Consider the LPP: Minimize \(Z = x + 2y\) subject to \(2x + y \ge 3,\ x + 2y \ge 6,\ x, y \ge 0\). The optimal feasible solution occurs at
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Both (6, 0) and (0, 3)
The two lines meet where:
\(y = 3 - 2x\)
\(x + 2(3 - 2x) = 6\)
\(x = 0,\quad y = 3\)
On the axes, the region starts at (0, 3) on the y-axis (both constraints give y ≥ 3) and at (6, 0) on the x-axis (x ≥ 6 from the second constraint). Between them it is bounded by the line x + 2y = 6, on which \(2x + y \ge 3\) holds automatically. So the corner points of the unbounded feasible region are (0, 3) and (6, 0).
\(Z(0, 3) = 0 + 6 = 6\)
\(Z(6, 0) = 6 + 0 = 6\)
The objective Z = x + 2y is the left side of the constraint \(x + 2y \ge 6\), so no feasible point gives Z < 6. The minimum 6 is reached at both corners (and at every point of the segment joining them).
NCERT (2026-27 edition):
Graphical Method of Solving Linear Programming Problems
