Feasible Sets

Non-empty and bounded. $x^2+y^2\le 1$. Every feasible point lies within distance 1 of the origin, so the set fits in a ball and is bounded.

Non-empty and unbounded. $y\ge x^2$. The point $(t,t^2)$ is feasible for every real $t$, so the set contains points arbitrarily far from the origin.

Empty. $x\ge 1$ together with $x\le 0$. No number is at once at least 1 and at most 0, so nothing survives and the problem is infeasible.

Non-empty, unbounded, yet the objective is limited. $x\ge 0$, minimizing $x$. The set runs on without limit, but the smallest permitted value is 0. An unbounded feasible set does not by itself mean an unbounded answer. The two questions are separate.

Linear. $x+y\le 4$, $x\ge 0$, $y\ge 0$. The set is a triangle: flat edges, and the segment between any two feasible points stays feasible.

Not linear. $xy\ge 1$, . The set is the region above one branch of a hyperbola. The points $(1,1)$ and $(4,0.25)$ are both feasible, yet their midpoint $(2.5,0.625)$ has $xy=1.5625\ge 1$, feasible here. But take $(0.5,2)$ and $(2,0.5)$: both feasible, midpoint $(1.25,1.25)$ with $xy=1.5625$, still feasible. Now $(0.1,10)$ and $(10,0.1)$: midpoint $(5.05,5.05)$, feasible again.

The hyperbola region happens to be convex. Change the restriction to $xy\le 1$ with $x,y\ge 0$ and it is not: $(4,0)$ and $(0,4)$ are feasible, their midpoint $(2,2)$ has and is not.

The lesson is that convexity is something to check, not something a feasible set has by nature. Linear restrictions guarantee it; general restrictions do not.

If the feasible set is unbounded, the problem has no optimal solution.