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An exposed point is a point uniquely selected by maximizing a linear functional. In ${\mathbb{R}}^d$, a point $p$ of a convex set $C$ is exposed if some vector $c$ satisfiesGeometrically, a supporting hyperplane touches the convex set only at $p$. Every exposed point is an extreme point: a convex combination that attains the maximum can put positive weight only on maximizers, and there is only one.^[ref]

The unit disk calculation in extreme point exposes every point of the unit circle. For a corner $(a,b)$ of the square, $c=(a,b)$ exposes it: $ax+by$ attains two only at $(a,b)$. However, maximizing $x$ exposes no single point on the right edge, since all of them have the same objective value. A linear functional may have several optimal points, even when some of them are extreme.

The distinction appears at a smooth join between a straight segment and a curved boundary. Consider the convex hullThe upper tangent from $P$ meets the unit circle at $T=(1/2,\sqrt{3}/2)$. The supporting line isEvery point of the unit disk has $T\cdot z\le 1$, with equality only at $T$. Hence the convex face of $C$ where equality holds is exactly the line segment $TP$.

The point $T$ is extreme. If a nontrivial convex combination of two points of $C$ equals $T$, both must belong to the convex face $TP$. Since $T$ is an endpoint of that line segment, both must equal $T$.

Yet $T$ is not exposed. A nonzero linear functional maximizing at $T$ on $C$ must also maximize there on the included unit disk. Its vector must be a positive multiple of $T$, by the equality condition in the Cauchy-Schwarz inequality. It then gives $P$ the same value as $T$. No linear functional can select $T$ alone. Being an extreme point is a property of possible mixtures; being an exposed point adds the ability to distinguish it by one linear objective.