Qu'est-ce qui se passe quand on utilise la méthode des multiplicateurs de Lagrange pour trouver des maxima ?
Single constraint

For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider the optimization problem maximize x, y f (x, y) subject to g (x, y) = 0. {\displaystyle {\begin{aligned}{\underset {x,y}{\text{maximize}}}\quad &f(x,y)\\{\text{subject to}}\quad &g(x,y)=0.\end{aligned}}} (Sometimes an additive constant is shown separately rather than being included in g {\displaystyle g}, in which case the constraint is written g (x, y) = c, {\displaystyle g(x,y)=c,} as in Figure 1.) We assume that both f {\displaystyle f} and g {\displaystyle g} have continuous first partial derivatives. We introduce a new variable (λ {\displaystyle \lambda }) called a Lagrange multiplier (or Lagrange undetermined multiplier) and study the Lagrange function (or Lagrangian or Lagrangian expression) defined by L (x, y, λ) = f (x, y) + λ ⋅ g (x, y), {\displaystyle {\mathcal {L}}(x,y,\lambda)=f(x,y)+\lambda \cdot g(x,y),} where the λ {\displaystyle \lambda } term may be either added or subtracted. If f (x 0, y 0) {\displaystyle f(x_{0},y_{0})} is a maximum of f (x, y) {\displaystyle f(x,y)} for the original constrained problem and ∇ g (x 0, y 0) ≠ 0, {\displaystyle \nabla g(x_{0},y_{0})\neq 0,} then there exists λ 0 {\displaystyle \lambda _{0}} such that (x 0, y 0, λ 0 {\displaystyle x_{0},y_{0},\lambda _{0}}) is a stationary point for the Lagrange function (stationary points are those points where the first partial derivatives of L {\displaystyle {\mathcal {L}}} are zero). The assumption ∇ g ≠ 0 {\displaystyle \nabla g\neq 0} is called constraint qualification. However, not all stationary points yield a solution of the original problem, as the method of Lagrange multipliers yields only a necessary condition for optimality in constrained problems. Sufficient conditions for a minimum or maximum also exist, but if a particular candidate solution satisfies the sufficient conditions, it is only guaranteed that that solution is the best one locally – that is, it is better than any permissible nearby points. The global optimum can be found by comparing the values of the original objective function at the points satisfying the necessary and locally sufficient conditions.

The method of Lagrange multipliers relies on the intuition that at a maximum, f (x, y) cannot be increasing in the direction of any such neighboring point that also has g = 0. If it were, we could walk along g = 0 to get higher, meaning that the starting point wasn't actually the maximum. Viewed in this way, it is an exact analogue to testing if the derivative of an unconstrained function is 0, that is, we are verifying that the directional derivative is 0 in any relevant (viable) direction.

We can visualize contours of f given by f (x, y) = d for various values of d, and the contour of g given by g (x, y) = c.

Suppose we walk along the contour line with g = c. We are interested in finding points where f almost does not change as we walk, since these points might be maxima.

There are two ways this could happen:

To check the first possibility (we touch a contour line of f), notice that since the gradient of a function is perpendicular to the contour lines, the tangents to the contour lines of f and g are parallel if and only if the gradients of f and g are parallel. Thus we want points (x, y) where g (x, y) = c and ∇ x, y f = λ ∇ x, y g, {\displaystyle \nabla _{x,y}f=\lambda \,\nabla _{x,y}g,} for some λ {\displaystyle \lambda }

where ∇ x, y f = (∂ f ∂ x, ∂ f ∂ y), ∇ x, y g = (∂ g ∂ x, ∂ g ∂ y) {\displaystyle \nabla _{x,y}f=\left({\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right),\qquad \nabla _{x,y}g=\left({\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right)} are the respective gradients. The constant λ {\displaystyle \lambda } is required because although the two gradient vectors are parallel, the magnitudes of the gradient vectors are generally not equal. This constant is called the Lagrange multiplier. (In some conventions λ {\displaystyle \lambda } is preceded by a minus sign).

Notice that this method also solves the second possibility, that f is level: if f is level, then its gradient is zero, and setting λ = 0 {\displaystyle \lambda =0} is a solution regardless of ∇ x, y g {\displaystyle \nabla _{x,y}g}.

To incorporate these conditions into one equation, we introduce an auxiliary function L (x, y, λ) ≡ f (x, y) + λ ⋅ g (x, y), {\displaystyle {\mathcal {L}}(x,y,\lambda)\equiv f(x,y)+\lambda \cdot g(x,y)\,,} and solve ∇ x, y, λ L (x, y, λ) = 0. {\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda)=0~.}

Note that this amounts to solving three equations in three unknowns. This is the method of Lagrange multipliers.

Note that ∇ λ L (x, y, λ) = 0 {\displaystyle \ \nabla _{\lambda }{\mathcal {L}}(x,y,\lambda)=0\ } implies g (x, y) = 0 , {\displaystyle \ g(x,y)=0\,} as the partial derivative of L {\displaystyle {\mathcal {L}}} with respect to λ {\displaystyle \lambda } is g (x, y). {\displaystyle \ g(x,y)~.}

To summarize ∇ x, y, λ L (x, y, λ) = 0 ⟺ { ∇ x, y f (x, y) = − λ ∇ x, y g (x, y) g (x, y) = 0 {\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda)=0\iff {\begin{cases}\nabla _{x,y}f(x,y)=-\lambda \,\nabla _{x,y}g(x,y)\\g(x,y)=0\end{cases}}}

The method generalizes readily to functions on n {\displaystyle n} variables ∇ x 1, …, x n, λ L (x 1, …, x n, λ) = 0 {\displaystyle \nabla _{x_{1},\dots,x_{n},\lambda }{\mathcal {L}}(x_{1},\dots,x_{n},\lambda)=0} which amounts to solving n + 1 equations in n + 1 unknowns.

The constrained extrema of f are critical points of the Lagrangian L {\displaystyle {\mathcal {L}}}, but they are not necessarily local extrema of L {\displaystyle {\mathcal {L}}} (see § Example 2 below).

One may reformulate the Lagrangian as a Hamiltonian, in which case the solutions are local minima for the Hamiltonian. This is done in optimal control theory, in the form of Pontryagin's minimum principle.

The fact that solutions of the method of Lagrange multipliers are not necessarily extrema of the Lagrangian, also poses difficulties for numerical optimization. This can be addressed by minimizing the magnitude of the gradient of the Lagrangian, as these minima are the same as the zeros of the magnitude, as illustrated in Example 5: Numerical optimization.
En optimisation mathématique, la méthode des multiplicateurs de Lagrange est une stratégie pour trouver les maxima et minima locaux d'une fonction soumise à des contraintes d'équation (c'est-à-dire, soumise à la condition qu'une ou plusieurs équations doivent être satisfaites exactement par les valeurs choisies des variables). Elle porte le nom du mathématicien Joseph-Louis Lagrange.
### Query ###
Qu'est-ce qui se passe quand on utilise la méthode des multiplicateurs de Lagrange pour trouver des maxima ?

### Main Text ###
Single constraint

For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider the optimization problem maximize x, y f (x, y) subject to g (x, y) = 0. {\displaystyle {\begin{aligned}{\underset {x,y}{\text{maximize}}}\quad &f(x,y)\\{\text{subject to}}\quad &g(x,y)=0.\end{aligned}}} (Sometimes an additive constant is shown separately rather than being included in g {\displaystyle g}, in which case the constraint is written g (x, y) = c, {\displaystyle g(x,y)=c,} as in Figure 1.) We assume that both f {\displaystyle f} and g {\displaystyle g} have continuous first partial derivatives. We introduce a new variable (λ {\displaystyle \lambda }) called a Lagrange multiplier (or Lagrange undetermined multiplier) and study the Lagrange function (or Lagrangian or Lagrangian expression) defined by L (x, y, λ) = f (x, y) + λ ⋅ g (x, y), {\displaystyle {\mathcal {L}}(x,y,\lambda)=f(x,y)+\lambda \cdot g(x,y),} where the λ {\displaystyle \lambda } term may be either added or subtracted. If f (x 0, y 0) {\displaystyle f(x_{0},y_{0})} is a maximum of f (x, y) {\displaystyle f(x,y)} for the original constrained problem and ∇ g (x 0, y 0) ≠ 0, {\displaystyle \nabla g(x_{0},y_{0})\neq 0,} then there exists λ 0 {\displaystyle \lambda _{0}} such that (x 0, y 0, λ 0 {\displaystyle x_{0},y_{0},\lambda _{0}}) is a stationary point for the Lagrange function (stationary points are those points where the first partial derivatives of L {\displaystyle {\mathcal {L}}} are zero). The assumption ∇ g ≠ 0 {\displaystyle \nabla g\neq 0} is called constraint qualification. However, not all stationary points yield a solution of the original problem, as the method of Lagrange multipliers yields only a necessary condition for optimality in constrained problems. Sufficient conditions for a minimum or maximum also exist, but if a particular candidate solution satisfies the sufficient conditions, it is only guaranteed that that solution is the best one locally – that is, it is better than any permissible nearby points. The global optimum can be found by comparing the values of the original objective function at the points satisfying the necessary and locally sufficient conditions.

The method of Lagrange multipliers relies on the intuition that at a maximum, f (x, y) cannot be increasing in the direction of any such neighboring point that also has g = 0. If it were, we could walk along g = 0 to get higher, meaning that the starting point wasn't actually the maximum. Viewed in this way, it is an exact analogue to testing if the derivative of an unconstrained function is 0, that is, we are verifying that the directional derivative is 0 in any relevant (viable) direction.

We can visualize contours of f given by f (x, y) = d for various values of d, and the contour of g given by g (x, y) = c.

Suppose we walk along the contour line with g = c. We are interested in finding points where f almost does not change as we walk, since these points might be maxima.

There are two ways this could happen:

To check the first possibility (we touch a contour line of f), notice that since the gradient of a function is perpendicular to the contour lines, the tangents to the contour lines of f and g are parallel if and only if the gradients of f and g are parallel. Thus we want points (x, y) where g (x, y) = c and ∇ x, y f = λ ∇ x, y g, {\displaystyle \nabla _{x,y}f=\lambda \,\nabla _{x,y}g,} for some λ {\displaystyle \lambda }

where ∇ x, y f = (∂ f ∂ x, ∂ f ∂ y), ∇ x, y g = (∂ g ∂ x, ∂ g ∂ y) {\displaystyle \nabla _{x,y}f=\left({\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right),\qquad \nabla _{x,y}g=\left({\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right)} are the respective gradients. The constant λ {\displaystyle \lambda } is required because although the two gradient vectors are parallel, the magnitudes of the gradient vectors are generally not equal. This constant is called the Lagrange multiplier. (In some conventions λ {\displaystyle \lambda } is preceded by a minus sign).

Notice that this method also solves the second possibility, that f is level: if f is level, then its gradient is zero, and setting λ = 0 {\displaystyle \lambda =0} is a solution regardless of ∇ x, y g {\displaystyle \nabla _{x,y}g}.

To incorporate these conditions into one equation, we introduce an auxiliary function L (x, y, λ) ≡ f (x, y) + λ ⋅ g (x, y), {\displaystyle {\mathcal {L}}(x,y,\lambda)\equiv f(x,y)+\lambda \cdot g(x,y)\,,} and solve ∇ x, y, λ L (x, y, λ) = 0. {\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda)=0~.}

Note that this amounts to solving three equations in three unknowns. This is the method of Lagrange multipliers.

Note that ∇ λ L (x, y, λ) = 0 {\displaystyle \ \nabla _{\lambda }{\mathcal {L}}(x,y,\lambda)=0\ } implies g (x, y) = 0 , {\displaystyle \ g(x,y)=0\,} as the partial derivative of L {\displaystyle {\mathcal {L}}} with respect to λ {\displaystyle \lambda } is g (x, y). {\displaystyle \ g(x,y)~.}

To summarize ∇ x, y, λ L (x, y, λ) = 0 ⟺ { ∇ x, y f (x, y) = − λ ∇ x, y g (x, y) g (x, y) = 0 {\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda)=0\iff {\begin{cases}\nabla _{x,y}f(x,y)=-\lambda \,\nabla _{x,y}g(x,y)\\g(x,y)=0\end{cases}}}

The method generalizes readily to functions on n {\displaystyle n} variables ∇ x 1, …, x n, λ L (x 1, …, x n, λ) = 0 {\displaystyle \nabla _{x_{1},\dots,x_{n},\lambda }{\mathcal {L}}(x_{1},\dots,x_{n},\lambda)=0} which amounts to solving n + 1 equations in n + 1 unknowns.

The constrained extrema of f are critical points of the Lagrangian L {\displaystyle {\mathcal {L}}}, but they are not necessarily local extrema of L {\displaystyle {\mathcal {L}}} (see § Example 2 below).

One may reformulate the Lagrangian as a Hamiltonian, in which case the solutions are local minima for the Hamiltonian. This is done in optimal control theory, in the form of Pontryagin's minimum principle.

The fact that solutions of the method of Lagrange multipliers are not necessarily extrema of the Lagrangian, also poses difficulties for numerical optimization. This can be addressed by minimizing the magnitude of the gradient of the Lagrangian, as these minima are the same as the zeros of the magnitude, as illustrated in Example 5: Numerical optimization.

### Supplementary Text ###
En optimisation mathématique, la méthode des multiplicateurs de Lagrange est une stratégie pour trouver les maxima et minima locaux d'une fonction soumise à des contraintes d'équation (c'est-à-dire, soumise à la condition qu'une ou plusieurs équations doivent être satisfaites exactement par les valeurs choisies des variables). Elle porte le nom du mathématicien Joseph-Louis Lagrange.

### Draft ###

Query: "Qu'est-ce qui se passe quand on utilise la méthode des multiplicateurs de Lagrange pour trouver des maxima ?"

Translation check: "What happens when using Lagrange multiplier method to find maxima?" → Standard mathematical terminology. No semantic ambiguity.

Task type: Information retrieval + conceptual explanation. Not complex reasoning.

### 1. Core Mechanism

Lagrange multipliers → constrained optimization technique.

Setup: Maximize f(x₁,x₂,...,xₙ) subject to g(x₁,x₂,...,xₙ) = c.

Key insight: At optimum, no "steepest ascent" direction exists along constraint boundary.

∇f ⊥ contour lines of f.
∇g ⊥ constraint boundary.
At optimum: ∇f || ∇g (parallel gradients)
∴ ∇f = λ∇g for some multiplier λ.

Lagrangian: L(x₁,x₂,...,xₙ,λ) = f(x₁,x₂,...,xₙ) + λ[g(x₁,x₂,...,xₙ) - c]

Critical points: ∇L = 0 → n+1 equations in n+1 unknowns.

### 2. Geometric Intuition

Constraint boundary acts as "shadow boundary" in objective function space.

Lagrange condition → normal vectors of objective and constraint surfaces are parallel at optimum.

λ = "intensity" of constraint force. Sign indicates whether constraint is binding.

### 3. Mathematical Properties

Necessary condition only. Sufficient conditions require second derivatives.

Stationary points ≠ necessarily maxima. Classification needed via:
- Second derivative test
- Sign analysis of λ
- Local vs global behavior

### 4. Generalizations

Single constraint: n equations in n+1 unknowns
Multiple constraints: m equations in n+1 unknowns

Equality constraints only. Inequality constraints need different formulation (KKT conditions).

### 5. Physical Interpretation

Method emerges naturally in physics/engineering:
- Equilibrium under forces
- Extremal principles
- Variational calculus

Lagrange multipliers → "shadow prices" in economics.

### 6. Limitations

Local maxima only. Global optimization requires additional techniques.

Constraint qualifications needed.

### 7. Applications

Applied everywhere constraints matter:
- Engineering design
- Economics (utility maximization)
- Physics (Hamilton-Jacobi theory)
- Operations research

● High confidence: Standard mathematical technique with clear geometric interpretation
◐ Partial: Sufficient conditions for global maxima can be subtle
○ Lower confidence: Numerical implementation details

Key insight: Method provides necessary conditions for constrained maxima through Lagrangian formulation and gradient orthogonality principle.

### Answer###
Lors de l'utilisation de la méthode des multiplicateurs de Lagrange pour trouver des maxima, plusieurs mécanismes fondamentaux entrent en jeu.

La méthode transforme le problème de maximisation d'une fonction soumise à contraintes en un problème d'optimisation sans contraintes en introduisant une variable auxiliaire appelée multiplicateur de Lagrange. Cette variable permet de formuler une condition nécessaire pour l'optimalité : le gradient de la fonction objectif doit être colinéaire au gradient de la contrainte, avec une constante proportionnalité appelée multiplicateur de Lagrange.

Concrètement, si vous avez une fonction f(x₁,x₂,...,xₙ) que vous voulez maximiser sous la contrainte g(x₁,x₂,...,xₙ) = c, la méthode introduit le lagrangien L(x₁,x₂,...,xₙ,λ) = f(x₁,x₂,...,xₙ) + λ[g(x₁,x₂,...,xₙ) - c]. Les points critiques de ce lagrangien correspondent aux solutions du système d'équations ∂L/∂xᵢ = 0 pour chaque variable xᵢ et ∂L/∂λ = 0 pour le multiplicateur.

Cette formulation géométrique repose sur l'intuition que, en un maximum, la fonction ne peut pas augmenter dans la direction de tout point voisin qui respecte la contrainte. Les courbes de niveau de la fonction et de la contrainte se comportent de manière particulière : les gradients sont parallèles aux tangentes des courbes de niveau, ce qui explique la condition de colinéarité des gradients.

Le multiplicateur de Lagrange λ représente l'intensité de l'influence de la contrainte sur le résultat optimal. Il peut être interprété comme le "prix ombre" de la contrainte : la variation du lagrangien par rapport à λ correspond exactement à la valeur de la contrainte.

Il est important de noter que les points stationnaires obtenus par cette méthode ne garantissent pas nécessairement des maxima locaux, seulement une condition nécessaire pour l'optimalité. Les conditions suffisantes nécessitent une analyse supplémentaire des dérivées secondes et peuvent être locales.