Consider the LPP: Max \(Z= 5x + 3y\) subject to \(3x+5y \le 15, 5x + 2y \le 10, x \ge 0, y \ge 0\)
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Objective function | (I) | \(3x+5y \ge 15\) |
| (B) | One constraint | (II) | \(x, y \ge 0\) |
| (C) | Non-negative restrictions | (III) | \(Z=5x+3y\) |
| (D) | Point (1, 2) does not lie in the region | (IV) | \(3x+5y \le 15\) |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A) Objective function
This matches (III) \(Z=5x+3y\), the function to be maximised.
(B) One constraint
This matches (IV) \(3x+5y \le 15\), the first constraint of the given LPP.
(C) Non-negative restrictions
This matches (II) \(x, y \ge 0\).
(D) Point (1, 2) does not lie in the region
This matches (I) \(3x+5y \ge 15\). At (1, 2):
\(3(1) + 5(2) = 13 \lt 15\)
So (1, 2) does not satisfy \(3x + 5y \ge 15\): it lies outside the region of this inequality. It does satisfy (II) and (IV), so (I) is the only entry whose region excludes (1, 2).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).