The maximum value of \(Z = 3x + 2y\) subject to the constraints \(x \le y\), \(y \le 2x\), \(x + y \le 12\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30
The constraints \(x \le y \le 2x\) keep the region between the lines \(y = x\) and \(y = 2x\), which both pass through the origin. The line \(x + y = 12\) meets them at
\(y = x\): \(2x = 12\), giving (6, 6)
\(y = 2x\): \(3x = 12\), giving (4, 8)
So the feasible region is the triangle with corner points (0, 0), (6, 6) and (4, 8). Values of Z:
\(Z(0, 0) = 0\)
\(Z(6, 6) = 18 + 12 = 30\)
\(Z(4, 8) = 12 + 16 = 28\)
So the maximum value is 30, at (6, 6).
Why the other options are incorrect
Option (2) 28 — Incorrect. This is Z at (4, 8), which is smaller than Z at (6, 6).
Option (3) 24 — Incorrect. This is Z at (0, 12), but there \(y = 12 \gt 2x = 0\), so (0, 12) is not feasible.
Option (4) 36 — Incorrect. This is Z at (12, 0), but there \(x = 12 \gt y = 0\), so (12, 0) is not feasible.