Minimise \(Z = x - 3y\) subject to \(x + 2y \ge 4\), \(y - x \le 2\), \(x \ge 0\), \(y \ge 0\), the minimum value of Z is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Z has no minimum value
The region lies on the far side of x + 2y = 4 from the origin and below the line y = x + 2. Nothing limits x from above, so the region is unbounded.
Its corner points are (4, 0), on x + 2y = 4, and (0, 2), where both lines meet the y-axis.
\(Z(4, 0) = 4\)
\(Z(0, 2) = -6\)
The smaller corner value −6 is the minimum only if no feasible point gives Z below −6. Take the points (t, t + 2) with \(t \ge 0\) on the line y = x + 2. They are feasible, since x + 2y = 3t + 4 ≥ 4, and there
\(Z = t - 3(t + 2)\)
\(= -2t - 6\)
For t = 2 this gives Z = −10 < −6, and Z keeps falling as t grows. So Z has no minimum value.
Why the other options are incorrect
Option (2) −6 — Incorrect. This is the smaller corner value, but the feasible point (2, 4) gives Z = −10.
Option (3) 4 — Incorrect. This is Z at the corner (4, 0), which is larger than Z(0, 2).
Option (4) 0 — Incorrect. This is Z at (0, 0), which is not feasible: there x + 2y = 0 < 4.