Singular Feasible Points That Escape the Equations

When the Lagrange Equations Are Not Enough

The Lagrange equations are a powerful way to find extrema subject to an equality constraint. But they do not apply at every feasible point. Their standard form requires the constraint gradient to be nonzero at the point being tested.

A feasible point where that gradient vanishes is called a singular feasible point. Such a point may be a constrained maximum or minimum even when it does not satisfy the Lagrange equations. It may also fail to be an extremum altogether. The right response is to inspect the objective on the feasible set near that point, rather than treating the equations as a complete test.

Definition: For a constraint \(g(x,y)=0\), a point \((a,b)\) is feasible if \(g(a,b)=0\). It is a singular feasible point if it is feasible and \(\nabla g(a,b)=\mathbf{0}\).

The Regularity Condition

For a differentiable objective \(f\) and a differentiable constraint \(g(x,y)=0\), the one-constraint Lagrange condition says that at a constrained local extremum \((a,b)\), if \(\nabla g(a,b)\ne\mathbf{0}\), then there is a number \(\lambda\) such that

\nabla f(a,b)=\lambda\nabla g(a,b).

The nonzero-gradient assumption matters. It ensures that, near the point, the constraint is a smooth curve with a well-defined tangent line. At a local extremum, the derivative of \(f\) in every feasible tangent direction must be zero. This forces \(\nabla f\) to be perpendicular to the tangent line, and hence parallel to \(\nabla g\).

At a singular feasible point, the displayed equations reduce to \(\nabla f(a,b)=\mathbf{0}\). But that is not a necessary condition for a constrained extremum there. The constraint may still restrict feasible motion in a way that creates an extremum, even though its gradient gives no usable tangent direction.

Why the Regular Argument Breaks Down

At a regular point, the tangent directions to the constraint are the vectors \(\mathbf{v}\) satisfying \(\nabla g\cdot\mathbf{v}=0\). At an extremum, the directional derivative of \(f\) in each such direction is zero, so \(\nabla f\cdot\mathbf{v}=0\) for every tangent vector \(\mathbf{v}\). In the plane, this means \(\nabla f\) is parallel to \(\nabla g\).

This reasoning depends on the regularity of the constraint curve. One way to see the problem at a singular point is to expand \(g\) near a feasible point \(\mathbf{p}\). Differentiability gives

g(\mathbf{p}+\mathbf{h}) = g(\mathbf{p})+\nabla g(\mathbf{p})\cdot\mathbf{h} +o(\|\mathbf{h}\|).

If \(\nabla g(\mathbf{p})=\mathbf{0}\), the linear term vanishes. The first-order equation then supplies no restriction on possible directions \(\mathbf{h}\). The actual feasible set is determined by higher-order terms, which the standard Lagrange equations do not capture.

Key takeaway: The Lagrange condition is a necessary condition at feasible points where \(\nabla g\ne\mathbf{0}\). It is not a complete test at singular feasible points. Check those points directly on the constraint.

A Cusp with a Missed Minimum

Worked Example: A Constrained Minimum at a Cusp

Find the minimum of \(f(x,y)=x\) subject to \(g(x,y)=y^2-x^3=0\), and check whether the Lagrange equations find it.

First determine what the constraint implies. On the feasible set,

$$ y^2=x^3. $$

Since \(y^2\ge 0\), every feasible point has \(x^3\ge 0\), hence \(x\ge 0\). The point \((0,0)\) is feasible because \(0^2-0^3=0\). Since the objective is \(f=x\), every feasible value satisfies \(f\ge 0\), and \(f(0,0)=0\). Therefore \((0,0)\) is a global constrained minimum.

Now compute the gradients:

\nabla f=(1,0), \qquad \nabla g=(-3x^2,2y).

At \((0,0)\), the constraint gradient is \((0,0)\), while the objective gradient is \((1,0)\). The Lagrange equations would require

(1,0)=\lambda(0,0)=(0,0),

which is impossible for every \(\lambda\). Thus the equations do not find this constrained minimum. The direct check is decisive: feasibility forces \(x\ge 0\), and the origin attains \(x=0\).

A Singular Point Need Not Be an Extremum

Worked Example: A Crossing Where the Objective Has No Extremum

Consider \(f(x,y)=x^2-y^2\) subject to \(g(x,y)=xy=0\). Determine whether the singular feasible point \((0,0)\) is a constrained extremum.

The point is feasible because \(0\cdot 0=0\). The gradients are

\nabla f=(2x,-2y), \qquad \nabla g=(y,x).

At the origin, both gradients vanish. In particular, the Lagrange equations hold there for every value of \(\lambda\), so they do not decide whether the point is an extremum.

Inspect feasible points on each axis. For any nonzero \(t\), the point \((t,0)\) is feasible and

$$ f(t,0)=t^2>0=f(0,0). $$

The point \((0,t)\) is also feasible, and

$$ f(0,t)=-t^2<0=f(0,0). $$

These values occur arbitrarily close to the origin, for both positive and negative \(t\). Hence the origin is neither a constrained local minimum nor a constrained local maximum.

A Compact Constraint with Missed Global Extrema

Worked Example: Finding the Global Extrema by Inspecting Singular Points

Find the global extrema of \(f(x,y)=x\) on the constraint

$$ g(x,y)=y^2-x^2(1-x^2)^3=0. $$

On the constraint, \(y^2=x^2(1-x^2)^3\). The left side is nonnegative, so the right side must be nonnegative. Because \(x^2\ge 0\), this requires \(|x|\le 1\) (with \(x=0\) also feasible). Conversely, for every \(x\in[-1,1]\), the right side is nonnegative and there are feasible \(y\)-values. Thus the feasible \(x\)-coordinates fill \([-1,1]\), and the objective \(f=x\) has global minimum \(-1\) and global maximum \(1\) on the constraint.

The minimum is attained at \((-1,0)\), and the maximum at \((1,0)\), since the constraint gives \(y=0\) when \(x=\pm1\). To check regularity, differentiate:

\nabla f=(1,0), \qquad \nabla g= \left( -2x(1-x^2)^2(1-4x^2),\,2y \right).

At \((1,0)\), both components of \(\nabla g\) are zero. At \((-1,0)\), they are also zero. These global extrema are therefore singular feasible points.

In fact, the regular Lagrange equations yield no candidates on the regular part of this constraint. At a regular feasible point, \(y\ne0\): if \(y=0\), the constraint forces \(x=0\) or \(x=\pm1\), and the gradient formula shows that each such feasible point is singular. The second component of \(\nabla f=\lambda\nabla g\) would require \(0=2\lambda y\). Since \(y\ne0\), this implies \(\lambda=0\). But the first component would then require \(1=0\), a contradiction. So the regular equations find neither global extremum; direct inspection finds both.

A Reliable Candidate-Checking Procedure

A Lagrange calculation is only one part of a constrained-extrema search. The following procedure keeps the regularity condition visible.

Identify feasible points.
Start with the constraint equation \(g=0\), and determine which points belong to the feasible set.

Separate regular and singular points.
Compute \(\nabla g\). Points with \(\nabla g\ne\mathbf{0}\) are regular; feasible points with \(\nabla g=\mathbf{0}\) require separate attention.

Use the equations only where justified.
Solve \(\nabla f=\lambda\nabla g\) for regular feasible points, subject to the original constraint.

Inspect singular feasible points directly.
Use the constraint and objective values near each such point to decide whether it is an extremum. Do not infer an extremum merely from singularity.

Compare the candidates when global extrema are sought.
Include both regular candidates and singular candidates in the comparison. For a global conclusion, also ensure the feasible set and existence conditions support that conclusion.

A Common Mistake

A common error is to apply the Lagrange equations at every feasible point and treat their solutions as the complete list of candidates. For the cusp example, this calculation would say that the origin cannot be a constrained extremum because

(1,0)\ne\lambda(0,0) \quad\text{for every }\lambda.

The calculation is correct, but the conclusion is not: the regularity assumption fails at the origin. The correction is not to alter the equations or search for a special multiplier. Instead, separate the singular feasible point and inspect the objective on the feasible set. Here, \(y^2=x^3\) forces \(x\ge0\), so \(f=x\) has a minimum of zero at the origin.

Why Singular Points Matter

The Lagrange equations translate a geometric fact about a smooth constraint into an algebraic system. When the constraint gradient vanishes, that translation is no longer reliable: it can miss a genuine extremum, or it can leave the extremum question undecided. This matters whenever a constraint set has a cusp, crossing, isolated point, or other singular feature.

A safe conclusion is therefore limited but useful: every constrained local extremum at a regular feasible point must satisfy the Lagrange equations. Singular feasible points are not covered by that necessity statement. They must be considered separately, and their status depends on the objective’s actual values on nearby feasible points.

Check Your Understanding

For each question, distinguish what the gradient equations establish from what must be checked directly.

What condition on \(\nabla g\) is required to use the standard one-constraint Lagrange condition as a necessary condition?
For \(g(x,y)=y^2-x^3=0\), why does the constraint imply \(x\ge0\)?
In the cusp example, why does the Lagrange equation fail at the origin, and why is the origin nevertheless a constrained minimum of \(f=x\)?
For the constraint \(xy=0\) and objective \(x^2-y^2\), which feasible points show that the origin is not a local extremum?
When searching for global extrema, why must singular feasible points be included even after solving the regular Lagrange equations?