The minimum value of the objective function \(Z = x + 2y\) of an L.P.P. subject to constraints \(2x + y \ge 3, \displaystyle\frac{x}{2} + 2y \ge 6, x \ge 0, y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
The constraint lines and their intercepts:
\(2x + y = 3\): through (0, 3) and \(\left(\frac{3}{2}, 0\right)\)
\(\displaystyle \frac{x}{2} + 2y = 6\), that is \(x + 4y = 12\): through (0, 3) and (12, 0)
Both lines pass through (0, 3). For x > 0, every point on \(x + 4y = 12\) also satisfies \(2x + y \ge 3\) (there \(2x + y = 3 + \frac{7x}{4}\)), so \(2x + y \ge 3\) is redundant. The feasible region is unbounded, with corner points
A(0, 3), B(12, 0)
\(Z(0, 3) = 0 + 6 = 6\)
\(Z(12, 0) = 12\)
The smaller value is 6. Since the region is unbounded, check that the half-plane \(x + 2y \lt 6\) has no point in common with it. For x ≥ 0 with \(x + 4y \ge 12\) and \(x + 2y \lt 6\), subtracting gives
\(2y \gt 6,\quad y \gt 3\)
\(x \lt 6 - 2y \lt 0\)
which contradicts x ≥ 0. So the minimum value of Z is 6, at (0, 3).
Why the other options are incorrect
Option (1) 12 — Incorrect. This is Z at the other corner (12, 0); the minimum is the smaller corner value, 6.
Option (3) \(\displaystyle\frac{3}{2}\) — Incorrect. This is Z at \(\left(\frac{3}{2}, 0\right)\), where 2x + y = 3 meets the x-axis. That point is not feasible: it gives \(\frac{x}{2} + 2y = \frac{3}{4} \lt 6\).
Option (4) 0 — Incorrect. This is Z at the origin, which satisfies neither \(2x + y \ge 3\) nor \(\frac{x}{2} + 2y \ge 6\).