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Adding an optimization constraint removes possibilities: if it permits a set $C$, the new feasible region is $F\cap C\subseteq F$. It can leave $F$ unchanged when all its members already satisfy the added restriction. For a fixed objective function, restricting the feasible set cannot lower a minimization problem's infimum. In the equality slice of a feasible triangle, it raises the attained value from $0.5$ to $1$.

Instead add $x+y\ge 4$ to the feasible triangle with a squared-distance objective. No point can have both $x+y\le 3$ and $x+y\ge 4$. The resulting feasible set is the empty set, so there is no feasible point and no optimizing decision. Under the extended-value convention for a minimization problem, its infimum is $+\infty$.^[ref]

A nonempty feasible set can also lack an optimizing decision. With domain $D=\mathbb{R}$, minimize $f(x)=x$ subject to the strict inequality $x>0$. The feasible region $(0,\infty )$ is an open set. Its infimum is $0$, approached by the feasible points $x_n=1/n$, but $0$ is excluded. Every feasible point $x$ is beaten by $x/2$, so no minimum is attained. Replacing $x>0$ by $x\ge 0$ includes the missing optimum. Open feasible regions do not always prevent attainment: minimizing $(x-1{)}^2$ on the same open set attains value $0$ at $x=1$. A useful sufficient condition is instead the extreme value theorem: a continuous objective function on a nonempty compact set attains its minimum.

Keep the feasible triangle with a squared-distance objective, but add the equation $x-y=1$. An equality requires an exact relation, whereas an inequality allows a range. Here the equality selects a line, and intersecting that line with the original feasible region leaves just the line segment from $(1,0)$ to $(2,1)$:The lower bound comes from $y\ge 0$; the upper bound comes from $x+y=2t+1\le 3$. This is a nonempty closed set even though it has empty interior in the plane. An equality can reduce the dimension of the allowed decisions without making the feasible region empty.

The old optimum $(1.5,1.5)$ violates the added equality. On the surviving line segment, the objective function isThe derivative shows that the unique global minimum is at $t=1$, giving $(x,y)=(2,1)$ and value $1$. The nearest point on the entire equality line would be $(2.5,1.5)$, at $t=1.5$ with value $0.5$, but it violates $x+y\le 3$. Both the equality and the original inequalities must hold simultaneously.

Use $g=x+y-3$ and $h=x-y-1$ to fix the multiplier sign convention in the optimization Lagrangian $L=f+\lambda g+\nu h$; the Lagrange multipliers of $-x$ and $-y$ are again zero. At $(2,1)$, Karush-Kuhn-Tucker conditions hold with $\lambda =1$ and $\nu =-1$:The Lagrange multiplier of an equality has no nonnegativity restriction. For an independent global certificate, completing the square gives $f+g-h=(x-2{)}^2+(y-1{)}^2+1$. On the feasible set, $h=0$ and $g\le 0$, so $f\ge 1$, with equality exactly at $(2,1)$.