For the linear programing problem,
Minimize \((Z) = 60x + 30y\)
subject to: \(2x - y \ge -5;\ 3x + y \ge 3;\ 2x-3y \le 12;\ x, y \ge 0\)
the optimal value of \(Z\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 60
The boundary lines meet the axes (in the first quadrant) at
\(2x - y = -5\): (0, 5)
\(3x + y = 3\): (1, 0) and (0, 3)
\(2x - 3y = 12\): (6, 0)
The feasible region lies on or below \(y = 2x + 5\), on or above \(3x + y = 3\), and on or above \(2x - 3y = 12\), in the first quadrant. It is unbounded, with corner points (0, 3), (0, 5), (1, 0) and (6, 0):
\(Z(0, 3) = 90\)
\(Z(0, 5) = 150\)
\(Z(1, 0) = 60\)
\(Z(6, 0) = 360\)
The smallest corner value is 60. Because the region is unbounded, check the open half-plane \(60x + 30y \lt 60\), that is \(2x + y \lt 2\). The line \(2x + y = 2\) joins (1, 0) and (0, 2), which lie on or below the boundary \(3x + y = 3\), so this half-plane has no point in common with the feasible region.
So the minimum value of Z is 60, at (1, 0).
Why the other options are incorrect
Option (1) 360 — Incorrect. This is the value at the corner (6, 0), not the smallest one.
Option (2) 150 — Incorrect. This is the value at the corner (0, 5).
Option (4) No Minimum value of Z — Incorrect. An unbounded region can still have a minimum; here the half-plane check confirms 60.