Lagrangian Function¶

Definition¶

For a constrained optimization problem with objective $f$ and equality constraints $h_j(\boldsymbol{x})=c_j$ $c_{j}$ , a Lagrangian function is an auxiliary function that combines the objective and the constraints with multiplier variables. Under the course convention for equality constraints,

L(\boldsymbol{x},\boldsymbol{\mu})=f(\boldsymbol{x})+\sum_{j=1}^m \mu_j(c_j-h_j(\boldsymbol{x})).

For inequality constraints, the same idea is used with sign conventions chosen so that multiplier restrictions match the direction of the inequality.

Interpretation¶

The Lagrangian rewrites a constrained problem as one expression whose stationary points encode both the objective trade-off and the constraint equations.

Circular level sets of f and the constraint line h(x) = c. At x star the gradients of f and h are parallel. At another point on the line they are not, so moving along the line still improves f.

At the illustrated candidate $\boldsymbol{x}^\ast$ , the gradients of $f$ and $h$ are parallel. At the orange point they are not, so moving along $h(\boldsymbol{x})=c$ can improve $f$ . The differentiability and rank assumptions for the multiplier test are stated in the equality-constraint method.

Setting all partial derivatives of $L(x,y,\mu)$ to zero recovers both the tangency condition and the constraint itself.

The Lagrangian is a function of both the decision variables and the multipliers. Partial differentiation with respect to a multiplier recovers its constraint residual; differentiation with respect to a decision variable gives the corresponding stationarity equation.

Key Results¶

Differentiating with respect to the multiplier variables recovers the constraints.
Differentiating with respect to the choice variables gives the first-order balance between objective gradients and constraint gradients.
Changing the sign convention changes multiplier signs, but not the underlying candidate points when handled consistently.

Examples¶

For $\max xy$ subject to $x+y=10$ $10$ , one convention uses $L(x,y,\mu)=xy+\mu(10-x-y)$ $y)$ .
For $\min x^2+y^2$ subject to $x+y=1$ $1$ , one can use $L(x,y,\mu)=x^2+y^2+\mu(1-x-y)$ $y)$ .

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 4, Lecture 2
Math IV: Multivariate Calculus · Week 5, Lecture 1
Math IV: Multivariate Calculus · Week 7, Lecture 1

Aliases: Lagrangian, Lagrange function, constrained optimization Lagrangian, Lagrangean Keywords: ordinary Lagrangian Symbols: mu, sum

Last updated: 26 October 2026 00:09