The feasible region of a linear programming problem is the quadrilateral with vertices \((0, 0)\), \((5, 0)\), \((3, 3)\) and \((0, 4)\), including its boundary. Besides \(x \ge 0\) and \(y \ge 0\), the constraints that give this region are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3x + 2y \le 15\), \(x + 3y \le 12\)
The two slanting sides of the region are the edges from (5, 0) to (3, 3) and from (3, 3) to (0, 4). The line \(3x + 2y = 15\) passes through both (5, 0) and (3, 3):
\(3(5) + 2(0) = 15\)
\(3(3) + 2(3) = 15\)
The line \(x + 3y = 12\) passes through both (3, 3) and (0, 4):
\(3 + 3(3) = 12\)
\(0 + 3(4) = 12\)
The region contains the origin, so both are of the “\(\le\)” type.
Why the other options are incorrect
Option (2) \(2x + 3y \le 15\), \(x + 3y \le 12\) — Incorrect. The line \(2x + 3y = 15\) does not pass through (5, 0), since \(2(5) = 10\).
Option (3) \(3x + 2y \le 15\), \(3x + y \le 12\) — Incorrect. The line \(3x + y = 12\) does not pass through (0, 4), since \(0 + 4 = 4\).
Option (4) \(3x + 2y \ge 15\), \(x + 3y \le 12\) — Incorrect. The origin is in the region, but \(0 \ge 15\) is false.