The maximum value of \(Z = 4x + y\) subject to the constraints \(x + 2y \le 8\), \(3x + 2y \le 12\), \(x - y \le 2\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 14
On the x-axis, \(x - y \le 2\) gives \(x \le 2\), so (2, 0) is a corner. On the y-axis, \(x + 2y \le 8\) gives \(y \le 4\), so (0, 4) is a corner.
Where \(x - y = 2\) meets \(3x + 2y = 12\): putting \(x = y + 2\),
\(3y + 6 + 2y = 12\)
\(y = 1.2, \; x = 3.2\)
Where \(3x + 2y = 12\) meets \(x + 2y = 8\): subtracting,
\(2x = 4\)
\(x = 2, \; y = 3\)
Both points satisfy the third constraint, so the corner points are (0, 0), (2, 0), (3.2, 1.2), (2, 3) and (0, 4). Values of Z:
\(Z(0, 0) = 0\)
\(Z(2, 0) = 8\)
\(Z(3.2, 1.2) = 12.8 + 1.2 = 14\)
\(Z(2, 3) = 11\)
\(Z(0, 4) = 4\)
So the maximum value is 14, at the corner (3.2, 1.2) with fractional coordinates.
Why the other options are incorrect
Option (1) 11 — Incorrect. This is Z at (2, 3), which is smaller than Z at (3.2, 1.2).
Option (2) 12 — Incorrect. This is Z at (3, 0), but there \(x - y = 3 \gt 2\), so (3, 0) is not feasible.
Option (4) 16 — Incorrect. This is Z at (4, 0), where \(3x + 2y \le 12\) holds but \(x - y = 4 \gt 2\).