Consider the linear programming problem: Maximise \(Z = 3x + 2y\) subject to \(x + y \le k\), \(x \le 4\), \(y \le 5\), \(x \ge 0\), \(y \ge 0\). Match each value of k in List-I with the maximum value of Z in List-II.
| List-I | List-II | ||
|---|---|---|---|
| Value of k | Maximum value of Z | ||
| (A) | \(k = 2\) | (I) | 22 |
| (B) | \(k = 4\) | (II) | 6 |
| (C) | \(k = 6\) | (III) | 16 |
| (D) | \(k = 10\) | (IV) | 12 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Each region is bounded, so the maximum is at a corner point. The corners change with k.
(A) \(k = 2\): the bounds \(x \le 4\), \(y \le 5\) play no part, and the corners are (0, 0), (2, 0), (0, 2).
\(Z(2, 0) = 6\), \(Z(0, 2) = 4\)
This matches (II) 6.
(B) \(k = 4\): the corners are (0, 0), (4, 0), (0, 4).
\(Z(4, 0) = 12\), \(Z(0, 4) = 8\)
This matches (IV) 12.
(C) \(k = 6\): \(x + y = 6\) meets \(x = 4\) at (4, 2) and \(y = 5\) at (1, 5), so the corners are (0, 0), (4, 0), (4, 2), (1, 5), (0, 5).
\(Z(4, 0) = 12\), \(Z(4, 2) = 16\)
\(Z(1, 5) = 13\), \(Z(0, 5) = 10\)
This matches (III) 16.
(D) \(k = 10\): every point of the rectangle \(0 \le x \le 4\), \(0 \le y \le 5\) has \(x + y \le 9 \lt 10\), so this constraint is redundant and the corners are (0, 0), (4, 0), (4, 5), (0, 5).
\(Z(4, 5) = 12 + 10 = 22\)
This matches (I) 22.