The maximum value of \(Z = 3x + 4y\) subject to the constraints \(x + y \le 5\), \(-1 \le x - y \le 3\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 18
The double inequality gives two constraints, \(x - y \le 3\) and \(x - y \ge -1\). The region is the part of the triangle \(x + y \le 5\), \(x, y \ge 0\) between the parallel lines \(x - y = 3\) and \(x - y = -1\).
\(x + y = 5\) and \(x - y = 3\): (4, 1)
\(x + y = 5\) and \(x - y = -1\): (2, 3)
\(x - y = 3\) on the x-axis: (3, 0)
\(x - y = -1\) on the y-axis: (0, 1)
The corner points are (0, 0), (3, 0), (4, 1), (2, 3) and (0, 1). Values of Z:
\(Z(0, 0) = 0\), \(Z(3, 0) = 9\)
\(Z(4, 1) = 16\), \(Z(2, 3) = 18\)
\(Z(0, 1) = 4\)
The maximum value is 18, at (2, 3).
Why the other options are incorrect
Option (1) 20 — Incorrect. This is Z at (0, 5), but there \(x - y = -5 \lt -1\), so (0, 5) is not feasible.
Option (2) 16 — Incorrect. This is Z at the corner (4, 1), which is less than 18.
Option (4) 15 — Incorrect. This is Z at (5, 0), but there \(x - y = 5 \gt 3\), so (5, 0) is not feasible.