Candidate Testing for Absolute Optima

Why Candidate Testing Needs Conditions

Finding a point where the derivative is zero is not enough to establish an absolute maximum or minimum. A constrained problem also has feasible boundaries, and an interior critical point may be only a local optimum—or neither kind of optimum. To decide which value is largest or smallest across the entire feasible domain, we need a complete candidate list and a valid comparison.

The method has an important condition: the objective function must be continuous on a closed, bounded feasible interval. Without continuity, an absolute optimum might not exist, even if we check the endpoints and every interior critical point. We will state the result precisely, explain why the candidates are sufficient, and work through examples.

Absolute Optima and Candidate Points

Let \(f\) be an objective function defined on a feasible set. An absolute maximum occurs at a feasible input \(c\) if \(f(c)\geq f(x)\) for every feasible \(x\). An absolute minimum occurs at \(c\) if \(f(c)\leq f(x)\) for every feasible \(x\). The corresponding values are \(f(c)\).

For a feasible interval \([a,b]\), the endpoints \(a\) and \(b\) are part of the domain. A critical point in the interior \((a,b)\) is a point where \(f'(x)=0\) or where \(f'(x)\) does not exist. Endpoints are included separately, whether or not the derivative exists there.

The theorem guarantees that the maximum and minimum exist, and that at least one of them occurs among the listed candidates. If there are finitely many candidates, we can evaluate each one and compare the resulting numbers directly. If the candidate set is infinite, the theorem still identifies where an extremum must occur, but comparing infinitely many values may require further analysis.

In a constrained model, first determine the feasible interval from the restrictions of the problem. A physical quantity might be restricted to nonnegative values and to a maximum allowed amount. Its endpoints are then genuine candidates if the constraints allow equality there.

Why the Candidate List Is Sufficient

The argument combines two facts. The Extreme Value Theorem gives existence of absolute extrema for continuous functions on closed, bounded intervals. Fermat’s theorem explains why a differentiable interior extremum must have derivative zero.

To see the interior condition, suppose \(c\in(a,b)\) is a local maximum and \(f'(c)\) exists. For sufficiently small positive \(h\), the points \(c+h\) and \(c-h\) are in the domain. Since \(c\) is a local maximum, \(f(c+h)-f(c)\leq 0\) and \(f(c-h)-f(c)\leq 0\). Therefore the difference quotient from the right is nonpositive, while the difference quotient from the left is nonnegative:

If the two-sided derivative exists, its value must be both nonpositive and nonnegative. Hence \(f'(c)=0\). For a local minimum, the inequalities reverse, and the same conclusion follows. Thus an interior absolute extremum is either a point where the derivative is zero or a point where the derivative does not exist. If an absolute extremum is at the boundary, it is one of the endpoints already included.

Worked Example: Comparing Endpoints and Critical Points

Interior Points Where the Derivative Does Not Exist

A complete candidate list must not be limited to solutions of \(f'(x)=0\). An interior point where the derivative does not exist can still be an absolute extremum. For example, the graph of an absolute-value function has a corner at its minimum. Such a point is critical even though its derivative is undefined there.

Worked Example: Apply the Test to a Constrained Model

Boundaries, Feasible Sets, and Limits

In constrained optimization, a boundary is a point where the restrictions stop allowing movement in at least one direction. For a one-variable feasible interval, these boundaries are its endpoints. If the constraints produce several separate closed intervals, apply the same test to each interval: include both endpoints of every interval and all its interior critical points, then compare every candidate value across all the intervals.

Do not include an endpoint that is not feasible. If the domain is \((a,b)\), neither \(a\) nor \(b\) belongs to the domain, so neither can be an attained optimum. The function may approach a value near an excluded boundary without ever attaining it. An open or unbounded feasible domain therefore needs a separate analysis of behavior near its boundaries and at infinity; the closed-interval theorem does not apply as stated.

Likewise, a closed, bounded domain alone is not sufficient if the objective is discontinuous. The continuity condition is essential to the existence guarantee and to the candidate test. A stationary-point calculation cannot compensate for a missing hypothesis.

Common Pitfall: Omitting Continuity

Consider the function on \([0,1]\) defined by \(h(x)=x\) for \(x<1\), and \(h(1)=0\). It is tempting to use the endpoint-and-critical-point procedure without checking continuity.

For every interior point \(x\in(0,1)\), \(h'(x)=1\), so there are no interior points where the derivative is zero or undefined. The endpoint values are \(h(0)=0\) and \(h(1)=0\). A mistaken calculation would declare \(0\) to be the absolute maximum because it is the largest value in this incomplete list.

But for every \(x<1\), \(h(x)=x<1\), and these values can be made arbitrarily close to \(1\). No point in the domain has value \(1\), so \(h\) has no absolute maximum. The function is discontinuous at \(x=1\), and the continuity hypothesis fails.

A Reliable Candidate-Testing Routine

For a one-variable constrained problem, organize the work so that the domain, candidate list, and comparison are all visible.

A frequent computational error is to find the critical points correctly but forget to evaluate the objective at a feasible endpoint. Another is to compare the inputs rather than the function values: the largest \(x\) does not necessarily give the largest \(f(x)\). The final comparison must be among the values \(f(x)\), not among the candidate \(x\)-coordinates.