Match List-I with List-II (in each problem, \(x \ge 0\) and \(y \ge 0\) also hold).
| List-I (Problem) | List-II (Maximum value of Z) | ||
| (A) | Maximise \(Z = x + y\) subject to \(2x + y \le 6\) | (I) | 12 |
| (B) | Maximise \(Z = 3x + y\) subject to \(x + y \le 4\) | (II) | 18 |
| (C) | Maximise \(Z = 2x + 5y\) subject to \(x + 2y \le 8\) | (III) | 6 |
| (D) | Maximise \(Z = 4x + 3y\) subject to \(3x + 2y \le 12\) | (IV) | 20 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Each region is the triangle cut off from the first quadrant by one line, with corner points (0, 0) and the two intercepts. At (0, 0) every Z is 0, so the maximum is at one of the intercepts.
(A) Maximise \(Z = x + y\) subject to \(2x + y \le 6\)
This matches (III) 6.
\(Z(3, 0) = 3,\ Z(0, 6) = 6\)
(B) Maximise \(Z = 3x + y\) subject to \(x + y \le 4\)
This matches (I) 12.
\(Z(4, 0) = 12,\ Z(0, 4) = 4\)
(C) Maximise \(Z = 2x + 5y\) subject to \(x + 2y \le 8\)
This matches (IV) 20.
\(Z(8, 0) = 16,\ Z(0, 4) = 20\)
(D) Maximise \(Z = 4x + 3y\) subject to \(3x + 2y \le 12\)
This matches (II) 18.
\(Z(4, 0) = 16,\ Z(0, 6) = 18\)