Why Junctions Need Their Own Checks
A smooth boundary can be analyzed one arc at a time, but a boundary made of several joined arcs has places where one piece ends and another begins. At such a junction the boundary may change direction, so a derivative test along a single smooth arc does not decide whether the junction is a maximum or minimum. It must be checked directly.
There is another important qualification. The derivative test finds an interior extremum of a restricted function only when that restriction is differentiable there. If it is not differentiable, an extremum may occur without the derivative being zero. A complete boundary search therefore checks junctions, stationary points on each arc, and any interior points of an arc where the restriction is not differentiable.
Restricting the Objective to an Arc
Let an arc be parametrized by \(\mathbf{r}(t)=(x(t),y(t))\), where \(a\le t\le b\), and let the objective be \(f(x,y)\). Its restriction to the arc is the single-variable function
The arc endpoints, corresponding to \(t=a\) and \(t=b\), are boundary junctions or endpoints of the boundary piece. The interior of the arc corresponds to \(a<t<b\). If \(g\) is differentiable at an interior point where it has a local maximum or minimum, Fermat's theorem gives \(g'(t)=0\). If \(g\) is not differentiable at an interior point, that point must be considered separately: Fermat's theorem does not apply there.
For a finite collection of joined arcs, repeat this procedure for every arc. The same geometric junction may appear as the endpoint of two adjacent parametrizations; it is still only one point to evaluate. Including every junction ensures that a candidate is not lost just because it is an endpoint for the arc under consideration.
Why These Candidates Are Sufficient
The boundary of a compact region is compact. If \(f\) is continuous on the region, its restriction to the boundary is continuous, so it attains an absolute maximum and an absolute minimum there. The following argument explains why the boundary candidates listed above capture those extrema.
Since the boundary is covered by its arcs, a boundary extremum lies on at least one of them. Write the restricted function on that arc as \(g(t)\), for \(a\le t\le b\).
If the extremum occurs at \(t=a\) or \(t=b\), it is a junction or an endpoint of a boundary piece and is included in the junction checks.
If it occurs at \(a<t<b\), then either \(g\) is differentiable there or it is not. In the differentiable case, Fermat's theorem gives \(g'(t)=0\). In the non-differentiable case, the point is included among the non-differentiability candidates.
Thus every boundary maximum and minimum is in the candidate set. Conversely, each candidate is a point on the boundary, so evaluating \(f\) there gives an actual boundary value. Comparing the values at all candidates therefore gives the boundary extrema. For absolute extrema on the whole region, combine the boundary candidates with the interior candidates from the interior-and-boundary candidate principle.
Worked Example: A Rectangular Boundary
Worked Example: A Rectangular Boundary
Find the maximum and minimum values of \(f(x,y)=(x-1)^2+(y-\frac12)^2\) on the boundary of the rectangle \(0\le x\le2,\ 0\le y\le1\).
Parametrize the bottom edge by \((t,0)\), \(0\le t\le2\). Its restriction and derivative are
The derivative is zero at \(t=1\), giving the point \((1,0)\) and value \(f(1,0)=0+\frac14=\frac14\). The endpoints \(t=0,2\) are corners.
On the top edge, \((t,1)\), the restriction is \(g_2(t)=(t-1)^2+\frac14\), so \(g_2'(t)=2(t-1)\). The stationary point \(t=1\) gives \((1,1)\) and value \(\frac14\). The endpoints again are corners.
On the left edge, \((0,t)\), the restriction is \(g_3(t)=1+(t-\frac12)^2\), so \(g_3'(t)=2(t-\frac12)\). Its zero \(t=\frac12\) gives \((0,\frac12)\) and value \(1\). On the right edge, \((2,t)\), the restriction is also \(g_4(t)=1+(t-\frac12)^2\), with derivative \(2(t-\frac12)\). Its zero gives \((2,\frac12)\) and value \(1\).
Now check all four junctions. At \((0,0)\) and \((0,1)\), the value is \(1+\frac14=\frac54\). At \((2,0)\) and \((2,1)\), it is also \(1+\frac14=\frac54\). Thus the boundary minimum is \(\frac14\), attained at \((1,0)\) and \((1,1)\); the boundary maximum is \(\frac54\), attained at all four corners. Direct substitution into \(f\) confirms each listed value. The center \((1,\frac12)\) is an interior point, not a boundary candidate; its value is \(0\), so the distinction between boundary extrema and extrema on the whole region matters here.
Worked Example: A Circular Arc Joined to a Line
Worked Example: A Circular Arc Joined to a Line
Find the boundary maximum and minimum of \(f(x,y)=x+y\) on the upper half disk \(x^2+y^2\le1,\ y\ge0\). Its boundary consists of the upper semicircle and the diameter.
Parametrize the semicircle by \((\cos t,\sin t)\), \(0\le t\le\pi\). The restricted function is \(g(t)=\cos t+\sin t\), so \(g'(t)=-\sin t+\cos t\). Setting the derivative to zero gives \(\cos t=\sin t\). In this interval the solution is \(t=\frac{\pi}{4}\). The point is \((\frac{\sqrt2}{2},\frac{\sqrt2}{2})\), and its value is \(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2\).
The semicircle endpoints are junctions: \(t=0\) gives \((1,0)\), with value \(1\), and \(t=\pi\) gives \((-1,0)\), with value \(-1\).
Parametrize the diameter by \((-1+2s,0)\), \(0\le s\le1\). Its restriction is \(h(s)=-1+2s\), and \(h'(s)=2\), which has no zero. The endpoints give \(-1\) and \(1\), already checked as junctions. Both restrictions are differentiable throughout their arc interiors, so there are no additional non-differentiability candidates.
Comparing all candidates, the boundary minimum is \(-1\) at \((-1,0)\), and the boundary maximum is \(\sqrt2\) at \((\frac{\sqrt2}{2},\frac{\sqrt2}{2})\). Substitution into \(x+y\) verifies both values. Also, the interior partial derivatives are \(f_x=1\) and \(f_y=1\), so there is no interior critical point; the boundary results give the absolute extrema on this half disk.
Worked Example: A Nondifferentiable Arc Restriction
Worked Example: A Nondifferentiable Arc Restriction
Consider the square \(0\le x\le1,\ 0\le y\le1\) and the continuous objective \(f(x,y)=-|x-\frac12|+y\). Find its boundary maximum and minimum.
On the bottom edge \((t,0)\), the restriction is \(g(t)=-|t-\frac12|\). For \(t<\frac12\), this is \(t-\frac12\), with derivative \(1\). For \(t>\frac12\), it is \(\frac12-t\), with derivative \(-1\). The restriction is not differentiable at \(t=\frac12\), so include \((\frac12,0)\); its value is \(0\). The endpoints \((0,0)\) and \((1,0)\) each have value \(-\frac12\).
On the top edge \((t,1)\), the restriction is \(-|t-\frac12|+1\). It has the same derivatives \(1\) for \(t<\frac12\) and \(-1\) for \(t>\frac12\), and is not differentiable at \(t=\frac12\). Include \((\frac12,1)\), where the value is \(1\). The endpoints \((0,1)\) and \((1,1)\) each have value \(\frac12\).
On the left edge \((0,t)\), the restriction is \(-\frac12+t\), with derivative \(1\); on the right edge \((1,t)\), it is also \(-\frac12+t\), with derivative \(1\). Neither has a stationary point or an interior non-differentiability point. Their endpoint values are already included among the four corner values.
Therefore the boundary maximum is \(1\) at \((\frac12,1)\), and the boundary minimum is \(-\frac12\) at \((0,0)\) and \((1,0)\). Substitution verifies these values. If one checked only points with derivative zero and the corners, the top-edge maximum would be missed: on either side of \(t=\frac12\), the derivative is \(1\) or \(-1\), never zero.
A Common Mistake: Treating Continuity as Differentiability
A frequent error is to state that every boundary extremum in an arc interior must make the restricted derivative zero, based only on continuity of the objective. The square example shows why this fails. Its objective is continuous, and the top-edge restriction has an interior maximum at \(t=\frac12\), but that restriction has no derivative there.
The correction is to state the hypotheses of the derivative test precisely. If the restriction is differentiable at an interior extremum, its derivative is zero. If it is not differentiable, include the point as a separate candidate. Never apply the derivative-zero conclusion at a junction either: a junction is an endpoint of each adjoining arc, and is checked by evaluating the objective there.
A Practical Boundary Checklist
- List the smooth arcs that cover the boundary, including each boundary component.
- For each arc, write a parametrization and form the restricted function \(g(t)=f(x(t),y(t))\).
- Find every interior solution of \(g'(t)=0\) where \(g\) is differentiable.
- Find every interior point where \(g\) is not differentiable and add it as a candidate.
- Evaluate the objective at every junction and endpoint, then compare all candidate values.
- For extrema on the entire region, include the appropriate interior candidates as well.
Some points can be listed more than once, for example when two arc parametrizations share a junction. This causes no problem; record the point once in the final candidate list and evaluate its objective value once. A candidate set may also contain points that are not extrema. The method is designed to avoid missing extrema, not to predict in advance which candidates will win.
Check Your Understanding
Answer these questions using the boundary-candidate method.
- Why must a junction be checked even if the objective is differentiable in the plane?
- What condition lets you conclude that an arc-interior extremum has restricted derivative zero?
- What additional points must be checked when an arc restriction is continuous but not differentiable throughout its interior?
- For \(g(t)=|t-2|\) on \(1\le t\le3\), identify the interior candidate not found by solving \(g'(t)=0\).
- Why are boundary candidates alone insufficient when finding extrema on the whole region?