The maximum value of \(Z = 2x + 5y\) subject to the constraints \(x + y \le 4\), \(x + 3y \le 6\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11
The feasible region is bounded, so the maximum of Z is at a corner point.
The lines x + y = 4 and x + 3y = 6 meet where
\(2y = 2,\quad y = 1\)
\(x = 4 - 1 = 3\)
On the axes, x + y = 4 meets the x-axis at (4, 0) and x + 3y = 6 meets the y-axis at (0, 2); the other two intercepts, (6, 0) and (0, 4), break one of the constraints. So the corner points are (0, 0), (4, 0), (3, 1) and (0, 2).
Values of Z = 2x + 5y:
\(Z(0, 0) = 0\)
\(Z(4, 0) = 8\)
\(Z(3, 1) = 6 + 5 = 11\)
\(Z(0, 2) = 10\)
The maximum is 11, at (3, 1).
Why the other options are incorrect
Option (1) 10 — Incorrect. This is Z at the corner (0, 2), which is less than 11.
Option (3) 8 — Incorrect. This is Z at the corner (4, 0).
Option (4) 20 — Incorrect. This is Z at (0, 4), which is not feasible: there x + 3y = 12 > 6.