Method of Lagrange for Inequality-Constrained Maximization¶

Theorem¶

Let $A\subseteq\mathbb{R}^n$ be open, and let $f,g_1,\ldots,g_k\colon A\to\mathbb{R}$ be continuously differentiable. Consider

\begin{aligned} &\max_{\boldsymbol{x}\in A} f(\boldsymbol{x})\\ &\text{subject to}\quad g_i(\boldsymbol{x})\le b_i,\ i=1,\ldots,k. \end{aligned}

The course convention writes $b_i-g_i(\boldsymbol{x})\ge0$ and defines

L(\boldsymbol{x},\boldsymbol{\lambda})=f(\boldsymbol{x})+\sum_{i=1}^k\lambda_i(b_i-g_i(\boldsymbol{x})).

If $\boldsymbol{x}^\ast$ is a local constrained maximizer, then either nondegenerate constraint qualification (NDCQ) fails or there exists $\boldsymbol{\lambda}^\ast$ such that

\boldsymbol{\nabla}_xL(\boldsymbol{x}^\ast,\boldsymbol{\lambda}^\ast)=0,\qquad \lambda_i^\ast(g_i(\boldsymbol{x}^\ast)-b_i)=0,\qquad \lambda_i^\ast\ge0,\qquad g_i(\boldsymbol{x}^\ast)\le b_i.

The condition $\lambda_i^\ast(g_i(\boldsymbol{x}^\ast)-b_i)=0$ is complementary slackness.

Interpretation¶

Inequality constraints are directional, so multiplier signs and slackness matter. At a maximizer each constraint is either slack, and plays no role nearby, or binding, and acts as a one-sided equality. The sign condition $\lambda_i\ge0$ and complementary slackness handle both cases.

Feasible triangle x1 + x2 at most 3.8, with the unconstrained maximum of f outside it. At x star on the boundary the gradients of f and g are parallel and point outwards.

The unconstrained maximum of $f$ , the centre of the grey level circles, lies outside the shaded feasible side of $x_1+x_2\le3.8$ $3.8$ , so the constrained maximum $\boldsymbol{x}^\ast$ lies on the line. There the gradient $\boldsymbol{\nabla} f(\boldsymbol{x}^\ast)$ and $\boldsymbol{\nabla} g=(1,1)$ point the same way, outwards towards the infeasible peak, so $\boldsymbol{\nabla} f(\boldsymbol{x}^\ast)=\lambda\boldsymbol{\nabla} g(\boldsymbol{x}^\ast)$ with $\lambda>0$ $0$ : $f$ could only increase outside the feasible set.

Drag the peak of $f$ : if it is feasible the constraint is slack and $\lambda=0$ $0$ ; otherwise the optimum sits on the line and $\boldsymbol{\nabla} f=\lambda\boldsymbol{\nabla} g$ with $\lambda>0$ $0$ .

The sign condition encodes this direction. If $\boldsymbol{\nabla} f(\boldsymbol{x}^\ast)$ were a negative multiple of $\boldsymbol{\nabla} g_i(\boldsymbol{x}^\ast)$ , it would point into the feasible set, and a small step inwards would increase $f$ . Complementary slackness says that each constraint is binding or has $\lambda_i^\ast=0$ $0$ . NDCQ is checked for the binding constraints only.

The conditions are necessary only. For $\max xy$ subject to $x+y\le10$ from the Examples, both $(0,0)$ with $\lambda=0$ and $(5,5)$ with $\lambda=5$ satisfy them. The first is a saddle point of $xy$ . The second is a local maximizer but not a global one: $xy=t^2$ grows without bound along the feasible points $x=y=-t$ $- t$ , $t\to\infty$ $\infty$ .

Compared with equality constraints, whose multipliers may have either sign, these two conditions are new. Linear programming shows the same pattern: a resource with unused slack has shadow price $0$ .

Key Results¶

A binding constraint satisfies $g_i(\boldsymbol{x}^\ast)=b_i$ $b_{i}$ .
A nonbinding constraint has zero multiplier by complementary slackness.
The theorem gives necessary, not sufficient, conditions.

Examples¶

For $\max xy$ subject to $x+y\le10$ $10$ , the Lagrangian is $xy+\lambda(10-x-y)$ $y)$ .
If no inequality is binding at a local maximum, all inequality multipliers are zero and the unconstrained stationarity condition returns.

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 5, Lecture 1

Keywords: inequality FONC for maximization, KKT necessary conditions for max Symbols: element of, infinity, lambda, nabla gradient, real numbers, subset, sum

Last updated: 22 October 2026 19:04