The maximum value of \(Z = x + 3y\) subject to the constraints \(x + y \le 8\), \(x \ge 2\), \(y \le 5\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 18
The region lies to the right of x = 2, below y = 5 and below x + y = 8, above the x-axis. It is bounded.
Its corner points:
x = 2 with y = 0: (2, 0)
x + y = 8 with y = 0: (8, 0)
x + y = 8 with y = 5: (3, 5)
x = 2 with y = 5: (2, 5)
Values of Z = x + 3y:
\(Z(2, 0) = 2\)
\(Z(8, 0) = 8\)
\(Z(3, 5) = 3 + 15 = 18\)
\(Z(2, 5) = 2 + 15 = 17\)
The maximum is 18, at (3, 5).
Why the other options are incorrect
Option (2) 17 — Incorrect. This is Z at the corner (2, 5), which is less than 18.
Option (3) 24 — Incorrect. This is Z at (0, 8), which breaks both \(x \ge 2\) and \(y \le 5\).
Option (4) 8 — Incorrect. This is Z at the corner (8, 0).