Method of Lagrange Multipliers

Summary

Lagrange multipliers locate constrained extrema of $f$ subject to $g=\text{constant}$ by solving $\mathrm{\nabla }f=\lambda \mathrm{\nabla }g$ together with the constraint. Critical points of the Lagrangian must still be classified (max, min, or saddle-type on the constraint).

Prerequisites

Partial Derivatives, Maxima and Minima, gradients

Theorem / Procedure

To extremize $f(\mathbf{x})$ subject to $g(\mathbf{x})=c$ (one constraint), introduce $\lambda$ and solve

equivalently $\mathrm{\nabla }L=\mathbf{0}$ for

Conditions / Assumptions

• $f$ and $g$ are $C^1$ near the solution.
• $\mathrm{\nabla }g\ne \mathbf{0}$ at candidate points (constraint qualification).
• On a closed bounded constraint set (compact), extrema exist; on unbounded constraints, there may be no maximum or no minimum.

Worked Example

Optimize $f(x,y)=x^2+y^2$ subject to $x+y=1$ .

Solve $2x=\lambda$ , $2y=\lambda$ , $x+y=1$ . Then $x=y=1/2$ , so the only critical point is $(1/2,1/2)$ with $f=1/2$ .

On the line $x+y=1$ , $f(x,1-x)=x^2+(1-x{)}^2=2x^2-2x+1$ , whose graph is a parabola opening upward. Thus $(1/2,1/2)$ is a minimum, not a maximum. As $|x|\to \mathrm{\infty }$ along the line, $f\to \mathrm{\infty }$ , so no maximum exists.

A corrected “maximize” example: maximize $f(x,y)=xy$ on $x+y=1$ with $x,y\ge 0$ (compact segment) yields maximum $1/4$ at $(1/2,1/2)$ .

Common Mistakes

• Reporting a Lagrange critical point as a maximum without checking the constraint geometry.
• Ignoring points where $\mathrm{\nabla }g=\mathbf{0}$ .
• Forgetting that unbounded constraints need behavior at infinity.

Connections

• Unconstrained test: Maxima and Minima
• Geometry: $\mathrm{\nabla }f$ parallel to $\mathrm{\nabla }g$ means level curves of $f$ and $g$ share a tangent

References

Lagrange multipliers are treated in OpenStax Calculus Volume 3.^[1]