From a Model to a Complete Candidate Search
A constrained model asks how large or small an objective can be when its variables must remain in a prescribed region. A reliable solution does more than solve derivative equations: it accounts for the region’s interior, every boundary piece, and all endpoints or junctions, then compares the objective values that result.
The word complete matters. An interior calculation cannot detect an extremum that occurs on the boundary. A boundary calculation that omits a corner can miss the answer too. The first task is therefore to describe the feasible region precisely, and the last is to justify that every possible absolute extremum has been considered.
Existence and the Candidate-Set Principle
The existence statement follows from the extreme value theorem: a closed and bounded subset of \(\mathbb{R}^2\) is compact, and a continuous real-valued function on a nonempty compact set attains both extreme values. Notice that this conclusion requires continuity, not differentiability.
The derivative conditions have additional hypotheses. At an interior extremum, differentiability lets us apply the one-variable necessary condition along each coordinate direction. For a boundary piece, we apply that condition to the one-variable restriction \(f(r(t))\). Neither condition automatically applies at a point where the relevant function is not differentiable, or at an endpoint of the parameter interval.
A complete search therefore includes:
- Interior points satisfying \(f_x=f_y=0\), where \(f\) is differentiable.
- Stationary points of each differentiable boundary restriction that lie in the interior of its parameter interval.
- All boundary endpoints, corners, and junctions.
- Any other points where a restriction or the objective is not differentiable and an extremum could occur.
The candidate set need not be finite in every problem. If it is infinite, the search must still determine the extreme values on the relevant candidate sets. When the candidates are finite, comparing their function values finishes the calculation.
Worked Example: A Performance Score on a Triangular Feasible Region
Worked Example: A Performance Score on a Triangular Feasible Region
A system uses two nonnegative inputs \(x\) and \(y\), with total input at most \(3\). Its performance score is \(f(x,y)=12-(x-1)^2-(y-1)^2\). Find the absolute minimum and maximum score over the feasible region.
The feasible region is \(D=\{(x,y):x\geq0,\ y\geq0,\ x+y\leq3\}\). It is nonempty, closed, and bounded, and \(f\) is continuous, so both absolute extrema are attained. The objective is differentiable everywhere.
First find interior candidates. The partial derivatives are \(f_x=-2(x-1)\) and \(f_y=-2(y-1)\). Setting both equal to zero gives \(x=1\) and \(y=1\). This point is interior because \(1>0\), \(1>0\), and \(1+1<3\). Its score is \(f(1,1)=12\).
Next examine each boundary edge. On \(x=0\), with \(0\leq y\leq3\), the restriction is \(h(y)=f(0,y)=12-1-(y-1)^2=11-(y-1)^2\). Its derivative is \(h'(y)=-2(y-1)\), so the only stationary point in the open edge is \(y=1\), with \(f(0,1)=11\). The endpoints give \(f(0,0)=10\) and \(f(0,3)=7\).
On \(y=0\), with \(0\leq x\leq3\), the restriction is \(k(x)=f(x,0)=11-(x-1)^2\). Thus \(k'(x)=-2(x-1)=0\) at \(x=1\), giving \(f(1,0)=11\). Its endpoints give \(f(0,0)=10\) and \(f(3,0)=7\).
The third edge is \(y=3-x\), with \(0\leq x\leq3\). Its restriction is \(m(x)=12-(x-1)^2-(2-x)^2\). Differentiating gives \(m'(x)=-2(x-1)+2(2-x)=6-4x\). Hence \(m'(x)=0\) at \(x=\frac32\), which gives \(y=\frac32\). The score there is \(12-(\frac12)^2-(\frac12)^2=\frac{23}{2}\). At the edge endpoints, \(f(0,3)=7\) and \(f(3,0)=7\).
The complete candidate list consists of the interior point \((1,1)\), the three edge stationary points \((0,1)\), \((1,0)\), and \((\frac32,\frac32)\), and the three vertices \((0,0)\), \((0,3)\), and \((3,0)\). Their scores are, respectively, \(12,11,11,\frac{23}{2},10,7,7\). Therefore the absolute maximum is \(12\) at \((1,1)\), and the absolute minimum is \(7\) at \((0,3)\) and \((3,0)\).
As a check, the score cannot exceed \(12\), since it is \(12\) minus two squares. The value \(12\) is attained when both squares are zero, at \((1,1)\). The candidate comparison establishes the minimum on the whole feasible region, including its boundary.
Why the Boundary Calculation Is One-Dimensional
A boundary curve can be described by a parameter \(t\). Substituting its coordinates into the objective produces an ordinary one-variable function. For example, if a boundary piece is the graph \(y=g(x)\), then its restriction is \(q(x)=f(x,g(x))\). Wherever the required derivatives exist, the chain rule gives
An interior extremum of that restriction must satisfy \(q'(x)=0\). This condition concerns changes along the boundary, not independent changes in both coordinates. That is why boundary candidates are generated by restricting the objective rather than simply solving \(f_x=f_y=0\) again.
Endpoints require separate attention: the one-variable derivative test gives a necessary condition only at an interior parameter value. At a corner or boundary junction, the relevant point may be an endpoint of one or more pieces, so include it directly and evaluate the objective there.
Worked Example: A Region Between a Curve and a Line
Worked Example: A Region Between a Curve and a Line
Find the absolute extrema of \(f(x,y)=x+y\) on \(D=\{(x,y):0\leq x\leq2,\ x^2\leq y\leq4\}\).
The region is closed and bounded: \(x\) lies between \(0\) and \(2\), and \(y\) lies between \(x^2\) and \(4\). It is nonempty, and \(f\) is continuous, so absolute extrema exist. Since \(f_x=1\) and \(f_y=1\), there are no interior critical points.
On the left boundary \(x=0\), the parameter is \(0\leq y\leq4\), and the restriction is \(f(0,y)=y\). Its derivative is \(1\), so it has no stationary point in the open edge. Its endpoints give values \(0\) at \((0,0)\) and \(4\) at \((0,4)\).
On the top boundary \(y=4\), where \(0\leq x\leq2\), the restriction is \(f(x,4)=x+4\). Its derivative is \(1\), so there are no interior stationary points on this edge. The endpoints give \(4\) at \((0,4)\) and \(6\) at \((2,4)\).
On the curved boundary \(y=x^2\), where \(0\leq x\leq2\), the restriction is \(f(x,x^2)=x+x^2\). Its derivative is \(1+2x\), which is positive throughout the interval because \(x\geq0\). Thus there is no stationary point on the open edge. The endpoint values are \(0\) at \((0,0)\) and \(6\) at \((2,4)\).
All boundary junctions have been included: they are \((0,0)\), \((0,4)\), and \((2,4)\). The candidate values are \(0,4,6\). Hence the absolute minimum is \(0\) at \((0,0)\), and the absolute maximum is \(6\) at \((2,4)\). As a direct check, \(x+y\geq x+x^2\geq0\) on \(D\), while \(x\leq2\) and \(y\leq4\) imply \(x+y\leq6\); both bounds are attained at the stated points.
Worked Example: Continuity Does Not Guarantee Stationarity
Worked Example: Continuity Does Not Guarantee Stationarity
Consider \(f(x,y)=|x|\) on the square \(D=[-1,1]\times[-1,1]\). Determine its absolute extrema and identify why a differentiability condition matters.
The square is nonempty, closed, and bounded, and \(f\) is continuous. Therefore the absolute extrema are attained. Since \(0\leq|x|\leq1\) throughout the square, the absolute minimum is \(0\), attained at every point \((0,y)\) with \(-1\leq y\leq1\). The absolute maximum is \(1\), attained at every point with \(x=-1\) or \(x=1\).
For \(x>0\), the derivative with respect to \(x\) is \(1\); for \(x<0\), it is \(-1\). At \(x=0\), the derivative with respect to \(x\) does not exist. In particular, the interior minimum at \((0,0)\) does not satisfy a pair of partial-derivative equations: \(f_x(0,0)\) is undefined. The existence of an absolute extremum follows from continuity, but the differentiable interior test cannot be applied there.
The same caution applies to boundary restrictions. On the edge \(y=1\), the restriction is \(|x|\). It is not differentiable at the interior edge point \(x=0\), where it attains its minimum. A search that considered only differentiable stationary points on that edge would miss this candidate unless it also checked the nondifferentiable point. The boundary edges \(x=\pm1\) have constant value \(1\), confirming the maximum. Thus the candidate values \(0\) and \(1\) agree with the direct bounds and are attained as described.
A Careful Workflow and a Common Pitfall
Write its inequalities or boundary pieces, and identify endpoints, corners, and junctions.
Verify that the region is nonempty, closed, and bounded and that the objective is continuous on it.
Where the objective is differentiable, solve \(f_x=f_y=0\) and retain solutions that lie in the interior.
Substitute a parametrization into the objective, find differentiable stationary points in the open parameter interval, and include endpoints and any nondifferentiable possibilities.
Evaluate the original objective at every candidate and identify the smallest and largest values.
A common error is to reason, “The objective is continuous, so every interior extremum has \(f_x=f_y=0\).” That conclusion does not follow. For \(f(x,y)=|x|\), the function is continuous and has an interior minimum at \((0,0)\), but \(f_x(0,0)\) does not exist. The correction is to keep the claims separate: continuity on a nonempty compact region guarantees that extrema are attained; differentiability at an interior extremum is an additional hypothesis for the partial-derivative test. Likewise, stationarity along a boundary piece requires differentiability of its one-variable restriction and an extremum at an interior parameter value.
The candidate comparison is what turns derivative calculations into a global conclusion. A stationary point may be a minimum, a maximum, or neither, and a boundary point may beat every interior point. Only after the feasible region has been covered and all candidate values have been compared can the absolute extrema be identified.
Check Your Understanding
Use the distinctions between existence, differentiability, and boundary restrictions to answer these questions.
- What hypotheses guarantee that a continuous objective attains an absolute maximum and minimum on a region in \(\mathbb{R}^2\)?
- Why must an interior candidate satisfy \(f_x=f_y=0\) only when the objective is differentiable there?
- For a boundary graph \(y=g(x)\), what one-variable function should be differentiated to find stationary points along the boundary?
- Why must corners and endpoints be included even if no derivative equation is solved there?
- For \(f(x,y)=|x|\) on the square \([-1,1]\times[-1,1]\), where are the absolute extrema, and why does the interior minimum not satisfy the differentiable stationary-point test?