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The Interior-And-Boundary Candidate Principle

Use compactness and the interior derivative test to justify where absolute maxima and minima on a planar region must be found.

Intermediate 9 min read

What You'll Learn

  • State the interior-and-boundary candidate principle and its assumptions.
  • Explain why compactness guarantees absolute extrema and boundary extrema.
  • Derive why an interior extremum of a differentiable function must be a critical point.
  • Distinguish candidates from points that are guaranteed to be absolute extrema.
  • Apply the principle to functions on a closed disk.
  • Recognize why searching only for interior critical points can miss an absolute extremum.

Why the Boundary Must Be Included

A function can have an absolute maximum or minimum on a region even when it has no critical point in the region’s interior. The reason is simple: an extremum may occur at the boundary, where the function cannot be varied in every direction while staying inside the region.

This observation turns the search for absolute extrema into two related tasks: examine critical points in the interior, and examine extrema of the function restricted to the boundary. The principle is useful because it explains why neither search can generally be omitted.

Definitions and Assumptions

A planar region here is a nonempty subset \(K\) of \(\mathbb{R}^2\). It is closed and bounded when it is compact. Its interior, written \(\operatorname{int}(K)\), consists of points that have an open disk around them lying entirely in \(K\). Its boundary, written \(\partial K\), consists of points every open disk around which meets both \(K\) and its complement.

For a function \(f:K\to\mathbb{R}\), an absolute maximum at \(\mathbf{p}\in K\) means \(f(\mathbf{p})\geq f(\mathbf{q})\) for every \(\mathbf{q}\in K\). An absolute minimum is defined by reversing the inequality. An interior point \((a,b)\) is a critical point of \(f\) when both first partial derivatives exist there and are zero.

Definition: The restriction of \(f\) to the boundary is the function \(f|_{\partial K}:\partial K\to\mathbb{R}\) given by \(f|_{\partial K}(\mathbf{p})=f(\mathbf{p})\). A boundary extremum means an absolute maximum or absolute minimum of this restricted function.

The candidate principle requires that \(K\) be closed and bounded and that \(f\) be continuous on \(K\). To identify interior extrema as critical points, also require that \(f\) be differentiable at interior points. Differentiability can be replaced by the weaker requirement that the two first partial derivatives exist at any interior extremum being tested.

The Interior-And-Boundary Candidate Principle

Theorem: Suppose \(K\subset\mathbb{R}^2\) is nonempty, closed, and bounded, and \(f:K\to\mathbb{R}\) is continuous. Then \(f\) attains an absolute maximum and an absolute minimum on \(K\). If \(f\) is differentiable at every point of \(\operatorname{int}(K)\), then every absolute extremum occurs either at an interior critical point or at a boundary extremum of \(f|_{\partial K}\).

The existence statement is the extreme value theorem on a compact set. The location statement adds a local derivative argument: an interior extremum cannot have a nonzero first derivative in either coordinate direction.

Why an Interior Extremum Must Be Critical

Let \(\mathbf{p}=(a,b)\) be an interior point where \(f\) has a local maximum or minimum. Because \(\mathbf{p}\) is interior, there is a small disk centered at \(\mathbf{p}\) that lies inside \(K\). In particular, for all sufficiently small positive and negative \(t\), the points \((a+t,b)\) and \((a,b+t)\) remain in \(K\).

Consider the one-variable function \(g(t)=f(a+t,b)\). It has a local extremum at \(t=0\), so the one-variable derivative test gives \(g'(0)=0\). By the definition of the partial derivative, \(g'(0)=f_x(a,b)\). Applying the same reasoning to \(h(t)=f(a,b+t)\) gives \(f_y(a,b)=0\). Therefore the gradient vanishes at \(\mathbf{p}\).

$$ \nabla f(a,b)=\bigl(f_x(a,b),f_y(a,b)\bigr)=(0,0). $$

This reasoning depends on being able to move a small amount in both positive and negative coordinate directions. At a boundary point, that may not be possible while remaining in \(K\). The derivative argument therefore does not rule out a boundary extremum with a nonzero gradient.

Proof of the Candidate Principle

1
Existence.
Since \(K\) is closed and bounded in \(\mathbb{R}^2\), it is compact. The extreme value theorem says that the continuous function \(f\) attains an absolute maximum and an absolute minimum on \(K\).
2
Classify the location of an extremum.
Take either absolute extremum and call its location \(\mathbf{p}\). Every point of \(K\) lies either in \(\operatorname{int}(K)\) or on \(\partial K\).
3
Use the appropriate condition.
If \(\mathbf{p}\) is interior, the coordinate-direction argument shows that \(f_x(\mathbf{p})=f_y(\mathbf{p})=0\). If \(\mathbf{p}\) is on the boundary, its value is an absolute extremum of \(f|_{\partial K}\): no boundary point can have a larger value than a global maximum on \(K\), or a smaller value than a global minimum on \(K\).

Thus every absolute extremum is accounted for by one of the two candidate types. The principle does not say that every interior critical point is an absolute extremum, or that every boundary extremum is an extremum on all of \(K\). It says that the global extrema must be found among these candidates.

Worked Example: A Radial Function on the Unit Disk

Worked Example: A Radial Function on the Unit Disk

Let \(K=\{(x,y):x^2+y^2\leq 1\}\) and \(f(x,y)=x^2+y^2\). Find the absolute extrema by separating the interior and boundary candidates.

The partial derivatives are \(f_x=2x\) and \(f_y=2y\). Setting both equal to zero gives the only interior critical point \((0,0)\), with \(f(0,0)=0\).

On the boundary, \(x^2+y^2=1\), so \(f(x,y)=1\) at every boundary point. Consequently, the boundary maximum and boundary minimum are both \(1\).

For a direct check, every point of \(K\) satisfies \(0\leq x^2+y^2\leq 1\). The lower equality occurs at \((0,0)\), and the upper equality occurs at every point of the boundary. Therefore the absolute minimum is \(0\) at \((0,0)\), and the absolute maximum is \(1\) on the boundary. These values agree with the candidate search.

Worked Example: A Function with No Interior Critical Points

Worked Example: A Function with No Interior Critical Points

On the same unit disk, let \(f(x,y)=x+y\). First, \(f_x=1\) and \(f_y=1\), so the gradient is never zero. There are no interior critical points.

For every point in the disk, the inequality \((x+y)^2\leq 2(x^2+y^2)\) gives \((x+y)^2\leq 2\), since \(x^2+y^2\leq 1\). Hence \(-\sqrt{2}\leq f(x,y)\leq\sqrt{2}\).

The upper bound is attained at \((1/\sqrt{2},1/\sqrt{2})\): its squared coordinates sum to \(1\), and its function value is \(\sqrt{2}\). The lower bound is attained at \((-1/\sqrt{2},-1/\sqrt{2})\): its squared coordinates also sum to \(1\), and its function value is \(-\sqrt{2}\). Both points are on the boundary. Thus the absolute maximum is \(\sqrt{2}\) and the absolute minimum is \(-\sqrt{2}\), despite there being no interior critical points.

Worked Example: An Interior Minimum and a Boundary Maximum

Worked Example: An Interior Minimum and a Boundary Maximum

Let \(K\) again be the unit disk and take \(f(x,y)=x^2+y^2-x\). Its partial derivatives are \(f_x=2x-1\) and \(f_y=2y\). Solving \(2x-1=0\) and \(2y=0\) gives \((1/2,0)\), which lies in the interior because \(1/4<1\). Its value is

$$ f\left(\frac12,0\right)=\frac14-\frac12=-\frac14. $$

On the boundary, \(x^2+y^2=1\), so the restricted function has value \(f(x,y)=1-x\). Since \(-1\leq x\leq 1\), these boundary values lie between \(0\) and \(2\). The value \(0\) occurs at \((1,0)\), and \(2\) occurs at \((-1,0)\); both points lie on the unit circle.

For a global check, complete the square:

$$ f(x,y)=\left(x-\frac12\right)^2+y^2-\frac14. $$

The squared terms are nonnegative, so \(f\geq-1/4\), with equality at \((1/2,0)\). The boundary calculation gives a maximum of \(2\), and no interior point can exceed it: the candidate principle ensures a global maximum must be an interior critical point or a boundary extremum, while the only interior critical point has value \(-1/4\). Therefore the absolute minimum is \(-1/4\) at \((1/2,0)\), and the absolute maximum is \(2\) at \((-1,0)\). The candidate values and their locations are consistent with the direct minimum check.

A Common Mistake: Searching Only the Interior

A tempting but incorrect argument is: “There are no interior critical points, so there are no absolute extrema.” The second worked example disproves this. The function \(x+y\) on the unit disk has no critical points, yet it attains an absolute maximum and minimum at boundary points.

The correction is to use both parts of the principle. Interior critical points are necessary candidates for extrema that occur inside the region. Boundary extrema are necessary candidates for extrema that occur on the boundary. Compactness guarantees that absolute extrema exist, but it does not force them into the interior.

Key takeaway: The candidate principle is a completeness result, not a shortcut that eliminates checking. It identifies where absolute extrema must occur; it does not automatically identify which candidates give the global maximum or minimum.

Using the Principle Carefully

In an optimization problem, the principle provides the logical structure of the search. Find the interior critical points, determine the extrema of the boundary restriction, and compare the resulting function values. The candidate collection need not be finite: a boundary restriction may be constant on an entire boundary component, and an equation for critical points may have infinitely many solutions.

The assumptions matter. If the region is not compact or the function is not continuous, an absolute maximum or minimum may fail to exist. For example, \(f(x,y)=x\) on the open unit disk has no absolute maximum: values approach \(1\) near the boundary, but no point in the disk has value \(1\). The candidate principle cannot be applied because the region is not closed.

The principle also distinguishes a global problem from a local one. A critical point is identified using behavior in a neighborhood of an interior point. An absolute extremum compares function values across the entire region. A critical point can therefore be neither an absolute maximum nor an absolute minimum. Candidate values must be compared over the full candidate collection before a global conclusion is drawn.

Check Your Understanding

Use the assumptions and reasoning in this tutorial to answer the following.

  1. Why does continuity on a closed, bounded region guarantee that an absolute maximum and minimum exist?
  2. Why must the two first partial derivatives vanish at a differentiable interior extremum?
  3. What condition must a boundary point satisfy to be a boundary candidate for the theorem?
  4. Can an interior critical point fail to be an absolute extremum? Explain what the candidate principle does and does not claim.
  5. Why does the function \(x+y\) on the unit disk show that searching only for interior critical points is insufficient?

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