Why a Complete Candidate Search Matters
A region between two curves has a boundary made from distinct pieces. An absolute maximum or minimum can occur in the interior, at a critical point of a boundary restriction, or where boundary pieces meet. A reliable solution must check all of these possibilities before comparing function values.
The goal here is to carry out that complete search on a closed region between graphs. The method uses the extreme value theorem and Fermat’s theorem: a continuous function on a compact region attains its absolute extrema, and a differentiable function at an interior local extremum has zero gradient. Along a boundary curve, the corresponding one-variable condition is that the derivative of the restricted function vanishes.
In the examples below, the graphs meet at the ends of the interval, so the boundary consists of just the two graph arcs. If the ends do not meet, the vertical segments must also be checked. Any point where boundary pieces meet is a junction and must be included among the boundary candidates.
A Candidate Search with Precise Conditions
Suppose \(D\) is a closed, bounded region between continuous graphs, and \(f\) is continuous on \(D\). The extreme value theorem guarantees that \(f\) attains an absolute maximum and minimum on \(D\). To locate them by a finite candidate search, add the assumption that \(f\) is differentiable at every interior point of \(D\), and that each boundary restriction is differentiable except possibly at finitely many parameters.
For a lower boundary \(y=g(x)\), define \(G(x)=f(x,g(x))\). For an upper boundary \(y=h(x)\), define \(H(x)=f(x,h(x))\). If there are vertical boundary segments, parameterize and restrict \(f\) to each one as well. On every boundary piece, check the endpoints, any junctions, and every point where the restriction is not differentiable or its derivative is zero.
The qualification about differentiability is important. Continuity guarantees that extrema exist, but it does not guarantee that every interior extremum has zero gradient. A continuous function can have an interior extremum at a point where it is not differentiable.
Why the Search Works
Let \(p\) be a point where the absolute maximum or minimum occurs. Because \(D\) is compact and \(f\) is continuous, such a point exists. If \(p\) lies in the interior and \(f\) is differentiable there, then it is also a local extremum in an open neighborhood. Fermat’s theorem applied in each coordinate direction gives \(f_x(p)=0\) and \(f_y(p)=0\). Thus \(\nabla f(p)=\mathbf{0}\).
If \(p\) lies on a smooth boundary arc parameterized by \(\mathbf{r}(t)\), the restricted function \(q(t)=f(\mathbf{r}(t))\) has a local extremum at the parameter corresponding to \(p\), unless \(p\) is an endpoint or junction. When \(q\) is differentiable there, the one-variable Fermat theorem gives \(q'(t)=0\). If it is not differentiable, the point must still be checked directly. This proves that the candidate search includes every possible location of an absolute extremum.
Write its inequalities and identify every boundary piece, endpoint, and junction.
Find points where both partial derivatives vanish, and include any interior points where the objective is not differentiable.
Differentiate each one-variable restriction and find its critical or nondifferentiable points.
Evaluate the objective at all candidates and select the smallest and largest values.
Worked Example: A Polynomial on a Curved Region
Worked Example: Find the Absolute Extrema Between Two Curves
Find the absolute maximum and minimum of \(f(x,y)=x^2+(y-1)^2\) on \(D=\{(x,y):-2\le x\le1,\ x^2\le y\le2-x\}\).
First check that the lower graph lies below the upper graph. Their difference is \(2-x-x^2=(1-x)(x+2)\), which is nonnegative for \(-2\le x\le1\). The graphs meet at \(x=-2\) and \(x=1\), so there are two boundary arcs and no vertical end segments. The region is closed and bounded, and \(f\) is continuous on it; hence absolute extrema exist.
For an interior candidate, compute the partial derivatives: \(f_x=2x\) and \(f_y=2(y-1)\). Setting both equal to zero gives \(x=0\) and \(y=1\). This point is interior because \(0<1<2\) when \(x=0\). Its value is \(f(0,1)=0^2+(1-1)^2=0\).
On the lower boundary, substitute \(y=x^2\): \(G(x)=x^2+(x^2-1)^2=x^4-x^2+1\). Its derivative is \(G'(x)=4x^3-2x=2x(2x^2-1)\). Thus the boundary critical parameters are \(x=0\) and \(x=\pm1/\sqrt{2}\), all in \([-2,1]\). The endpoint parameters are \(-2\) and \(1\).
Evaluate the restriction at its candidates: \(G(0)=1\); \(G(1/\sqrt{2})=1/4-1/2+1=3/4\); \(G(-1/\sqrt{2})=1/4-1/2+1=3/4\); \(G(-2)=16-4+1=13\); and \(G(1)=1-1+1=1\).
On the upper boundary, substitute \(y=2-x\): \(H(x)=x^2+(1-x)^2=2x^2-2x+1\). Its derivative is \(H'(x)=4x-2\), so its only critical parameter is \(x=1/2\). The endpoints are again \(-2\) and \(1\). The values are \(H(1/2)=1/2\), \(H(-2)=4+9=13\), and \(H(1)=1\).
The complete candidate values are therefore \(0\) at the interior point, \(1,3/4,3/4,13,1\) on the lower boundary, and \(1/2,13,1\) on the upper boundary. The smallest is \(0\), and the largest is \(13\). Thus the absolute minimum is \(0\) at \((0,1)\), and the absolute maximum is \(13\) at \((-2,4)\).
Check the reported values in the original function: \(f(0,1)=0\), and \(f(-2,4)=(-2)^2+(4-1)^2=4+9=13\). The endpoint \((-2,4)\) belongs to both graph arcs, so its value is the same whichever restriction is used.
Worked Example: A Boundary Restriction That Is Constant
Worked Example: A Region with a Whole Arc of Candidates
Find the absolute extrema of \(f(x,y)=x^2-y\) on \(D=\{(x,y):0\le x\le2,\ x^2\le y\le2x\}\).
The graphs meet at \(x=0\) and \(x=2\), since \(x^2=2x\) gives \(x(x-2)=0\). On \([0,2]\), \(x^2\le2x\), so the region is closed and bounded with two boundary arcs. The function is continuous there, so it attains both absolute extrema.
In the interior, \(f_x=2x\) and \(f_y=-1\). Since \(f_y\) is never zero, there is no interior point where both partial derivatives vanish.
On the lower boundary \(y=x^2\), the restriction is \(G(x)=x^2-x^2=0\) for every \(x\in[0,2]\). Its derivative is \(G'(x)=0\) throughout the interval. This means every point on the lower arc is a boundary candidate, and the value on that entire arc is \(0\). In particular, its endpoints have value \(0\).
On the upper boundary \(y=2x\), the restriction is \(H(x)=x^2-2x\). Its derivative is \(H'(x)=2x-2\), which vanishes at \(x=1\). The candidates on this arc are \(x=0,1,2\), with values \(H(0)=0\), \(H(1)=1-2=-1\), and \(H(2)=4-4=0\).
Comparing the lower-arc value \(0\) with the upper-arc values \(0,-1,0\), the absolute maximum is \(0\), attained at every point \((x,x^2)\) on the lower boundary. The absolute minimum is \(-1\), attained at \((1,2)\). Direct substitution checks the reported values: \(f(x,x^2)=x^2-x^2=0\) and \(f(1,2)=1-2=-1\).
Worked Example: An Interior Extremum at a Nondifferentiable Point
Worked Example: Why Continuity Alone Is Not Enough for a Gradient Search
On the unit square \(D=[0,1]\times[0,1]\), consider \(f(x,y)=-\sqrt{(x-\tfrac12)^2+(y-\tfrac12)^2}\). Find its absolute maximum and explain why a search using only interior points with zero gradient would fail.
The square is compact and \(f\) is continuous on it, so both absolute extrema exist. Since a square root is nonnegative, \(f(x,y)\le0\) everywhere. Equality holds exactly when both squared terms inside the root are zero, which requires \(x=y=1/2\). Therefore the absolute maximum is \(0\) at \((1/2,1/2)\).
This objective is not differentiable at the center. For example, along the horizontal line through the center, \(f(1/2+t,1/2)=-|t|\). The right-hand derivative at \(t=0\) is \(-1\), while the left-hand derivative is \(1\); the derivative does not exist. Thus the center cannot be found by solving \(\nabla f=\mathbf{0}\).
The boundary also illustrates the need to check every piece. On the bottom edge, the restriction is \(q(x)=-\sqrt{(x-\tfrac12)^2+\tfrac14}\). Its derivative on \((0,1)\) is \(q'(x)=-(x-\tfrac12)/\sqrt{(x-\tfrac12)^2+\tfrac14}\), which vanishes at \(x=1/2\). The value there is \(-1/2\); at either endpoint it is \(-\sqrt{1/2}=-1/\sqrt2\). By the square’s symmetry, the same values occur on the top, left, and right edges: each edge midpoint has value \(-1/2\), and each corner has value \(-1/\sqrt2\).
A mistaken search that checks only differentiable interior stationary points and boundary candidates would compare the boundary values and report \(-1/2\) as the maximum. That is wrong: the actual value at the nondifferentiable interior center is \(0\), which is larger. The correction is to include interior nondifferentiability points. The minimum is \(-1/\sqrt2\) at the four corners, since \(1/\sqrt2>1/2\), so \(-1/\sqrt2<-1/2<0\). This verifies both extrema.
Common Mistakes to Avoid
A candidate search is only as reliable as its coverage. In particular, do not assume that checking the two graph restrictions automatically checks any vertical end segments, or that checking endpoints on one arc automatically covers all junctions. Trace the entire boundary before differentiating.
A second mistake is to apply the equation \(\nabla f=\mathbf{0}\) at every possible interior extremum without checking differentiability. That equation is required at an interior extremum only when \(f\) is differentiable there. For continuous but nondifferentiable objectives, include the nondifferentiable interior points as candidates; otherwise the search may miss an absolute extremum.
Finally, solving the derivative equations is not the last step. Evaluate the original objective at every candidate, retain exact values when possible, and compare all of them. A critical point of one boundary restriction need not be an absolute extremum on the whole region.
Why This Method Is Useful
A region between curves can have a curved boundary, so corner checks alone are insufficient. Restricting the objective to each boundary graph turns the boundary search into one-variable calculus, while the gradient condition handles differentiable interior extrema. The extreme value theorem supplies the guarantee that a maximum and minimum exist; the candidate argument explains where to look for them.
The method also makes omissions visible. Writing down the lower graph, upper graph, endpoints, and any additional boundary pieces creates a checklist. The final comparison then turns a collection of local candidates into a conclusion about the whole region.
Check Your Understanding
For each question, distinguish the existence of extrema from the method used to locate them.
- For a region between two graphs, what boundary pieces must be checked when the graphs do not meet at the ends of the interval?
- For a differentiable objective, what equations identify an interior stationary point?
- What candidates should be checked on a differentiable boundary restriction, in addition to its critical points?
- Why can a continuous objective have an interior extremum that is not found by solving \(\nabla f=\mathbf{0}\)?
- If a boundary restriction is constant on an entire arc, how should its candidate values be handled?