The feasible region of an LPP is given by the constraints \(x + y \le 10\), \(x \le 6\), \(y \le 8\), \(2x + y \le 20\), \(x \ge 0\), \(y \ge 0\). Which one of the following constraints can be removed without changing the feasible region?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2x + y \le 20\)
A constraint is redundant if every point satisfying the other constraints already satisfies it. For any point with \(x \le 6\) and \(x + y \le 10\):
\(2x + y = x + (x + y) \le 6 + 10 = 16\)
So \(2x + y \le 16 \lt 20\) holds automatically, and removing \(2x + y \le 20\) leaves the region unchanged. The region keeps its corner points (0, 0), (6, 0), (6, 4), (2, 8) and (0, 8).
Why the other options are incorrect
Option (2) \(x + y \le 10\) — Incorrect. Without it, (6, 8) satisfies all the other constraints, but \(6 + 8 = 14 \gt 10\), so the region would grow.
Option (3) \(x \le 6\) — Incorrect. Without it, (10, 0) satisfies all the other constraints, but \(10 \gt 6\).
Option (4) \(y \le 8\) — Incorrect. Without it, (0, 10) satisfies all the other constraints, but \(10 \gt 8\).