Boundary Critical Points and Endpoint Values

Why Boundary Endpoints Matter

To find absolute extrema on a region, it is not enough to check points in the interior. A point where the function is largest or smallest may lie on the boundary. When a boundary piece is a smooth arc, we can study the function along that arc by expressing its points with one parameter.

This turns the boundary search into a one-variable problem. Interior points of the parameter interval can be tested by setting the derivative of the restricted function equal to zero. The two endpoints require separate attention: an extremum can occur there even when that derivative is not zero.

Key idea: On a closed smooth boundary arc, check every interior parameter where the restricted derivative is zero, and check both endpoints. The endpoint checks are required even if the derivative is nonzero there.

Restriction to a Parametrized Arc

Let $f$ be a differentiable function of two variables, and let a boundary arc be parametrized by $\boldsymbol{\gamma}(t)=(x(t),y(t))$, where $a\le t\le b$. The function restricted to the arc is the one-variable function $g(t)=f(x(t),y(t))$. A parameter value $t$ therefore corresponds to the boundary point $\boldsymbol{\gamma}(t)$, and $g(t)$ is the value of $f$ at that point.

Definition: For a parametrized arc $\boldsymbol{\gamma}:[a,b]\to\mathbb{R}^2$, the restriction of $f$ to the arc, expressed in the parameter $t$, is $g:[a,b]\to\mathbb{R}$ defined by $g(t)=f(\boldsymbol{\gamma}(t))=f(x(t),y(t))$.

The chain rule gives the derivative of this restriction:

g'(t) = f_x(x(t),y(t))x'(t) + f_y(x(t),y(t))y'(t) = \nabla f(\boldsymbol{\gamma}(t))\cdot\boldsymbol{\gamma}'(t).

This formula measures how the function changes as we move along the arc. It is the derivative along the boundary path, not a requirement that both partial derivatives of $f$ vanish.

The Complete Candidate Rule

Assume that $\boldsymbol{\gamma}$ is continuous on the closed interval $[a,b]$, that $g=f\circ\boldsymbol{\gamma}$ is continuous there, and that $g$ is differentiable for $a<t<b$. The Extreme Value Theorem guarantees that $g$ attains a maximum and a minimum on $[a,b]$. If either extremum occurs at an interior parameter $t_0$, Fermat’s theorem for one-variable functions says that $g'(t_0)=0$. Otherwise, it occurs at an endpoint.

Theorem: Under these continuity and differentiability conditions, every maximum or minimum of $f$ restricted to the arc occurs at a point in the candidate set $$ \{\boldsymbol{\gamma}(a),\boldsymbol{\gamma}(b)\} \;\cup\; \{\boldsymbol{\gamma}(t):a<t<b,\;g'(t)=0\}. $$ The endpoints are included whether or not the derivative there is defined or equal to zero.

This statement gives a complete set of necessary candidates; it does not say that every candidate is an extremum. A zero derivative might correspond to neither a maximum nor a minimum. Evaluate $f$ at all candidates and compare the resulting values to determine the extrema on the arc.

Why the Rule Works

The justification is a direct application of the one-variable extreme-value and Fermat theorems. The key point is that an extremum strictly inside the parameter interval can be approached along the arc in both directions. At an endpoint, only one direction along the interval is available, so the derivative need not vanish.

Form the restricted function.
Set $g(t)=f(\boldsymbol{\gamma}(t))$. It is continuous on the closed interval $[a,b]$, so it attains its largest and smallest values there.

Consider an interior extremum.
If $t_0$ is an interior maximum or minimum of $g$, then $g$ has a local extremum at $t_0$. Since $g$ is differentiable there, Fermat’s theorem implies $g'(t_0)=0$.

Include the remaining possibilities.
If an extremum is not at an interior zero of $g'$, it must occur at $a$ or $b$. Thus both endpoint values belong on the candidate list.

When the arc is regular, meaning $\boldsymbol{\gamma}'(t)\ne\mathbf{0}$, the equation $g'(t)=0$ says that the directional change of $f$ along the tangent to the arc is zero. It does not require $\nabla f=\mathbf{0}$: the full gradient may be nonzero while its component along the arc’s tangent is zero.

Worked Examples

Worked Example: A Parabolic Arc with an Interior Candidate

Find the maximum and minimum of $f(x,y)=x^2+y$ on the arc $\boldsymbol{\gamma}(t)=(t,t^2)$, $-1\le t\le1$.

First restrict $f$ to the arc:

$$ g(t)=f(t,t^2)=t^2+t^2=2t^2. $$

Differentiate and solve for interior stationary parameters:

g'(t)=4t, \qquad 4t=0 \quad\Longrightarrow\quad t=0.

The full candidate parameter list is $t=-1,0,1$: the stationary parameter $0$ and both endpoints. Their points and function values are

\begin{aligned} t=-1 &: \quad \boldsymbol{\gamma}(-1)=(-1,1), & f(-1,1)&=1+1=2,\\ t=0 &: \quad \boldsymbol{\gamma}(0)=(0,0), & f(0,0)&=0+0=0,\\ t=1 &: \quad \boldsymbol{\gamma}(1)=(1,1), & f(1,1)&=1+1=2. \end{aligned}

Thus the minimum on this arc is $0$, attained at $(0,0)$, and the maximum is $2$, attained at both endpoints. As a check, $g(t)=2t^2\ge0$ for every $t$ in the interval, and $g(t)\le2$ when $-1\le t\le1$, confirming the candidate comparison.

Worked Example: Several Stationary Parameters

Find the maximum and minimum of $f(x,y)=(x^2-1)^2+y$ on the line segment $\boldsymbol{\gamma}(t)=(t,0)$, $-2\le t\le2$.

The restriction is

$$ g(t)=f(t,0)=(t^2-1)^2. $$

Differentiate by the chain rule:

$$ g'(t)=2(t^2-1)(2t)=4t(t^2-1)=4t(t-1)(t+1). $$

The interior stationary parameters are $t=-1,0,1$. Include the endpoints $t=-2$ and $t=2$. Evaluate $g$ at all five candidates:

\begin{aligned} g(-2)&=((-2)^2-1)^2=(4-1)^2=9,\\ g(-1)&=((-1)^2-1)^2=(1-1)^2=0,\\ g(0)&=(0^2-1)^2=(-1)^2=1,\\ g(1)&=(1^2-1)^2=(1-1)^2=0,\\ g(2)&=(2^2-1)^2=(4-1)^2=9. \end{aligned}

The minimum is $0$, attained at $(-1,0)$ and $(1,0)$; the maximum is $9$, attained at the two endpoints $(-2,0)$ and $(2,0)$. To check the stationary calculation, substitution into $g'(t)=4t(t-1)(t+1)$ gives zero at each of $-1,0,1$. The endpoint values are obtained directly from $g$, as required even though those parameters are not interior stationary points.

Worked Example: An Arc with No Interior Stationary Parameter

Find the maximum and minimum of $f(x,y)=x$ on the parabolic arc $\boldsymbol{\gamma}(t)=(t,t^2)$, $0\le t\le2$.

Restrict the function and differentiate:

g(t)=f(t,t^2)=t, \qquad g'(t)=1.

The equation $g'(t)=0$ has no solutions, so there are no interior stationary candidates. The endpoints must still be checked:

\begin{aligned} \boldsymbol{\gamma}(0)&=(0,0), & f(0,0)&=0,\\ \boldsymbol{\gamma}(2)&=(2,4), & f(2,4)&=2. \end{aligned}

Therefore the minimum on the arc is $0$ at $(0,0)$, and the maximum is $2$ at $(2,4)$. The check is immediate from $g(t)=t$: on $0\le t\le2$, its values range from $0$ to $2$. In particular, the absence of interior stationary parameters does not mean the arc has no extrema; both occur at endpoints.

A Common Mistake: Omitting the Endpoints

A frequent error is to solve $g'(t)=0$ and treat the resulting stationary parameters as the entire candidate list. The third example shows why that fails. There, $g'(t)=1$ everywhere, so the stationary-point equation has no solution. Concluding that there are no extrema would be wrong: the restricted function has a minimum at one endpoint and a maximum at the other.

Correction: For a restriction to $[a,b]$, use the interior equation $g'(t)=0$ only for $a<t<b$, and add $t=a$ and $t=b$ regardless. Endpoint extrema are one-sided possibilities and need not satisfy the derivative-zero condition.

Another related error is to solve $f_x=f_y=0$ as though every boundary extremum must be a critical point of $f$ in the full plane. The condition for an interior extremum along the arc is instead $g'(t)=0$, or equivalently $\nabla f(\boldsymbol{\gamma}(t))\cdot\boldsymbol{\gamma}'(t)=0$. These are different tests: the latter checks change along the boundary direction.

Using the Candidate List

For one closed smooth arc, the practical steps are:

Write its parametrization $\boldsymbol{\gamma}(t)$ and parameter interval $[a,b]$.
Form $g(t)=f(\boldsymbol{\gamma}(t))$.
Differentiate $g$, using the chain rule.
Solve $g'(t)=0$ for parameters strictly inside $(a,b)$.
Add both endpoint parameters $a$ and $b$.
Evaluate $f$ at the corresponding boundary points and compare the values.

The candidate set may contain repeated points or points that are not extrema; keeping such candidates is harmless. Also, the stationary set need not always be finite. If $g'$ vanishes at many parameter values, the rule still identifies the endpoint and stationary locations that must be considered, but solving or describing that set may require further analysis.

When a region’s boundary is made of several arcs, this procedure applies to each arc, with the endpoints of every arc included. A point shared by two arcs may appear more than once; it is still just one geometric candidate. The interior candidates of the region are a separate part of the absolute-extrema search.

Check Your Understanding

Use the restriction-to-an-arc rule to answer these questions.

For $g(t)=f(\boldsymbol{\gamma}(t))$ on $[a,b]$, which parameter values must be checked besides interior solutions of $g'(t)=0$?
Why does an interior extremum of the restricted function require $g'(t)=0$, while an endpoint extremum need not?
For $f(x,y)=x+y$ and $\boldsymbol{\gamma}(t)=(t,t^2)$ on $[-1,1]$, find $g(t)$, its interior stationary parameters, and the endpoint parameters.
Does $g'(t)=0$ imply that both partial derivatives $f_x$ and $f_y$ vanish? Explain using the chain-rule formula.
If $g'(t)$ has no zero in $(a,b)$, can $g$ still attain a maximum and minimum on $[a,b]$? Where must they occur?