Why the hypotheses matter
The Lagrange condition is a useful way to identify possible extrema of a function subject to an equality constraint. But it is not an unconditional rule. Its conclusion depends on the behavior of the constraint near the point being studied, and its conclusion is a necessary condition—not a guarantee that an extremum exists.
For a single constraint \(g(\mathbf{x})=c\), the key hypothesis is that the gradient of \(g\) is nonzero at the point. This says that the level set has a well-defined tangent direction structure there. If the gradient vanishes, a constrained extremum can fail to satisfy the usual Lagrange condition.
Regular points and constrained extrema
Let \(U\) be an open subset of \(\mathbb{R}^n\), let \(f,g:U\to\mathbb{R}\) be continuously differentiable, and fix a real number \(c\). The feasible set is the level set
A point \(\mathbf{a}\in S\) is a regular point of the constraint if \(\nabla g(\mathbf{a})\ne\mathbf{0}\). The condition is local: it concerns the gradient at this particular feasible point. A constraint may have regular points and singular points on the same level set.
The differentiability assumptions in the theorem below are sufficient for the argument: continuous first partial derivatives allow the implicit function theorem to describe the level set locally as a graph near a regular point. The theorem concerns local extrema; it does not assert that an extremum exists or that a point satisfying the condition is an extremum.
The equality says that, at a regular constrained extremum, the two gradients are parallel. It is a necessary condition: every regular local constrained extremum must satisfy it. The theorem does not say that every solution of this gradient equation is a maximum or a minimum.
Why the condition follows
The reason is that a constrained extremum cannot have a nonzero first-order change in any direction that stays tangent to the constraint. At a regular point, these tangent directions are exactly the vectors perpendicular to \(\nabla g(\mathbf{a})\). The following argument makes that relationship precise.
Since \(\nabla g(\mathbf{a})\ne\mathbf{0}\), at least one partial derivative of \(g\) at \(\mathbf{a}\) is nonzero. Relabel coordinates so that \(\frac{\partial g}{\partial x_n}(\mathbf{a})\ne 0\). The implicit function theorem then expresses the nearby constraint as \(x_n=\phi(x_1,\ldots,x_{n-1})\), for a continuously differentiable function \(\phi\).
Let \(\mathbf{v}=(v_1,\ldots,v_n)\) satisfy \(\nabla g(\mathbf{a})\cdot\mathbf{v}=0\). For small \(t\), vary the first \(n-1\) coordinates as \(a_i+t v_i\), and choose the last coordinate from the local graph. This defines a differentiable feasible curve \(\boldsymbol{\gamma}(t)\) with \(\boldsymbol{\gamma}(0)=\mathbf{a}\). Differentiating \(g(\boldsymbol{\gamma}(t))=c\) at \(t=0\) gives \(\nabla g(\mathbf{a})\cdot\boldsymbol{\gamma}'(0)=0\). The first \(n-1\) components of \(\boldsymbol{\gamma}'(0)\) are \(v_1,\ldots,v_{n-1}\); the equation and \(\frac{\partial g}{\partial x_n}(\mathbf{a})\ne0\) force its last component to be \(v_n\). Thus \(\boldsymbol{\gamma}'(0)=\mathbf{v}\).
The function \(f(\boldsymbol{\gamma}(t))\) has a local extremum at \(t=0\). Its derivative there must be zero. By the chain rule, \(0=\frac{d}{dt}f(\boldsymbol{\gamma}(t))\big|_{t=0}=\nabla f(\mathbf{a})\cdot\mathbf{v}\). This holds for every \(\mathbf{v}\) perpendicular to \(\nabla g(\mathbf{a})\).
Because \(\nabla g(\mathbf{a})\ne\mathbf{0}\), the vectors perpendicular to \(\nabla g(\mathbf{a})\) form a hyperplane. A vector perpendicular to every vector in that hyperplane must be a scalar multiple of \(\nabla g(\mathbf{a})\). Since \(\nabla f(\mathbf{a})\) is perpendicular to every such tangent vector, \(\nabla f(\mathbf{a})=\lambda\nabla g(\mathbf{a})\) for some real \(\lambda\).
This proof also explains the role of regularity. A nonzero constraint gradient allows the level set to be described locally as a smooth graph, so its tangent directions are captured by the equation \(\nabla g(\mathbf{a})\cdot\mathbf{v}=0\). When the gradient is zero, that reasoning no longer applies.
Worked Example: Checking the condition at a constrained maximum
Worked Example: Checking the condition at a constrained maximum
Let \(f(x,y)=x\) and impose \(g(x,y)=x^2+y^2=1\). Check the theorem's condition at \(\mathbf{a}=(1,0)\), then verify that this point is a constrained maximum.
First, the point is feasible because \(g(1,0)=1^2+0^2=1\). The gradients are
At \((1,0)\), \(\nabla f(1,0)=(1,0)\) and \(\nabla g(1,0)=(2,0)\). The constraint is regular there because \((2,0)\ne(0,0)\). The condition holds with \(\lambda=\frac12\):
To check the extremum directly, any feasible point satisfies \(x^2+y^2=1\), so \(x^2\le1\), hence \(-1\le x\le1\). Since \(f(x,y)=x\), its value is at most \(1=f(1,0)\). Equality \(x=1\) forces \(y=0\), so \((1,0)\) is in fact the global constrained maximum. This direct check—not the gradient condition alone—establishes the maximum.
Why the condition is not sufficient
An equation that every regular constrained extremum must satisfy can still have solutions that are not extrema. A solution of the gradient condition is a candidate that needs further analysis. To decide whether it is a local maximum, a local minimum, or neither, one must examine the objective along the feasible set.
Worked Example: A regular candidate that is not an extremum
Let \(f(x,y)=y^3\), with constraint \(g(x,y)=x=0\), and consider \((0,0)\).
The point is feasible since \(g(0,0)=0\). The gradients are \(\nabla f(x,y)=(0,3y^2)\) and \(\nabla g(x,y)=(1,0)\). At the origin, these become \(\nabla f(0,0)=(0,0)\) and \(\nabla g(0,0)=(1,0)\), so the constraint is regular. The Lagrange condition holds with \(\lambda=0\):
But the feasible points include \((0,t)\) for every real \(t\), and along them \(f(0,t)=t^3\). For every \(\delta>0\), choosing \(t=\delta/2\) gives \(0<t<\delta\) and \(f(0,t)>0=f(0,0)\); choosing \(t=-\delta/2\) gives \(-\delta<t<0\) and \(f(0,t)<0=f(0,0)\). Thus every neighborhood of the origin on the constraint contains points with both greater and smaller function values. The origin is neither a local maximum nor a local minimum.
The calculation illustrates a common mistake: concluding that a point is an extremum simply because its gradients satisfy the Lagrange condition. In this example the condition holds, but the feasible values change from below to above the value at the point. The correction is to treat the condition as a way to find necessary candidates, then test those candidates on the constraint.
What can fail at a singular constraint point
The nonzero-gradient hypothesis is essential. At a point where \(\nabla g=\mathbf{0}\), the level set may not have the local graph and tangent description used in the proof. A constrained extremum can then fail to satisfy the Lagrange equation.
Worked Example: A singular constraint that hides extrema
Set \(f(x,y)=x\) and \(g(x,y)=x^2+y^2\), with constraint \(g(x,y)=0\). The only feasible point is \((0,0)\): a sum of two squares is zero only when both squares are zero. Therefore \(f\) takes the value \(0\) at every feasible point. In particular, the origin is both a local constrained maximum and a local constrained minimum, using the definition of extrema relative to the feasible set.
However, \(\nabla f(x,y)=(1,0)\) and \(\nabla g(x,y)=(2x,2y)\). At the origin, \(\nabla g(0,0)=(0,0)\), so the constraint is not regular. There is no real \(\lambda\) for which
The feasible point is an extremum, but it does not satisfy the Lagrange condition. This does not contradict the theorem: the theorem's regularity hypothesis fails exactly at that point.
Using the theorem responsibly
For a problem with one equality constraint, the theorem gives a conditional, one-way implication: if the functions are continuously differentiable near a feasible point, the constraint gradient is nonzero there, and the point is a local constrained extremum, then the gradients are parallel. The reverse implication is not guaranteed. Nor does the theorem say that all extrema must be found by this condition if the regularity hypothesis fails at some feasible points.
| Situation | What follows |
|---|---|
| Regular feasible point is a local constrained extremum | The gradient condition must hold. |
| Regular feasible point satisfies the gradient condition | It is a candidate; more analysis is needed to classify it. |
| Feasible point has \(\nabla g=\mathbf{0}\) | The theorem gives no conclusion; the point must be examined separately. |
The distinction matters in applications: a list of points satisfying the gradient equation is not, by itself, a list of proven maxima and minima. Each candidate needs an appropriate check, and singular feasible points must not be silently discarded. The theorem is powerful because its hypotheses identify where the geometric argument is valid; respecting those hypotheses is part of using the method correctly.
Check Your Understanding
Use the hypotheses and conclusions carefully when answering these questions.
- What does it mean for a feasible point to be regular for the constraint \(g(\mathbf{x})=c\)?
- State the differentiability, feasibility, and gradient hypotheses in the one-constraint Lagrange theorem.
- Why does a regular constrained extremum have zero first derivative along every tangent direction?
- Why does satisfying the gradient condition not prove that a point is an extremum?
- In the constraint \(x^2+y^2=0\), why does the Lagrange theorem not apply at the feasible point?