Why Comparing Candidates Gives Absolute Extrema
For an equality constraint, solving the Lagrange equations identifies possible extrema at regular feasible points. But those equations alone do not usually tell us which candidate is a maximum or minimum, and they can miss feasible points where the constraint is singular.
The reliable method is to establish that absolute extrema exist, find every point that could be an extremum, and compare the objective values there. Compactness supplies the existence guarantee; the Lagrange condition and a separate check of singular points supply the candidate search.
The requirement \(\nabla g(\mathbf{x})\ne\mathbf{0}\) is part of the definition of a regular Lagrange candidate. A singular feasible point is not a regular Lagrange candidate, even if the multiplier equation happens to hold there.
The Complete Candidate Principle
Suppose \(S\) is compact, and \(f\) is continuous on \(S\). Then the Extreme Value Theorem guarantees that \(f\) attains both an absolute minimum and an absolute maximum on \(S\). If, in addition, \(f\) and \(g\) are continuously differentiable near \(S\), every regular point at which either absolute extremum occurs satisfies the one-constraint Lagrange condition.
This principle does not say that every candidate is an extremum. The Lagrange condition is necessary at a regular extremum, not sufficient. Candidates must be compared, and a candidate can be neither an absolute maximum nor an absolute minimum.
Why the Candidate Principle Works
Here is the argument that makes the search valid. Compactness and continuity first guarantee actual points where the largest and smallest values occur. Each such point belongs to \(S\), so it is either regular or singular. At a regular constrained extremum, the Lagrange theorem applies: the gradients are parallel, so the point appears among the regular Lagrange candidates. A singular point need not satisfy that theorem's hypothesis, so every singular feasible point must be added separately. Thus neither kind of possible extremum is omitted.
Verify that \(S\) is compact and \(f\) is continuous on \(S\). Then absolute extrema exist.
Solve \(g(\mathbf{x})=c\) together with \(\nabla f(\mathbf{x})=\lambda\nabla g(\mathbf{x})\), retaining only points where \(\nabla g(\mathbf{x})\ne\mathbf{0}\).
Solve \(g(\mathbf{x})=c\) and \(\nabla g(\mathbf{x})=\mathbf{0}\) separately.
Evaluate \(f\) at every point in the complete candidate list. The smallest and largest values are the absolute minimum and maximum.
The method is complete only if the search finds all the relevant points. If the candidate set is infinite, it is not enough to evaluate a few of its members: one must determine the minimum and maximum values over that set by some further argument. In the examples below, the candidate sets are finite.
Worked Example: Comparing Candidates on a Quartic Constraint
Worked Example: Comparing Candidates on a Quartic Constraint
Find the absolute extrema of \(f(x,y)=x^2+2y^2\) subject to \(g(x,y)=x^4+y^4=1\).
The feasible set is closed because it is the level set of a continuous function. It is bounded because \(x^4\le 1\) and \(y^4\le 1\), so \(|x|\le 1\) and \(|y|\le 1\). Hence it is compact. The objective is continuous, so both absolute extrema exist.
The gradients are \(\nabla f=(2x,4y)\) and \(\nabla g=(4x^3,4y^3)\). On the feasible set, \(x\) and \(y\) cannot both be zero, so \(\nabla g\ne\mathbf{0}\). There are no singular feasible points. The Lagrange equations are
If \(x=0\), feasibility gives \(y^4=1\), so \(y=\pm1\). At either point, \(f=0+2=2\). If \(y=0\), then \(x=\pm1\), and \(f=1+0=1\).
If both \(x\ne0\) and \(y\ne0\), divide the first equation by \(2x\) and the second by \(4y\). This gives
Thus \(\lambda=1/(2x^2)=1/y^2\), which implies \(y^2=2x^2\). Substitution into the constraint gives
There are four points with these squared coordinates, one for each choice of signs. At each, \(f=1/\sqrt{5}+4/\sqrt{5}=\sqrt{5}\). The candidate values are therefore \(1\), \(2\), and \(\sqrt{5}\). Since \(1<2<\sqrt{5}\), the absolute minimum is \(1\) at \((1,0)\) and \((-1,0)\); the absolute maximum is \(\sqrt{5}\) at the four points with \(x^2=1/\sqrt{5}\), \(y^2=2/\sqrt{5}\).
Check: the axis candidates satisfy \(x^4+y^4=1\). For each of the four other points, \(x^4=1/5\) and \(y^4=4/5\), so their sum is \(1\). Also \(\lambda=\sqrt{5}/2\) satisfies both multiplier equations there, since \(2\lambda x^2=1\) and \(\lambda y^2=1\). The objective values used in the comparison are thus attained at feasible candidates.
Singular Points Need a Separate Check
When \(\nabla g=\mathbf{0}\), the regular Lagrange theorem does not apply. The multiplier equation cannot be used to certify that all extrema have been found. The next example shows why the singular points must be treated as a separate part of the candidate search.
Worked Example: A Singular Point Alongside Regular Candidates
Find the absolute extrema of \(f(x,y)=x+y^2\) subject to \(g(x,y)=(x^2+y^2)(x^2+y^2-1)=0\).
Write \(q=x^2+y^2\). The constraint is \(q(q-1)=0\), so the feasible set consists of the origin and the unit circle. This set is compact, and \(f\) is continuous, so absolute extrema exist. Differentiating \(g=q(q-1)\) gives \(\nabla g=(2x(2q-1),2y(2q-1))\). At the origin, \(\nabla g=\mathbf{0}\), making it singular. On the unit circle, \(q=1\), so \(\nabla g=(2x,2y)\ne\mathbf{0}\).
On the circle, \(\nabla f=(1,2y)\), and the Lagrange equations become
If \(y=0\), the constraint gives \(x=\pm1\). Both points satisfy the first equation with \(\lambda=1/(2x)\), and their objective values are \(f(1,0)=1\) and \(f(-1,0)=-1\). If \(y\ne0\), the second equation gives \(\lambda=1\); the first then gives \(x=1/2\). The constraint yields \(y^2=3/4\), so \(y=\pm\sqrt{3}/2\), and both objective values are \(1/2+3/4=5/4\).
The singular point at the origin must also be evaluated directly: it is feasible and \(f(0,0)=0\). Comparing all values, \(-1<0<1<5/4\). Therefore the absolute minimum is \(-1\) at \((-1,0)\), and the absolute maximum is \(5/4\) at \((1/2,\sqrt{3}/2)\) and \((1/2,-\sqrt{3}/2)\).
Check: each circle candidate has \(x^2+y^2=1\), and the origin satisfies the constraint because \(q=0\). For the two candidates with \(y\ne0\), \(\lambda=1\) gives \(2\lambda x=1\) and \(2\lambda y=2y\); for \((\pm1,0)\), \(\lambda=1/(2x)\) gives \(2\lambda x=1\). The singular point was checked by feasibility and direct evaluation, not by treating it as a regular Lagrange candidate.
When Every Feasible Point Is Singular
Sometimes the entire feasible set consists of singular points. In that case there may be no regular Lagrange candidates at all. The extrema still exist if the set is compact and the objective is continuous, but they must be found by examining the singular feasible set itself.
Worked Example: A Constraint with Only Singular Feasible Points
Find the absolute extrema of \(f(x,y)=x^2+3x\) subject to \((x^2-1)^2+y^2=0\).
Both terms on the left side of the constraint are nonnegative. Their sum is zero exactly when each is zero. Thus \(x^2=1\) and \(y=0\), so the feasible set is \(\{(1,0),(-1,0)\}\). It is finite and hence compact; the continuous objective attains its extrema there.
For \(g(x,y)=(x^2-1)^2+y^2\), the gradient is \(\nabla g=(4x(x^2-1),2y)\). At both feasible points, this gradient equals \(\mathbf{0}\). Both feasible points are singular, so there are no regular Lagrange candidates to search for. Evaluate the objective directly:
Therefore the absolute maximum is \(4\) at \((1,0)\), and the absolute minimum is \(-2\) at \((-1,0)\). Check: both points satisfy the constraint, since \(x^2-1=0\) and \(y=0\); the displayed objective values follow by substitution.
Common Pitfall: Stopping at the Multiplier Equations
A common error is to solve the multiplier equations, compare the values found, and conclude that the largest and smallest are the absolute extrema without checking for singular feasible points. In the second example, restricting attention to regular candidates would give the values \(-1\), \(1\), and \(5/4\). That list happens to contain the actual extrema, but it is not a complete candidate list: it omits the singular feasible point \((0,0)\), where \(f=0\).
The omission does not change the final maximum or minimum in that particular example. That coincidence is not a justification for the omission. A singular point could instead have an objective value larger or smaller than every regular candidate. The correction is to solve \(g=c\) together with \(\nabla g=\mathbf{0}\) as a separate step, then evaluate \(f\) at every resulting feasible point.
A second issue is assuming that an equality constraint automatically describes a compact set. It does not: a level set can be unbounded. Before using candidate comparison to claim absolute extrema, check compactness and continuity. Without the existence guarantee, a complete list of stationary points alone may not show that a global maximum or minimum is attained.
Check Your Understanding
Use the compactness check, the regular Lagrange equations, and a separate singular-point check to answer these questions.
- What two conditions guarantee that the objective attains an absolute maximum and minimum on the feasible set?
- State the regularity condition required for a regular Lagrange candidate.
- Why must singular feasible points be checked separately from the regular Lagrange equations?
- In the quartic-constraint example, which candidate values determine the absolute minimum and maximum?
- If the regular candidate list is infinite, why is evaluating only a few of its points insufficient?