The feasible region determined by the constraints \(x + y \ge 4\), \(x \ge 0\), \(y \ge 0\) is unbounded. Which one of the following constraints, when added to these, makes the feasible region bounded?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(x + 2y \le 10\)
A region is bounded when it can be enclosed in a circle, that is, neither x nor y can become as large as we like. A constraint that limits both x and y in the first quadrant is needed.
With \(x + 2y \le 10\), \(x \ge 0\), \(y \ge 0\):
\(x \le 10\) and \(y \le 5\)
The region now lies between the lines \(x + y = 4\) and \(x + 2y = 10\). Its corner points are (4, 0), (10, 0), (0, 5) and (0, 4), so it is a bounded quadrilateral.
Why the other options are incorrect
Option (1) \(x \ge 1\) — Incorrect. Points (t, 0) with t as large as we like still satisfy all the constraints.
Option (2) \(y \le 5\) — Incorrect. y is limited, but x is not: (t, 0) with large t is still feasible.
Option (3) \(x - y \le 2\) — Incorrect. Points (t, t) with large t satisfy \(x - y = 0 \le 2\) and \(x + y \ge 4\), so the region still extends without limit.