The maximum value of the objective function \(Z=2x+y\) of an LPP, subject to the constraints \(x \le 6, y \le 2, x-y \le 0, x \ge 0, y \ge 0\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
The constraint \(x - y \le 0\) means \(x \le y\). Together with \(y \le 2\) it gives \(x \le 2\), so the constraint \(x \le 6\) is redundant.
The feasible region is the triangle bounded by x = 0, y = 2 and y = x, with corner points:
O(0, 0), P(0, 2), Q(2, 2)
(Q is where y = x meets y = 2.) The values of Z = 2x + y at the corners:
\(Z(0, 0) = 0\)
\(Z(0, 2) = 2\)
\(Z(2, 2) = 4 + 2 = 6\)
The region is bounded, so the maximum is the largest corner value: 6, at (2, 2).
Why the other options are incorrect
Option (1) 14 — Incorrect. This is Z at (6, 2), where x = 6 meets y = 2. That point does not satisfy \(x \le y\), so it is not in the feasible region.
Option (3) 12 — Incorrect. This is Z at (6, 0), which also does not satisfy \(x \le y\).
Option (4) 8 — Incorrect. Z is at most 6 on the feasible region; 8 is not the value at any corner.