Why an Inequality Describes More Than Its Equality
An equality constraint such as \(g(x,y)=c\) restricts points to a level set. By contrast, an inequality constraint such as \(g(x,y)\leq c\) generally admits a whole region of points. Replacing the inequality with the equality keeps only points where the constraint is active; it can remove feasible points where an extremum occurs.
This distinction matters in constrained optimization. An equality-constrained calculation can help analyze the active level set, but it does not by itself settle optimization over the entire inequality-feasible region. We will see this directly in examples, without introducing inequality multiplier methods.
A global minimum of \(f\) on \(D\) is a point \(\mathbf{x}_*\in D\) such that \(f(\mathbf{x}_*)\leq f(\mathbf{x})\) for every \(\mathbf{x}\in D\); a global maximum reverses the inequality. A local extremum uses the same comparison only for feasible points sufficiently close to \(\mathbf{x}_*\). The distinction between local and global does not change the central issue: an extremum may occur at an inactive point.
What Continuity Says About the Boundary
For a continuous constraint function, every boundary point of the feasible set must satisfy the constraint with equality. The converse is not guaranteed: equality points need not be boundary points. The following argument establishes the precise one-way statement.
Here \(\partial D\) denotes the geometric boundary of \(D\): a point is on the boundary if every open ball centered at it meets both \(D\) and its complement. To prove the inclusion, take \(\mathbf{p}\in\partial D\). Since \(g\) is continuous, \(D\), the inverse image of the closed interval \((-\infty,c]\), is closed. Thus \(\mathbf{p}\in D\), so \(g(\mathbf{p})\leq c\). If \(g(\mathbf{p})<c\), continuity gives a neighborhood of \(\mathbf{p}\) on which \(g<c\). That neighborhood is contained in \(D\), making \(\mathbf{p}\) an interior point, contrary to \(\mathbf{p}\in\partial D\). Therefore \(g(\mathbf{p})=c\), as claimed.
To see why the reverse inclusion can fail, let \(g(x,y)=0\) everywhere and take \(c=0\). Then \(D=\mathbb{R}^2\), whose boundary is empty, while every point satisfies \(g(x,y)=c\). So the active set is all of \(\mathbb{R}^2\), not the boundary. The words “active point” and “boundary point” should therefore not be treated as interchangeable without additional assumptions.
Worked Example: A Disk with an Interior Minimum
Worked Example: A Disk with an Interior Minimum
Find the global minimum and maximum of \(f(x,y)=x^2+y^2\) on \(D=\{(x,y):x^2+y^2\leq 1\}\). Then compare these values with what an equality-only calculation finds.
Every point in \(D\) satisfies \(x^2+y^2\geq0\). The origin is feasible because \(0^2+0^2=0\leq1\), and \(f(0,0)=0\). Thus the global minimum is \(0\), attained at \((0,0)\). This point is inactive: its constraint value is \(0<1\).
For every feasible point, \(f(x,y)=x^2+y^2\leq1\). Equality is attained, for example, at \((1,0)\), so the global maximum is \(1\). The full feasible set is closed and bounded, and \(f\) is continuous, so the extreme value theorem also guarantees that both extrema are attained.
Now impose only the equality \(x^2+y^2=1\). On this circle, \(f(x,y)=1\) at every point. The equality set recovers the maximum but gives no information about the true minimum \(0\), because it has discarded the origin. Substitution verifies the claimed extrema: \(f(0,0)=0\), while \(f(1,0)=1\).
Why the Equality-Only Search Can Miss an Extremum
When an equality set \(A\) is contained in the feasible set \(D\), optimizing on \(A\) examines only a subset of the points allowed by the inequality. If the extrema exist, then
These comparisons follow directly from the definitions: every point of \(A\) is among the feasible points in \(D\), so a maximum over all of \(D\) cannot be smaller than a maximum over \(A\), and a minimum over all of \(D\) cannot be larger than a minimum over \(A\). Equality may occur, as in the disk example for the maximum, but it need not. A calculation on \(A\) cannot rule out better values at points in \(D\setminus A\).
Interior points deserve particular attention. If \(f\) is differentiable and a point is a local extremum in the interior of a region in \(\mathbb{R}^n\), then the ordinary first-derivative necessary condition is \(\nabla f=\mathbf{0}\). This condition is not sufficient to prove an extremum, and it does not apply in the same way at a boundary point, where feasible directions may be restricted. The practical lesson here is modest: a boundary-only search must be accompanied by a check of feasible interior candidates.
Worked Example: An Interior Maximum in a Half-Plane
Worked Example: An Interior Maximum in a Half-Plane
Maximize \(f(x,y)=-(x^2+y^2)\) subject to \(x\leq1\), and compare the answer with optimization on the active line \(x=1\).
The feasible set is \(D=\{(x,y):x\leq1\}\). Since \(x^2+y^2\geq0\), we have \(f(x,y)\leq0\) throughout \(D\). The point \((0,0)\) is feasible and gives \(f(0,0)=0\), so the global maximum is \(0\), attained at the inactive point \((0,0)\), where \(0<1\).
On the active set \(x=1\), the objective becomes \(f(1,y)=-(1+y^2)\). Because \(y^2\geq0\), this is at most \(-1\), with equality at \(y=0\). Thus the maximum on the active set is \(-1\), attained at \((1,0)\). It is strictly less than the maximum \(0\) on the feasible half-plane. Substitution checks both values: \(f(0,0)=0\) and \(f(1,0)=-1\).
This unbounded feasible set has no global minimum: along feasible points \((0,t)\), \(f(0,t)=-t^2\), which decreases without bound as \(|t|\) increases. The example concerns the maximum, which does exist, and shows that restricting to the active set loses it.
When Boundary Equality Does Capture an Extremum
Replacing an inequality by its active equality is not always wrong; it is incomplete unless its relevance is established. An extremum may genuinely occur on the active set. The next example has both a minimum and a maximum on the boundary, so the equality set happens to find them. It also illustrates why this outcome is a feature of that objective and feasible set, not a general rule.
Worked Example: A Linear Objective on a Disk
Find the global extrema of \(f(x,y)=x\) on \(D=\{(x,y):x^2+y^2\leq4\}\), and compare with the circle \(x^2+y^2=4\).
For every feasible point, \(x^2\leq x^2+y^2\leq4\), so \(-2\leq x\leq2\). The points \((-2,0)\) and \((2,0)\) are feasible because each satisfies \(x^2+y^2=4\). They give \(f(-2,0)=-2\) and \(f(2,0)=2\). Therefore the global minimum is \(-2\), and the global maximum is \(2\).
On the equality circle, the same bounds \(-2\leq x\leq2\) hold, and the same two points attain them. Thus equality-only optimization captures both extrema in this case. The direct bounds and substitutions verify the result. Unlike the previous examples, this objective has no better value in the interior: the coordinate \(x\) cannot exceed the endpoints already attained on the circle.
A Reliable Way to Use an Equality Calculation
An equality calculation can be useful as part of a larger argument, but it should be interpreted as a calculation on the active set, not automatically as a solution of the inequality-constrained problem. The following sequence keeps the distinction clear.
Record the inequality itself, for example \(D=\{\mathbf{x}:g(\mathbf{x})\leq c\}\), rather than silently replacing it by an equality.
The equation \(g(\mathbf{x})=c\) describes points where the inequality is active. For continuous \(g\), the boundary is contained in this level set, but the two sets can differ.
For differentiable objectives, interior local extrema must satisfy \(\nabla f=\mathbf{0}\). This is a necessary condition, not a guarantee that a candidate is an extremum.
Check all relevant candidates and justify that no other feasible points can give a better value. If the feasible set is compact and the objective continuous, global extrema are guaranteed to exist.
Common Mistake: Assuming Every Active Point Is on the Boundary
A tempting shortcut is to write, “The boundary is \(g(\mathbf{x})=c\), so optimize there.” This can make two errors at once: it identifies the boundary with the active set without justification, and it ignores feasible inactive points. For example, if \(g(x,y)=0\) and \(c=0\), then \(g=c\) everywhere, but \(D=\mathbb{R}^2\) and \(\partial D=\varnothing\). Thus the asserted boundary equation would incorrectly label every point as a boundary point.
The correction is to state only what is warranted: for continuous \(g\), \(\partial D\subseteq\{\mathbf{x}:g(\mathbf{x})=c\}\). Then analyze the feasible region, including its inactive points. The two disk examples show why: an equality-only calculation found the maximum in one case and missed an interior minimum; in another, it missed the interior maximum.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- For continuous \(g\) and \(D=\{\mathbf{x}:g(\mathbf{x})\leq c\}\), what inclusion relates \(\partial D\) to the level set \(g=c\), and why does continuity matter?
- For \(x^2+y^2\leq1\) and \(f(x,y)=x^2+y^2\), which extremum is missed if you optimize only on \(x^2+y^2=1\)?
- In the half-plane example, what is the maximum on the feasible region, and what is the maximum on the active line?
- Give an example in which optimizing on the active equality does capture the extrema over the whole feasible set.
- Why does the equation \(g(\mathbf{x})=c\) not always describe the geometric boundary of \(D\)?