The maximum value of \(Z = 2x + 3y\) subject to the constraints \(x - y \le 3\), \(x + 2y \le 12\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 21
The region lies above the line x − y = 3 (the origin side) and below the line x + 2y = 12, in the first quadrant. It is bounded.
On the x-axis, x − y = 3 gives (3, 0). On the y-axis, x + 2y = 12 gives (0, 6).
The two lines meet where
\(x = y + 3\)
\(y + 3 + 2y = 12\)
\(y = 3,\ x = 6\)
The corner points are (0, 0), (3, 0), (6, 3) and (0, 6).
\(Z(0, 0) = 0\)
\(Z(3, 0) = 6\)
\(Z(6, 3) = 12 + 9 = 21\)
\(Z(0, 6) = 18\)
The maximum is 21, at (6, 3).
Why the other options are incorrect
Option (1) 18 — Incorrect. This is Z at the corner (0, 6), which is less than 21.
Option (2) 6 — Incorrect. This is Z at the corner (3, 0).
Option (4) 24 — Incorrect. This is Z at (12, 0), which is not feasible: there x − y = 12 > 3.