For the linear programming problem: Maximise \(Z = y\) subject to \(2x + y \le 14\), \(y - x \le 2\), \(x \ge 1\), \(y \ge 0\), the maximum value of Z is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
Here Z is just the y-coordinate, so the task is to find the highest point of the region. The region lies to the right of \(x = 1\), above the x-axis, below \(2x + y = 14\) and below \(y = x + 2\). Its corner points are:
\(x = 1\) and \(y = 0\): (1, 0)
\(x = 1\) and \(y = x + 2\): (1, 3)
\(y = x + 2\) and \(2x + y = 14\): \(3x + 2 = 14\), so (4, 6)
\(2x + y = 14\) and \(y = 0\): (7, 0)
The values of \(Z = y\) at these corners are 0, 3, 6 and 0, so the maximum value is 6, at (4, 6).
Why the other options are incorrect
Option (1) 14 — Incorrect. This is y at (0, 14) on \(2x + y = 14\), but \(x = 0 \lt 1\) and \(y - x = 14 \gt 2\) there.
Option (3) 12 — Incorrect. This is y at (1, 12) on \(2x + y = 14\), but there \(y - x = 11 \gt 2\).
Option (4) 3 — Incorrect. This is y at the corner (1, 3); the corner (4, 6) is higher.