Match List-I with List-II (in each problem, \(x \ge 0\) and \(y \ge 0\) also hold).
| List-I | List-II | ||
|---|---|---|---|
| Linear programming problem | Result | ||
| (A) | Maximise \(Z = x + y\) subject to \(x + y \ge 6\), \(2x + 3y \le 10\) | (I) | No feasible solution |
| (B) | Maximise \(Z = x + 2y\) subject to \(x + y \ge 2\) | (II) | A unique optimal solution |
| (C) | Maximise \(Z = 2x + 3y\) subject to \(x + y \le 4\) | (III) | Infinitely many optimal solutions |
| (D) | Maximise \(Z = 2x + 2y\) subject to \(x + y \le 5\), \(x \le 3\) | (IV) | Feasible region is not empty, but Z has no maximum value |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(A) Maximise \(Z = x + y\) subject to \(x + y \ge 6\), \(2x + 3y \le 10\)
This matches (I) No feasible solution. For \(x, y \ge 0\) with \(x + y \ge 6\):
\(2x + 3y \ge 2x + 2y = 2(x + y) \ge 12\)
So \(2x + 3y \le 10\) can never hold together with \(x + y \ge 6\): the feasible region is empty.
(B) Maximise \(Z = x + 2y\) subject to \(x + y \ge 2\)
This matches (IV). The region is unbounded: every point (0, y) with \(y \ge 2\) is feasible, and there
\(Z = 2y\), which grows without limit
(C) Maximise \(Z = 2x + 3y\) subject to \(x + y \le 4\)
This matches (II) A unique optimal solution. The corner points are (0, 0), (4, 0), (0, 4):
\(Z(0, 0) = 0\)
\(Z(4, 0) = 8\)
\(Z(0, 4) = 12\)
The maximum 12 occurs only at (0, 4).
(D) Maximise \(Z = 2x + 2y\) subject to \(x + y \le 5\), \(x \le 3\)
This matches (III) Infinitely many optimal solutions. The corner points are (0, 0), (3, 0), (3, 2), (0, 5):
\(Z(0, 0) = 0\)
\(Z(3, 0) = 6\)
\(Z(3, 2) = 10\)
\(Z(0, 5) = 10\)
The maximum 10 occurs at two adjacent corners, so every point of the edge joining (3, 2) and (0, 5) is optimal. Here \(Z = 2(x + y)\) is parallel to the constraint \(x + y \le 5\).