Lagrange Multipliers

The central idea

To optimize differentiable $f:\mathbb R^n\to\mathbb R$ subject to a regular constraint $g(\mathbf x)=c$, candidate constrained extrema satisfy $\nabla f=\lambda\nabla g$ with $g=c$, provided $\nabla g\ne0$. With several independent constraints, add one multiplier per constraint.

Definitions, hypotheses, and notation

The regularity condition is essential: if $\nabla g=0$, the constraint may not have a smooth tangent space and the multiplier equation can miss or overproduce candidates. Constraints with endpoints, corners, or inequalities require boundary decomposition or Karush–Kuhn–Tucker ideas beyond the basic theorem.

For a compact constraint and continuous objective, absolute extrema exist by the Extreme Value Theorem. Lagrange multipliers then generate candidates, but global classification comes only from evaluating all candidates. The sign and size of $\lambda$ should not be mistaken for max/min classification.

Conceptual meaning

At a constrained extremum, every feasible tangent direction has zero directional change in $f$. Thus $\nabla f$ is normal to the constraint surface, as is $\nabla g$, making them parallel. The multiplier records sensitivity to the constraint level under suitable conditions.

A dependable method and decision rule

1. Check compactness or other conditions if absolute extrema are requested.
2. Compute both gradients and note singular constraint points separately.
3. Solve $\nabla f=\lambda\nabla g$ together with the constraint.
4. Include all solution branches, especially zero-coordinate cases.
5. Evaluate $f$ at every candidate and compare values.

Fully worked example

Graphical or geometric meaning

Level curves $xy=c$ slide until tangent to the constraint circle. At tangency their normal vectors, the two gradients, are parallel. The four symmetric contacts account for both maximum and minimum pairs.

Common mistakes and why they fail

Verification and reasonableness checks

• Substitute every candidate into the constraint.
• Evaluate objective values rather than classifying from $\lambda$ alone.
• Use symmetry or a direct inequality as an independent check.

Include regularity and every candidate

At a constrained extremum of $f$ on $g=c$, with $\nabla g\ne0$, gradients are parallel: $\nabla f=\lambda\nabla g$. This condition identifies candidates because the objective's level surface is tangent to the constraint. It is necessary under regularity, not a guarantee that every solution is an extremum. Solve the multiplier equations together with the constraint, then compare objective values. Also examine points where $\nabla g=0$, nonsmooth parts, endpoints, or boundary intersections separately because the standard equation can miss them. On a compact constraint, continuity guarantees absolute extrema, making the final value comparison decisive. Keep the multiplier as an auxiliary unknown rather than dividing by a component that might be zero.

Practice

1. Maximize $x+y$ on $x^2+y^2=1$.
2. What must hold at a regular constrained extremum?
3. Does the method alone guarantee an absolute extremum?

Connections and next steps

• Multivariable Optimization
• Gradient
• Tangent Planes