Consider the feasible region determined by the constraints \(2x + y \le 16\), \(x + 3y \le 18\), \(x \ge 0\), \(y \ge 0\).
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Objective function | Maximum value on the region | ||
| (A) | \(Z = x + y\) | (I) | 14 |
| (B) | \(Z = 3x + y\) | (II) | 24 |
| (C) | \(Z = x + 2y\) | (III) | 32 |
| (D) | \(Z = 2x + 5y\) | (IV) | 10 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
The line \(2x + y = 16\) meets the x-axis at (8, 0), and \(x + 3y = 18\) meets the y-axis at (0, 6). The two lines meet where, putting \(y = 16 - 2x\),
\(x + 48 - 6x = 18\)
\(x = 6, \; y = 4\)
The region is bounded, with corner points (0, 0), (8, 0), (6, 4) and (0, 6). At (0, 0) every Z below is 0, so only the other three corners matter.
(A) \(Z = x + y\)
This matches (IV) 10:
\(Z(8, 0) = 8, \; Z(6, 4) = 10, \; Z(0, 6) = 6\)
(B) \(Z = 3x + y\)
This matches (II) 24:
\(Z(8, 0) = 24, \; Z(6, 4) = 22, \; Z(0, 6) = 6\)
(C) \(Z = x + 2y\)
This matches (I) 14:
\(Z(8, 0) = 8, \; Z(6, 4) = 14, \; Z(0, 6) = 12\)
(D) \(Z = 2x + 5y\)
This matches (III) 32:
\(Z(8, 0) = 16, \; Z(6, 4) = 32, \; Z(0, 6) = 30\)