For the feasible region of an LPP as shown, if the equations of OA and BC are \(y-2x=0\) and \(y-2x=4\) respectively, then constraints for LPP are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y\ge 2x;\ y-2x\le 4;\ x\le 6;\ x, y\ge 0\)
The shaded region is the strip between the two parallel lines OA (y = 2x) and BC (y = 2x + 4), to the left of the vertical line x = 6 through A and B, in the first quadrant.
A convenient test point inside the region is (1, 4):
\(y-2x = 4-2 = 2\)
Since \(0 \le 2 \le 4\), the region satisfies \(y-2x \ge 0\) (i.e. \(y \ge 2x\), above OA) and \(y-2x \le 4\) (below BC). With \(x \le 6\) and \(x, y \ge 0\), these are the constraints of Option (1).
Why the other options are incorrect
Option (2) \(y\le 2x;\ y-2x\le 4;\ x\le 6;\ x, y\ge 0\) — Incorrect. \(y \le 2x\) is the side below OA, outside the strip.
Option (3) \(y\ge 2x;\ y-2x\ge 4;\ x\le 6;\ x, y\ge 0\) — Incorrect. \(y-2x \ge 4\) is the side above BC, outside the strip.
Option (4) \(y\le 2x;\ y-2x\ge 4;\ x\le 6;\ x, y\ge 0\) — Incorrect. Both signs are reversed; no point satisfies \(y-2x \le 0\) and \(y-2x \ge 4\) together.