The Feasible Region of a Linear Program

Each case is settled by checking the constraints, not by looking at the drawing above.

Bounded. $x_1+x_2\le 4$, $x_1\ge 0$, $x_2\ge 0$. Every feasible point has $0\le x_1\le 4$ and $0\le x_2\le 4$, because each variable is nonnegative and their sum is at most $4$. The region fits inside a finite ball, so it is bounded, and it is the triangle with corners $(0,0)$, $(4,0)$ and $(0,4)$.

Unbounded. $x_1-x_2\le 1$, $x_1\ge 0$, $x_2\ge 0$. The point $(0,t)$ satisfies all three for every $t\ge 0$, since $0-t=-t\le 1$. As $t$ may be arbitrarily large, no ball contains the region.

Empty. $x_1+x_2\le 1$, $x_1+x_2\ge 3$, $x_1\ge 0$, $x_2\ge 0$. The first two require $x_1+x_2$ to be at most $1$ and at least $3$ at once. No number is both, so no point is feasible.

Bounded, unbounded and empty are the three possibilities, and which one holds is a property of the constraints alone.

These two properties are often merged, but they answer different questions. The region is determined by the constraints; whether a best value exists also depends on the objective and the direction you optimize in.

Take the same unbounded region throughout:

A finite optimum exists. Minimise $x_1+x_2$. The objective is nonnegative on the region and equals $0$ at $(0,0)$, which is feasible. The minimum is $0$, attained, despite the region being unbounded.

No finite optimum exists. Maximise $x_2$ over the same region. The points $(0,t)$ are feasible for every $t\ge 0$, so the objective grows without limit. The program is unbounded.

The region did not change between these two cases. What changed was the objective.

The reliable order is: classify the region from the constraints first, then ask separately whether the objective attains a finite best value on it. An unbounded region is a necessary condition for an unbounded objective value, never a sufficient one.

If the feasible region of a linear program is unbounded, the optimal objective value must also be unbounded.