Lagrange Multipliers and Constraint Geometry

At an extremum, allowed directions cannot improve the value

On a regular constraint $g(x,y)=c$ , tangent directions are perpendicular to $\nabla g$ . At a differentiable constrained extremum, $\nabla f$ is also perpendicular to those tangent directions. Thus

\nabla f=\lambda\nabla g,

g=c,

provided $\nabla g\ne0$ . The multiplier is an additional unknown, not a replacement for the constraint. The equations produce candidates that still require comparison or classification.

Visual guide

VISUAL GUIDEAt an extremum, the level line is tangent

Constraint circle
Maximum level x + 2y = 5

Maximizing x + 2y on x² + y² = 5 puts the maximum at (1, 2). The level line x + 2y = 5 is tangent to the circle there. Its normal is parallel to the constraint gradient, expressing ∇f = λ∇g.

Worked example: a linear objective on a circle

Maximize and minimize $f=x+2y$ on $x^2+y^2=5$ . The equations are $1=2\lambda x$ , $2=2\lambda y$ , so $y=2x$ . Substituting into the circle gives $5x^2=5$ , yielding $(1,2)$ and $(-1,-2)$ .

Their values are $5$ and $-5$ . Compactness of the circle and continuity of $f$ ensure extrema exist, and the constraint gradient never vanishes there. Therefore these candidates supply the absolute maximum and minimum.

Worked example: a singular constraint

Take $g(x,y)=x^2+y^2=0$ . The feasible set consists only of the origin, so every objective reaches both its feasible maximum and minimum there. But $\nabla g(0,0)=0$ . For $f=x$ , the multiplier equation $(1,0)=\lambda(0,0)$ has no solution.

The method missed a genuine feasible extremum because the constraint qualification failed. Always inspect points where the constraint gradient vanishes. For several constraints, their gradients must be linearly independent for the standard multiplier conclusion; then $\nabla f=\sum_j\lambda_j\nabla g_j$ .

A multiplier can also describe sensitivity. Under suitable smoothness of the optimizing branch, the derivative of the optimal value with respect to the constraint level $c$ equals $\lambda$ with the convention $\nabla f=\lambda\nabla g$ . This interpretation depends on the chosen scaling of $g$ and is not itself a proof of optimality.

Practice

Maximize $xy$ on $x^2+y^2=2$ .
Why must $g=c$ be included in the equation system?
Can a finite candidate list alone prove an extremum exists on an unbounded feasible set?

Show worked solutions

The maximum is $1$ at $(1,1)$ and $(-1,-1)$ , since $2xy\le x^2+y^2=2$ ; multiplier equations recover these points.
Otherwise a gradient relation may hold away from the feasible set.
No. Behavior at infinity must also be controlled; the objective may be unbounded or approach an unattained limit.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.