Why Noncompact Constraints Need Extra Analysis
The Lagrange equations are a method for finding points where a differentiable objective may have a constrained local maximum or minimum. They do not, by themselves, determine whether an absolute maximum or minimum exists.
The distinction matters when the feasible set is unbounded. A continuous function on a compact feasible set attains both an absolute maximum and an absolute minimum. On a noncompact set, that conclusion is unavailable: the objective may be unbounded, or it may approach a limiting value without attaining it. Even when the Lagrange equations produce candidates, those points must be checked against the behavior of the objective elsewhere on the constraint.
The Lagrange Condition and Its Role
Consider an objective \(f:\mathbb{R}^n\to\mathbb{R}\) subject to one equality constraint \(g(\mathbf{x})=0\), where \(f\) and \(g\) are differentiable. A feasible point \(\mathbf{p}\) satisfies the constraint. If \(\nabla g(\mathbf{p})\ne\mathbf{0}\), the constraint is regular at \(\mathbf{p}\).
The regularity condition matters: at a point where \(\nabla g=\mathbf{0}\), this theorem does not provide the necessary condition. The equations also say nothing about points that escape to infinity, since those points are not finite feasible points at which the equations can be solved.
Why the Lagrange Condition Holds
At a regular point of the constraint, tangent directions describe the directions in which one can move while remaining on the constraint, at least locally. If \(\mathbf{v}\) is such a direction, differentiating the constraint along a feasible curve through \(\mathbf{p}\) gives \(\nabla g(\mathbf{p})\cdot\mathbf{v}=0\). Thus the tangent directions are orthogonal to the constraint gradient.
If \(\mathbf{p}\) is a constrained local extremum, the objective cannot have a nonzero first-order change in any feasible tangent direction. Differentiating \(f\) along the same feasible curve gives \(\nabla f(\mathbf{p})\cdot\mathbf{v}=0\). Therefore \(\nabla f(\mathbf{p})\) is orthogonal to every tangent direction. For one regular equality constraint, the vectors orthogonal to all tangent directions are precisely the scalar multiples of \(\nabla g(\mathbf{p})\), which proves the Lagrange condition.
This argument establishes a necessary condition at a regular constrained extremum. It does not compare the objective value at \(\mathbf{p}\) with values at distant feasible points. That is why a separate global analysis is needed on a noncompact set.
Worked Example: Stationary Points on a Hyperbola
Worked Example: Stationary Points on a Hyperbola
Find the stationary candidates for \(f(x,y)=x+y\) subject to \(xy=1\), and decide whether the constrained objective has absolute extrema.
Write the constraint as \(g(x,y)=xy-1=0\). Its gradient is \(\nabla g=(y,x)\), which is nonzero at every feasible point because \(xy=1\) forces both coordinates to be nonzero. The Lagrange equations are
Thus \(1=\lambda y\) and \(1=\lambda x\). These equations imply \(x=y\). Substituting into \(xy=1\) gives \(x^2=1\), so the candidates are \((1,1)\) and \((-1,-1)\). At \((1,1)\), \(\lambda=1\); at \((-1,-1)\), \(\lambda=-1\). The objective values are \(2\) and \(-2\), respectively.
To check the behavior on the whole constraint, set \(x=t\) and \(y=1/t\), where \(t\ne0\). Then \(f=t+1/t\). For \(t>0\), this is at least \(2\), since
Equality holds at \(t=1\). For \(t<0\), the value is at most \(-2\), since
Equality holds at \(t=-1\). These inequalities verify that \((1,1)\) is a local minimum on its hyperbola branch and \((-1,-1)\) is a local maximum on its branch. However, along \(t=n\), where \(n\) is a positive integer, \(f=n+1/n\) grows without bound. Along \(t=-n\), \(f=-n-1/n\) decreases without bound. The objective therefore has neither an absolute maximum nor an absolute minimum on the full constraint. The parameterization also confirms that its stationary points occur where \(1-1/t^2=0\), namely \(t=\pm1\).
A Minimum Can Exist Without a Maximum
Noncompactness does not mean that no extrema exist. An objective may attain one absolute extremum while remaining unbounded in the other direction.
Worked Example: A Paraboloid with an Absolute Minimum
Find the absolute extrema of \(f(x,y,z)=z\) subject to \(x^2+y^2-z=0\).
Let \(g(x,y,z)=x^2+y^2-z\), so \(\nabla g=(2x,2y,-1)\), which is nonzero everywhere. Since \(\nabla f=(0,0,1)\), the Lagrange equations are
The coordinate equations give \(0=2\lambda x\), \(0=2\lambda y\), and \(1=-\lambda\). Thus \(\lambda=-1\). Substituting this into the first two equations gives \(x=0\) and \(y=0\). The constraint then gives \(z=0\), so the only candidate is \((0,0,0)\), with objective value \(0\).
On the constraint, \(z=x^2+y^2\ge0\), so the candidate is an absolute minimum. To test for an absolute maximum, take the feasible path \((x,y,z)=(t,0,t^2)\). It satisfies the constraint for every real \(t\), and \(f=t^2\) grows without bound as \(t\) grows. There is no absolute maximum. As a final check, the candidate satisfies the constraint and gives \(f(0,0,0)=0\), as found above.
When There Are No Stationary Candidates
A noncompact constraint may have no Lagrange candidates at all. That does not mean the objective has extrema. The equations might have no solution precisely because the objective continues changing along every feasible direction.
Worked Example: An Unbounded Objective with No Lagrange Solution
Consider \(f(x,y,z)=x\) subject to \(g(x,y,z)=x^2+y^2-z=0\). Here \(\nabla f=(1,0,0)\) and \(\nabla g=(2x,2y,-1)\). The Lagrange equations would require
The third coordinate gives \(0=-\lambda\), so \(\lambda=0\). The first coordinate would then give \(1=0\), a contradiction. There are no solutions to the Lagrange equations.
To determine whether extrema exist, use the feasible path \((x,y,z)=(t,0,t^2)\). The constraint holds because \(t^2+0-t^2=0\), and the objective value is \(t\). As \(t\) increases without bound, the objective increases without bound; as \(t\) decreases without bound, the objective decreases without bound. Hence there is no absolute maximum or minimum. The contradiction in the multiplier equations is consistent with this behavior, but the path analysis supplies the global conclusion.
A Practical Strategy for Noncompact Constraints
For a regular equality constraint, the Lagrange equations are a useful first step, not the whole solution. A complete analysis separates the search for stationary candidates from the question of global behavior.
Verify that the constraint gradient is nonzero at the feasible points where the Lagrange theorem is being applied.
Find every feasible solution of \(\nabla f=\lambda\nabla g\), and evaluate the objective at each one.
Use a parameterization, a feasible path, or another direct argument to see what objective values occur far from the candidate points.
Decide whether the objective attains an absolute maximum or minimum, and justify the claim over the entire feasible set.
A path whose objective values tend to \(+\infty\) proves that there is no absolute maximum. A path whose values tend to \(-\infty\) proves that there is no absolute minimum. When neither kind of path settles the question, a full argument is still needed: a few paths cannot rule out extrema elsewhere on the constraint.
A Common Mistake: Choosing the Best Candidate Value
A tempting but incorrect conclusion in the first example is: “The stationary values are \(2\) and \(-2\), so the absolute maximum is \(2\) and the absolute minimum is \(-2\).” This compares only the stationary candidates. It does not compare them with all feasible points.
The correction is to test the full constraint. At \(x=n\), \(y=1/n\), the constraint \(xy=1\) holds and \(f=n+1/n\), which grows without bound. At \(x=-n\), \(y=-1/n\), the constraint still holds and \(f=-n-1/n\), which decreases without bound. The candidate values are stationary values, not absolute extrema. In particular, a largest or smallest value among a finite list of candidates is not necessarily a global extremum on a noncompact set.
Check Your Understanding
Use the distinction between stationary candidates and global behavior to answer these questions.
- What regularity condition on a single constraint is required for the stated Lagrange condition?
- For \(f(x,y)=x+y\) subject to \(xy=1\), what are the stationary candidates, and why are neither absolute extrema?
- For \(f(x,y,z)=z\) subject to \(z=x^2+y^2\), why is the origin an absolute minimum but not a maximum?
- Give a feasible path for the constraint \(z=x^2+y^2\) along which the objective \(f(x,y,z)=x\) is unbounded above and below.
- Why is it insufficient to choose the largest and smallest objective values among the Lagrange candidates on a noncompact constraint?