Lagrange multiplier

The basic idea is to convert a constrained problem into a form such that the derivative test of an unconstrained problem can still be applied. The relationship between the gradient of the function and gradients of the constraints rather naturally leads to a reformulation of the original problem, known as the Lagrangian function or Lagrangian.{{cite book |first1=Brian |last1=Beavis |first2=Ian M. |last2=Dobbs |chapter=Static Optimization |title=Optimization and Stability Theory for Economic Analysis |location=New York |publisher=Cambridge University Press |year=1990 |isbn=0-521-33605-8 |page=40 |chapter-url=https://books.google.com/books?id=L7HMACFgnXMC&pg=PA40 }} In the general case, the Lagrangian is defined as

$$\backslash mathcal\{L\}(x,\; \backslash lambda)\; \backslash equiv\; f(x)\; +\; \backslash langle\; \backslash lambda,\; g(x)\backslash rangle$$

for functions $f,\; g$; the notation $\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle$ denotes an inner product. The value $\backslash lambda$ is called the Lagrange multiplier.

In simple cases, where the inner product is defined as the dot product, the Lagrangian is

$$\backslash mathcal\{L\}(x,\; \backslash lambda)\; \backslash equiv\; f(x)\; +\; \backslash lambda\backslash cdot\; g(x)$$

The method can be summarized as follows: in order to find the maximum or minimum of a function $f$ subject to the equality constraint $g(x)\; =\; 0$, find the stationary points of $\backslash mathcal\{L\}$ considered as a function of $x$ and the Lagrange multiplier $\backslash lambda\; ~$. This means that all partial derivatives should be zero, including the partial derivative with respect to $\backslash lambda\; ~$.{{cite book |first1=Murray H. |last1=Protter |author-link=Murray H. Protter |first2=Charles B. Jr. |last2=Morrey |author-link2=Charles B. Morrey Jr. |year=1985 |title=Intermediate Calculus |place=New York, NY |publisher=Springer |edition=2nd |page=267 |isbn=0-387-96058-9 }}

{{block indent | em = 1.5 | text = $\backslash frac\{\; \backslash partial\; \backslash mathcal\{L\}\; \}\{\backslash partial\; x\}\; =\; 0$ {{pad|2em}} and {{pad|2em}} $\backslash frac\{\backslash \; \backslash partial\; \backslash mathcal\{L\}\backslash \; \}\{\backslash partial\; \backslash lambda\}\; =\; 0\backslash \; ;$}}

or equivalently

{{block indent | em = 1.5 | text = $\backslash frac\{\; \backslash partial\; f(x)\; \}\{\backslash partial\; x\}\; +\; \backslash lambda\backslash cdot\; \backslash frac\{\; \backslash partial\; g(x)\; \}\{\backslash partial\; x\}\; =\; 0$ {{pad|2em}} and {{pad|2em}}$g(x)\; =\; 0\; ~.$}}

The solution corresponding to the original constrained optimization is always a saddle point of the Lagrangian function,{{cite book |first=G.R. |last=Walsh |year=1975 |title=Methods of Optimization |place=New York, NY |publisher=John Wiley & Sons |isbn=0-471-91922-5 |chapter=Saddle-point Property of Lagrangian Function |pages=39–44 |chapter-url=https://books.google.com/books?id=K0EZAQAAIAAJ&pg=PA39 }}{{cite journal |first=Dan |last=Kalman |year=2009 |title=Leveling with Lagrange: An alternate view of constrained optimization |journal=Mathematics Magazine |volume=82 |issue=3 |pages=186–196 |doi=10.1080/0025570X.2009.11953617 |jstor=27765899 |s2cid=121070192 }} which can be identified among the stationary points from the definiteness of the bordered Hessian matrix.{{cite book |first1=Eugene |last1=Silberberg |first2=Wing |last2=Suen |year=2001 |title=The Structure of Economics: A Mathematical Analysis |location=Boston |publisher=Irwin McGraw-Hill |edition=Third |isbn=0-07-234352-4 |pages=134–141 }}

The great advantage of this method is that it allows the optimization to be solved without explicit parameterization in terms of the constraints. As a result, the method of Lagrange multipliers is widely used to solve challenging constrained optimization problems. Further, the method of Lagrange multipliers is generalized by the Karush–Kuhn–Tucker conditions, which can also take into account inequality constraints of the form $h(\backslash mathbf\{x\})\; \backslash leq\; c$ for a given constant $c$.

The following is known as the Lagrange multiplier theorem.{{cite book |first=Angel |last={{nobr|de la Fuente}} |title=Mathematical Methods and Models for Economists |location=Cambridge |publisher=Cambridge University Press |year=2000 |doi=10.1017/CBO9780511810756 |isbn=978-0-521-58512-5 |page=[https://archive.org/details/mathematicalmeth00fuen/page/n288 285] |url=https://archive.org/details/mathematicalmeth00fuen |url-access=limited }}

Let $f:\; \backslash mathbb\{R\}^n\; \backslash to\; \backslash mathbb\{R\}$ be the objective function and let $g:\; \backslash mathbb\{R\}^n\; \backslash to\; \backslash mathbb\{R\}^c$ be the constraints function, both belonging to $C^1$ (that is, having continuous first derivatives). Let $x\_\backslash star$ be an optimal solution to the following optimization problem such that, for the matrix of partial derivatives $\backslash Bigl[\; \backslash operatorname\{D\}g(x\_\backslash star)\; \backslash Bigr]\_\{j,k\}\; =\; \backslash frac\{\backslash \; \backslash partial\; g\_j\backslash \; \}\{\backslash partial\; x\_k\}$, $\backslash operatorname\{rank\}\; (\backslash operatorname\{D\}g(x\_\backslash star))\; =\; c\; \backslash le\; n$:

$$\backslash begin\{align\}$$

& \text{maximize } f(x) \\

& \text{subject to: } g(x) = 0

\end{align}

Then there exists a unique Lagrange multiplier $\backslash lambda\_\backslash star\; \backslash in\; \backslash mathbb\{R\}^c$ such that $\backslash operatorname\{D\}f(x\_\backslash star)\; =\; \backslash lambda\_\backslash star^\{\backslash mathsf\{T\}\}\backslash operatorname\{D\}g(x\_\backslash star)\; ~.$ (Note that this is a somewhat conventional thing where $\backslash lambda\_\backslash star$ is clearly treated as a column vector to ensure that the dimensions match. But, we might as well make it just a row vector without taking the transpose.){{Tone inline|date=April 2025}}

The Lagrange multiplier theorem states that at any local maximum (or minimum) of the function evaluated under the equality constraints, if constraint qualification applies (explained below), then the gradient of the function (at that point) can be expressed as a linear combination of the gradients of the constraints (at that point), with the Lagrange multipliers acting as coefficients.{{cite book |first=David G. |last=Luenberger |author-link=David Luenberger |year=1969 |title=Optimization by Vector Space Methods |location=New York |publisher=John Wiley & Sons |pages=188–189 }} This is equivalent to saying that any direction perpendicular to all gradients of the constraints is also perpendicular to the gradient of the function. Or still, saying that the directional derivative of the function is {{math|0}} in every feasible direction.

Image:LagrangeMultipliers2D.svg

For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider the optimization problem

$$\backslash begin\{aligned\}$$

\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 $g$, in which case the constraint is written $g(x,y)\; =\; c\; ,$ as in Figure 1.) We assume that both $f$ and $g$ have continuous first partial derivatives. We introduce a new variable ($\backslash lambda$) called a Lagrange multiplier (or Lagrange undetermined multiplier) and study the Lagrange function (or Lagrangian or Lagrangian expression) defined by

$$\backslash mathcal\{L\}(x,y,\backslash lambda)\; =\; f(x,y)\; +\; \backslash lambda\backslash cdot\; g(x,y),$$

where the $\backslash lambda$ term may be either added or subtracted. If $f(x\_0,\; y\_0)$ is a maximum of $f(x,y)$ for the original constrained problem and $\backslash nabla\; g(x\_0,y\_0)\; \backslash ne\; 0\; ,$ then there exists $\backslash lambda\_0$ such that ($x\_0,\; y\_0,\; \backslash lambda\_0$) is a stationary point for the Lagrange function (stationary points are those points where the first partial derivatives of $\backslash mathcal\{L\}$ are zero). The assumption $\backslash nabla\; g\; \backslash ne\; 0$ 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.{{cite book | last = Bertsekas | first = Dimitri P. | author-link = Dimitri P. Bertsekas | year = 1999 | title = Nonlinear Programming | edition = Second | publisher = Athena Scientific | location = Cambridge, MA | isbn = 1-886529-00-0 }}{{springer |id=Lagrange_multipliers |title=Lagrange multipliers |first=I.B. |last=Vapnyarskii }}.{{cite book |last=Lasdon |first=Leon S. |year=2002 |orig-date=1970 |title=Optimization Theory for Large Systems |publisher=Dover |location=Mineola, New York, NY |edition=reprint |mr=1888251 |isbn=0-486-41999-1 }}{{cite book |last1=Hiriart-Urruty |first1=Jean-Baptiste |last2=Lemaréchal |first2=Claude |author2-link=Claude Lemaréchal |year=1993 |chapter=Chapter XII: Abstract duality for practitioners |title=Convex analysis and minimization algorithms |id=Volume II: Advanced theory and bundle methods |series=Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences] |volume=306 |publisher=Springer-Verlag |place=Berlin, DE |isbn=3-540-56852-2 |mr=1295240 |pages=136–193 (and Bibliographical comments pp. 334–335)}}{{cite conference |last=Lemaréchal |first=Claude |author-link=Claude Lemaréchal |date=15–19 May 2000 |title=Lagrangian relaxation |editor1-last=Jünger |editor1-first=Michael |editor2-last=Naddef |editor2-first=Denis |publication-date=2001 |book-title=Computational combinatorial optimization: Papers from the Spring School held in Schloß Dagstuhl |conference=Spring School held in Schloß Dagstuhl, {{nobr|May 15–19, 2000}} |publisher=Springer-Verlag |isbn=3-540-42877-1 |series=Lecture Notes in Computer Science |volume=2241 |place=Berlin, DE |pages=112–156 |doi=10.1007/3-540-45586-8_4 |mr=1900016 |s2cid=9048698}} 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, {{math|f(x, y)}} cannot be increasing in the direction of any such neighboring point that also has {{math|1= g = 0}}. If it were, we could walk along {{math|1= 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 {{math|0}}, that is, we are verifying that the directional derivative is 0 in any relevant (viable) direction.

We can visualize contours of {{mvar|f}} given by {{math|1= f(x, y) = d}} for various values of {{mvar|d}}, and the contour of {{mvar|g}} given by {{math|1= g(x, y) = c}}.

Suppose we walk along the contour line with {{nobr|{{math|1= g = c}} .}} We are interested in finding points where {{mvar|f}} almost does not change as we walk, since these points might be maxima.

There are two ways this could happen:

1. We could touch a contour line of {{mvar|f}}, since by definition {{mvar|f}} does not change as we walk along its contour lines. This would mean that the tangents to the contour lines of {{mvar|f}} and {{mvar|g}} are parallel here.
2. We have reached a "level" part of {{mvar|f}}, meaning that {{mvar|f}} does not change in any direction.

To check the first possibility (we touch a contour line of {{mvar|f}}), notice that since the gradient of a function is perpendicular to the contour lines, the tangents to the contour lines of {{mvar|f}} and {{mvar|g}} are parallel if and only if the gradients of {{mvar|f}} and {{mvar|g}} are parallel. Thus we want points {{math|(x, y)}} where {{nobr| {{math| g(x, y) {{=}} c}} }} and

$$\backslash nabla\_\{x,y\}\; f\; =\; \backslash lambda\; \backslash ,\; \backslash nabla\_\{x,y\}\; g,$$

for some $\backslash lambda$ where

$$\backslash nabla\_\{x,y\}\; f\; =\; \backslash left(\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\},\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; y\}\; \backslash right),\; \backslash qquad\; \backslash nabla\_\{x,y\}\; g\; =\; \backslash left(\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; x\},\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; y\}\; \backslash right)$$

are the respective gradients. The constant $\backslash lambda$ 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 $\backslash lambda$ is preceded by a minus sign).

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

To incorporate these conditions into one equation, we introduce an auxiliary function

$$\backslash mathcal\{L\}(x,y,\backslash lambda)\; \backslash equiv\; f(x,y)\; +\; \backslash lambda\backslash cdot\; g(x,y)\backslash ,\; ,$$

and solve

$$\backslash nabla\_\{x,y,\backslash lambda\}\; \backslash mathcal\{L\}(x,\; y,\; \backslash lambda)\; =\; 0\; ~.$$Note that this amounts to solving three equations in three unknowns. This is the method of Lagrange multipliers.

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

To summarize

\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 $n$ variables

$$\backslash nabla\_\{x\_1,\; \backslash dots,\; x\_n,\backslash lambda\}\; \backslash mathcal\{L\}(x\_1,\; \backslash dots,\; x\_n,\; \backslash lambda)\; =\; 0$$

which amounts to solving {{math|n + 1}} equations in {{math|n + 1}} unknowns.

The constrained extrema of {{mvar|f}} are critical points of the Lagrangian $\backslash mathcal\{L\}$, but they are not necessarily local extrema of $\backslash mathcal\{L\}$ (see {{slink||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 maximum 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.

Image:As wiki lgm parab.svg

Image:As wiki lgm levelsets.svg

The method of Lagrange multipliers can be extended to solve problems with multiple constraints using a similar argument. Consider a paraboloid subject to two line constraints that intersect at a single point. As the only feasible solution, this point is obviously a constrained extremum. However, the level set of $f$ is clearly not parallel to either constraint at the intersection point (see Figure 3); instead, it is a linear combination of the two constraints' gradients. In the case of multiple constraints, that will be what we seek in general: The method of Lagrange seeks points not at which the gradient of $f$ is a multiple of any single constraint's gradient necessarily, but in which it is a linear combination of all the constraints' gradients.

Concretely, suppose we have $M$ constraints and are walking along the set of points satisfying $g\_i(\backslash mathbf\{x\})\; =\; 0,\; i=1,\; \backslash dots,\; M\; \backslash ,.$ Every point $\backslash mathbf\{x\}$ on the contour of a given constraint function $g\_i$ has a space of allowable directions: the space of vectors perpendicular to $\backslash nabla\; g\_i(\backslash mathbf\{x\})\; \backslash ,\; .$ The set of directions that are allowed by all constraints is thus the space of directions perpendicular to all of the constraints' gradients. Denote this space of allowable moves by $\backslash \; A\backslash$ and denote the span of the constraints' gradients by $S\; \backslash ,.$ Then $A\; =\; S^\{\backslash perp\}\backslash ,\; ,$ the space of vectors perpendicular to every element of $S\; \backslash ,.$

We are still interested in finding points where $f$ does not change as we walk, since these points might be (constrained) extrema. We therefore seek $\backslash mathbf\{x\}$ such that any allowable direction of movement away from $\backslash mathbf\{x\}$ is perpendicular to $\backslash nabla\; f(\backslash mathbf\{x\})$ (otherwise we could increase $f$ by moving along that allowable direction). In other words, $\backslash nabla\; f(\backslash mathbf\{x\})\; \backslash in\; A^\{\backslash perp\}\; =\; S\; \backslash ,.$ Thus there are scalars $\backslash lambda\_1,\; \backslash lambda\_2,\backslash \; \backslash dots,\; \backslash lambda\_M$ such that

$$\backslash nabla\; f(\backslash mathbf\{x\})\; =\; \backslash sum\_\{k=1\}^M\; \backslash lambda\_k\; \backslash ,\; \backslash nabla\; g\_k\; (\backslash mathbf\{x\})\; \backslash quad\; \backslash iff\; \backslash quad\; \backslash nabla\; f(\backslash mathbf\{x\})\; -\; \backslash sum\_\{k=1\}^M\; \{\backslash lambda\_k\; \backslash nabla\; g\_k\; (\backslash mathbf\{x\})\}\; =\; 0\; ~.$$

These scalars are the Lagrange multipliers. We now have $M$ of them, one for every constraint.

As before, we introduce an auxiliary function

$$\backslash mathcal\{L\}\backslash left(\; x\_1,\backslash ldots\; ,\; x\_n,\; \backslash lambda\_1,\; \backslash ldots,\; \backslash lambda\; \_M\; \backslash right)\; =\; f\backslash left(\; x\_1,\; \backslash ldots,\; x\_n\; \backslash right)\; -\; \backslash sum\backslash limits\_\{k=1\}^M\; \{\backslash lambda\_k\; g\_k\backslash left(\; x\_1,\; \backslash ldots\; ,\; x\_n\; \backslash right)\}\backslash$$

and solve

$$\backslash nabla\_\{x\_1,\; \backslash ldots\; ,\; x\_n,\; \backslash lambda\_1,\; \backslash ldots,\; \backslash lambda\; \_M\}\; \backslash mathcal\{L\}(x\_1,\; \backslash ldots\; ,\; x\_n,\; \backslash lambda\_1,\; \backslash ldots,\; \backslash lambda\; \_M)=0\; \backslash iff$$

\begin{cases}

\nabla f(\mathbf{x}) - \sum_{k=1}^M {\lambda_k \, \nabla g_k (\mathbf{x})} = 0\\

g_1(\mathbf{x}) = \cdots = g_M(\mathbf{x}) = 0

\end{cases}

which amounts to solving $n+M$ equations in $\backslash \; n+M\backslash$ unknowns.

The constraint qualification assumption when there are multiple constraints is that the constraint gradients at the relevant point are linearly independent.

The problem of finding the local maxima and minima subject to constraints can be generalized to finding local maxima and minima on a differentiable manifold $\backslash \; M\; ~.${{cite book |last=Lafontaine |first=Jacques |year=2015 |title=An Introduction to Differential Manifolds |publisher=Springer |isbn=978-3-319-20735-3 |page=70 |url=https://books.google.com/books?id=KNhJCgAAQBAJ&pg=PA70}} In what follows, it is not necessary that $M$ be a Euclidean space, or even a Riemannian manifold. All appearances of the gradient $\backslash \; \backslash nabla\backslash$ (which depends on a choice of Riemannian metric) can be replaced with the exterior derivative $\backslash \; \backslash operatorname\{d\}\; ~.$

Let $\backslash \; M\backslash$ be a smooth manifold of dimension $\backslash \; m\; ~.$ Suppose that we wish to find the stationary points $\backslash \; x\backslash$ of a smooth function $\backslash \; f:M\; \backslash to\; \backslash mathbb\{R\}\backslash$ when restricted to the submanifold $\backslash \; N\backslash$ defined by $\backslash \; g(x)\; =\; 0\backslash \; ,$ where $\backslash \; g:M\; \backslash to\; \backslash mathbb\{R\}\backslash$ is a smooth function for which {{math|0}} is a regular value.

Let $\backslash \; \backslash operatorname\{d\}f\backslash$ and $\backslash \; \backslash operatorname\{d\}g\backslash$ be the exterior derivatives of $\backslash \; f\backslash$ and $\backslash \; g\backslash$. Stationarity for the restriction $\backslash \; f|\_\{N\}\backslash$ at $\backslash \; x\backslash in\; N\backslash$ means $\backslash \; \backslash operatorname\{d\}(f|\_N)\_x=0\; ~.$ Equivalently, the kernel $\backslash \; \backslash ker(\backslash operatorname\{d\}f\_x)\backslash$ contains $\backslash \; T\_x\; N\; =\; \backslash ker(\backslash operatorname\{d\}g\_x)\; ~.$ In other words, $\backslash \; \backslash operatorname\{d\}f\_x\backslash$ and $\backslash \; \backslash operatorname\{d\}g\_x\backslash$ are proportional 1-forms. For this it is necessary and sufficient that the following system of $\backslash \; \backslash tfrac\{1\}\{2\}m(m-1)\backslash$ equations holds:

$$\backslash operatorname\{d\}f\_x\; \backslash wedge\; \backslash operatorname\{d\}g\_x\; =\; 0\; \backslash in\; \backslash Lambda^2(T^\{\backslash ast\}\_x\; M)$$

where $\backslash \; \backslash wedge\backslash$ denotes the exterior product. The stationary points $\backslash \; x\backslash$ are the solutions of the above system of equations plus the constraint $\backslash \; g(x)\; =\; 0\; ~.$ Note that the $\backslash \; \backslash tfrac\{1\}\{2\}\; m(m-1)\backslash$ equations are not independent, since the left-hand side of the equation belongs to the subvariety of $\backslash \; \backslash Lambda^\{2\}(T^\{\backslash ast\}\_x\; M)\backslash$ consisting of decomposable elements.

In this formulation, it is not necessary to explicitly find the Lagrange multiplier, a number $\backslash \; \backslash lambda\backslash$ such that $\backslash \; \backslash operatorname\{d\}f\_x\; =\; \backslash lambda\; \backslash cdot\; \backslash operatorname\{d\}g\_x\; ~.$

Let $\backslash \; M\backslash$ and $\backslash \; f\backslash$ be as in the above section regarding the case of a single constraint. Rather than the function $g$ described there, now consider a smooth function $\backslash \; G:M\backslash to\; \backslash R^p\; (p>1)\backslash \; ,$ with component functions $\backslash \; g\_i:\; M\; \backslash to\; \backslash R\backslash \; ,$ for which $0\backslash in\backslash R^p$ is a regular value. Let $N$ be the submanifold of $\backslash \; M\backslash$ defined by $\backslash \; G(x)=0\; ~.$

$\backslash \; x\backslash$ is a stationary point of $f|\_\{N\}$ if and only if $\backslash \; \backslash ker(\; \backslash operatorname\{d\}f\_x\; )\backslash$ contains $\backslash \; \backslash ker(\; \backslash operatorname\{d\}G\_x\; )\; ~.$ For convenience let $\backslash \; L\_x\; =\; \backslash operatorname\{d\}f\_x\backslash$ and $\backslash \; K\_x\; =\; \backslash operatorname\{d\}G\_x\backslash \; ,$ where $\backslash \; \backslash operatorname\{d\}G$ denotes the tangent map or Jacobian $\backslash \; TM\; \backslash to\; T\backslash R^p\; ~$ ($\backslash \; T\_x\backslash R^p$ can be canonically identified with $\backslash \; \backslash R^p$). The subspace $\backslash ker(K\_x)$ has dimension smaller than that of $\backslash ker(L\_x)$, namely $\backslash \; \backslash dim(\backslash ker(L\_x))\; =\; n-1\backslash$ and $\backslash \; \backslash dim(\backslash ker(K\_x))\; =\; n-p\; ~.$ $\backslash ker(K\_x)$ belongs to $\backslash \; \backslash ker(L\_x)\backslash$ if and only if $L\_x\; \backslash in\; T^\{\backslash ast\}\_x\; M$ belongs to the image of $\backslash \; K^\{\backslash ast\}\_x:\; \backslash R^\{p\backslash ast\}\backslash to\; T^\{\backslash ast\}\_x\; M\; ~.$ Computationally speaking, the condition is that $L\_x$ belongs to the row space of the matrix of $\backslash \; K\_x\backslash \; ,$ or equivalently the column space of the matrix of $K^\{\backslash ast\}\_x$ (the transpose). If $\backslash \; \backslash omega\_x\; \backslash in\; \backslash Lambda^\{p\}(T^\{\backslash ast\}\_x\; M)\backslash$ denotes the exterior product of the columns of the matrix of $\backslash \; K^\{\backslash ast\}\_x\backslash \; ,$ the stationary condition for $\backslash \; f|\_\{N\}\backslash$ at $\backslash \; x\backslash$ becomes

$$L\_x\; \backslash wedge\; \backslash omega\_x\; =\; 0\; \backslash in\; \backslash Lambda^\{p+1\}\; \backslash left\; (T^\{\backslash ast\}\_x\; M\; \backslash right\; )$$

Once again, in this formulation it is not necessary to explicitly find the Lagrange multipliers, the numbers $\backslash \; \backslash lambda\_1,\; \backslash ldots,\; \backslash lambda\_p\backslash$ such that

$$\backslash \; \backslash operatorname\{d\}f\_x\; =\; \backslash sum\_\{i=1\}^p\; \backslash lambda\_i\; \backslash operatorname\{d\}(g\_i)\_x\; ~.$$

In this section, we modify the constraint equations from the form $g\_i(\{\backslash bf\; x\})\; =\; 0$ to the form $\backslash \; g\_i(\{\backslash bf\; x\})\; =\; c\_i\backslash \; ,$ where the $\backslash \; c\_i\backslash$ are {{mvar|m}} real constants that are considered to be additional arguments of the Lagrangian expression $\backslash mathcal\{L\}$.

Often the Lagrange multipliers have an interpretation as some quantity of interest. For example, by parametrising the constraint's contour line, that is, if the Lagrangian expression is

\begin{align}

& \mathcal{L}(x_1, x_2, \ldots;\lambda_1, \lambda_2, \ldots; c_1, c_2, \ldots) \\[4pt]

= {} & f(x_1, x_2, \ldots) + \lambda_1(c_1-g_1(x_1, x_2, \ldots))+\lambda_2(c_2-g_2(x_1, x_2, \dots))+\cdots

\end{align}

then

$$\backslash \; \backslash frac\{\backslash partial\; \backslash mathcal\{L\}\}\{\backslash partial\; c\_k\}\; =\; \backslash lambda\_k\; ~.$$

So, {{math|λ_k}} is the rate of change of the quantity being optimized as a function of the constraint parameter.

As examples, in Lagrangian mechanics the equations of motion are derived by finding stationary points of the action, the time integral of the difference between kinetic and potential energy. Thus, the force on a particle due to a scalar potential, {{math|F {{=}} −∇V}}, can be interpreted as a Lagrange multiplier determining the change in action (transfer of potential to kinetic energy) following a variation in the particle's constrained trajectory.

In control theory this is formulated instead as costate equations.

Moreover, by the envelope theorem the optimal value of a Lagrange multiplier has an interpretation as the marginal effect of the corresponding constraint constant upon the optimal attainable value of the original objective function: If we denote values at the optimum with a star ($\backslash star$), then it can be shown that

$$\backslash frac\{\backslash \; \backslash operatorname\{d\}\; f\backslash left(\backslash \; x\_\{1\; \backslash star\; \}(c\_1,\; c\_2,\; \backslash dots),\backslash \; x\_\{2\; \backslash star\; \}(c\_1,\; c\_2,\; \backslash dots),\backslash \; \backslash dots\backslash \; \backslash right)\backslash \; \}\{\; \backslash operatorname\{d\}\; c\_k\; \}\; =\; \backslash lambda\_\{\backslash star\; k\}\; ~.$$

For example, in economics the optimal profit to a player is calculated subject to a constrained space of actions, where a Lagrange multiplier is the change in the optimal value of the objective function (profit) due to the relaxation of a given constraint (e.g. through a change in income); in such a context $\backslash \; \backslash lambda\_\{\backslash star\; k\}\backslash$ is the marginal cost of the constraint, and is referred to as the shadow price.{{cite book |first=Avinash K. |last=Dixit |author-link=Avinash Dixit |chapter=Shadow Prices |title=Optimization in Economic Theory |location=New York |publisher=Oxford University Press |edition=2nd |year=1990 |pages=40–54 |isbn=0-19-877210-6 |chapter-url=https://books.google.com/books?id=dHrsHz0VocUC&pg=PA40 }}

{{Main|Bordered Hessian}}

Sufficient conditions for a constrained local maximum or minimum can be stated in terms of a sequence of principal minors (determinants of upper-left-justified sub-matrices) of the bordered Hessian matrix of second derivatives of the Lagrangian expression.{{cite book|author-link=Alpha Chiang |last=Chiang |first= Alpha C. |title=Fundamental Methods of Mathematical Economics |url=https://archive.org/details/fundamentalmetho0000chia_h4v2 |url-access=registration |publisher=McGraw-Hill |edition=Third |date=1984 |page=[https://archive.org/details/fundamentalmetho0000chia_h4v2/page/386 386] |isbn=0-07-010813-7 }}

Image:Lagrange very simple.svg

Suppose we wish to maximize $\backslash \; f(x,y)\; =\; x+y\backslash$ subject to the constraint $\backslash \; x^2\; +\; y^2\; =\; 1\; ~.$ The feasible set is the unit circle, and the level sets of {{mvar|f}} are diagonal lines (with slope −1), so we can see graphically that the maximum occurs at $\backslash \; \backslash left(\backslash tfrac\{1\}\{\backslash sqrt\{2\}\},\; \backslash tfrac\{1\}\{\backslash sqrt\{2\}\}\backslash right)\backslash \; ,$ and that the minimum occurs at $\backslash \; \backslash left(-\backslash tfrac\{1\}\{\backslash sqrt\{2\}\},\; -\backslash tfrac\{1\}\{\backslash sqrt\{2\}\}\backslash right)\; ~.$

For the method of Lagrange multipliers, the constraint is

$$g(x,y)\; =\; x^2\; +\; y^2-1\; =\; 0\backslash \; ,$$

hence the Lagrangian function,

$$\backslash begin\{align\}$$

\mathcal{L}(x, y, \lambda) &= f(x,y) + \lambda \cdot g(x,y) \\[4pt]

&= x + y + \lambda (x^2 + y^2 - 1)\ ,

\end{align}

is a function that is equivalent to $\backslash \; f(x,y)\backslash$ when $\backslash \; g(x,y)\backslash$ is set to {{math|0}}.

Now we can calculate the gradient:

$$\backslash begin\{align\}$$

\nabla_{x,y,\lambda} \mathcal{L}(x , y, \lambda) &= \left( \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial \lambda} \right ) \\[4pt]

&= \left ( 1 + 2 \lambda x, 1 + 2 \lambda y, x^2 + y^2 -1 \right)

\ \color{gray}{,}

\end{align}

and therefore:

$$\backslash nabla\_\{x,y,\backslash lambda\}\; \backslash mathcal\{L\}(x\; ,\; y,\; \backslash lambda)=0\; \backslash quad\; \backslash Leftrightarrow\; \backslash quad\; \backslash begin\{cases\}\; 1\; +\; 2\; \backslash lambda\; x\; =\; 0\; \backslash \backslash \; 1\; +\; 2\; \backslash lambda\; y\; =\; 0\; \backslash \backslash \; x^2\; +\; y^2\; -1\; =\; 0\; \backslash end\{cases\}$$

Notice that the last equation is the original constraint.

The first two equations yield

$$x\; =\; y\; =\; -\; \backslash frac\{1\}\{2\backslash lambda\},\; \backslash qquad\; \backslash lambda\; \backslash neq\; 0\; ~.$$

By substituting into the last equation we have:

$$\backslash frac\{1\}\{4\backslash lambda^2\}\; +\; \backslash frac\{1\}\{4\backslash lambda^2\}\; -\; 1\; =\; 0\backslash \; ,$$

so

$$\backslash lambda\; =\; \backslash pm\; \backslash frac\{1\}\{\backslash sqrt\{2\backslash \; \}\}\backslash \; ,$$

which implies that the stationary points of $\backslash mathcal\{L\}$ are

$$\backslash left(\backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\},\; \backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\},\; -\backslash tfrac\{1\}\{\backslash sqrt\{2\backslash \; \}\}\backslash right),\; \backslash qquad\; \backslash left(-\backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\},\; -\backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\},\; \backslash tfrac\{1\}\{\backslash sqrt\{2\backslash \; \}\}\; \backslash right)\; ~.$$

Evaluating the objective function {{mvar|f}} at these points yields

$$f\backslash left(\backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\},\; \backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\}\backslash right)\; =\; \backslash sqrt\{2\backslash \; \}\backslash \; ,\; \backslash qquad\; f\backslash left(-\backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\},\; -\backslash tfrac\{\backslash sqrt\{2\backslash \; \}\}\{2\}\; \backslash right)\; =\; -\backslash sqrt\{2\backslash \; \}\; ~.$$

Thus the constrained maximum is $\backslash \; \backslash sqrt\{2\backslash \; \}\backslash$ and the constrained minimum is $-\backslash sqrt\{2\}$.

Image:Lagrange very simple-1b.svg

Now we modify the objective function of Example 1 so that we minimize $\backslash \; f(x,y)\; =\; (x\; +\; y)^2\backslash$ instead of $\backslash \; f(x,y)\; =\; x\; +\; y\backslash \; ,$ again along the circle $\backslash \; g(x,y)=x^2+y^2-1\; =\; 0\; ~.$ Now the level sets of $f$ are still lines of slope −1, and the points on the circle tangent to these level sets are again $\backslash \; (\backslash sqrt\{2\}/2,\backslash sqrt\{2\}/2)\backslash$ and $\backslash \; (-\backslash sqrt\{2\}/2,-\backslash sqrt\{2\}/2)\; ~.$ These tangency points are maxima of $\backslash \; f\; ~.$

On the other hand, the minima occur on the level set for $\backslash \; f\; =\; 0\backslash$ (since by its construction $\backslash \; f\backslash$ cannot take negative values), at $\backslash \; (\backslash sqrt\{2\}/2,-\backslash sqrt\{2\}/2)\backslash$ and $\backslash \; (-\backslash sqrt\{2\}/2,\; \backslash sqrt\{2\}/2)\backslash \; ,$ where the level curves of $\backslash \; f\backslash$ are not tangent to the constraint. The condition that $\backslash \; \backslash nabla\_\{x,y,\backslash lambda\}\backslash left(f(x,y)+\backslash lambda\backslash cdot\; g(x,y)\; \backslash right)=0\backslash$ correctly identifies all four points as extrema; the minima are characterized in by $\backslash \; \backslash lambda\; =0\backslash$ and the maxima by $\backslash \; \backslash lambda\; =\; -2\; ~.$

Image:Lagrange simple.svg

This example deals with more strenuous calculations, but it is still a single constraint problem.

Suppose one wants to find the maximum values of

$$f(x,\; y)\; =\; x^2\; y$$

with the condition that the $\backslash \; x\backslash$- and $\backslash \; y\backslash$-coordinates lie on the circle around the origin with radius $\backslash \; \backslash sqrt\{3\backslash \; \}\; ~.$ That is, subject to the constraint

$$g(x,y)\; =\; x^2\; +\; y^2\; -\; 3\; =\; 0\; ~.$$

As there is just a single constraint, there is a single multiplier, say $\backslash \; \backslash lambda\; ~.$

The constraint $\backslash \; g(x,y)\backslash$ is identically zero on the circle of radius $\backslash \; \backslash sqrt\{3\backslash \; \}\; ~.$ Any multiple of $\backslash \; g(x,y)\backslash$ may be added to $\backslash \; g(x,y)\backslash$ leaving $\backslash \; g(x,y)\backslash$ unchanged in the region of interest (on the circle where our original constraint is satisfied).

Applying the ordinary Lagrange multiplier method yields

$$\backslash begin\{align\}$$

\mathcal{L}(x, y, \lambda) &= f(x,y) + \lambda \cdot g(x, y) \\

&= x^2y + \lambda (x^2 + y^2 - 3)\ ,

\end{align}

from which the gradient can be calculated:

$$\backslash begin\{align\}$$

\nabla_{x,y,\lambda} \mathcal{L}(x , y, \lambda) &= \left ( \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial \lambda} \right) \\

&= \left ( 2 x y + 2 \lambda x, x^2 + 2 \lambda y, x^2 + y^2 -3 \right) ~.

\end{align}

And therefore:

$$\backslash nabla\_\{x,y,\backslash lambda\}\; \backslash mathcal\{L\}(x\; ,\; y,\; \backslash lambda)=0\; \backslash quad\; \backslash iff\; \backslash quad\; \backslash begin\{cases\}$$

2 x y + 2 \lambda x = 0 \\

x^2 + 2 \lambda y = 0 \\

x^2 + y^2 - 3 = 0

\end{cases} \quad \iff \quad \begin{cases}

x (y + \lambda) = 0 & \text{(i)} \\

x^2 = -2 \lambda y & \text{(ii)} \\

x^2 + y^2 = 3 & \text{(iii)}

\end{cases}

(iii) is just the original constraint. (i) implies $\backslash \; x=0\backslash$ or $\backslash \; \backslash lambda=-y\; ~.$ If $x=0$ then $\backslash \; y\; =\; \backslash pm\; \backslash sqrt\{3\backslash \; \}\backslash$ by (iii) and consequently $\backslash \; \backslash lambda=0\backslash$ from (ii). If $\backslash \; \backslash lambda=-y\backslash \; ,$ substituting this into (ii) yields $\backslash \; x^2\; =\; 2y^2\; ~.$ Substituting this into (iii) and solving for $\backslash \; y\backslash$ gives $\backslash \; y=\backslash pm1\; ~.$ Thus there are six critical points of $\backslash \; \backslash mathcal\{L\}\backslash \; :$

$$(\backslash sqrt\{2\backslash \; \},1,-1);\; \backslash quad\; (-\backslash sqrt\{2\backslash \; \},1,-1);\; \backslash quad\; (\backslash sqrt\{2\backslash \; \},-1,1);\; \backslash quad\; (-\backslash sqrt\{2\backslash \; \},-1,1);\; \backslash quad\; (0,\backslash sqrt\{3\backslash \; \},\; 0);\; \backslash quad\; (0,-\backslash sqrt\{3\backslash \; \},\; 0)\; ~.$$

Evaluating the objective at these points, one finds that

$$f(\backslash pm\backslash sqrt\{2\backslash \; \},1)\; =\; 2;\; \backslash quad\; f(\backslash pm\backslash sqrt\{2\backslash \; \},-1)\; =\; -2;\; \backslash quad\; f(0,\backslash pm\; \backslash sqrt\{3\backslash \; \})\; =\; 0\; ~.$$

Therefore, the objective function attains the global maximum (subject to the constraints) at $\backslash \; (\backslash pm\backslash sqrt\{2\backslash \; \},1\backslash \; )$ and the global minimum at $\backslash \; (\backslash pm\backslash sqrt\{2\backslash \; \},-1)\; ~.$ The point $\backslash \; (0,\backslash sqrt\{3\backslash \; \})\backslash$ is a local minimum of $\backslash \; f\backslash$ and $\backslash \; (0,-\backslash sqrt\{3\backslash \; \})\backslash$ is a local maximum of $\backslash \; f\backslash \; ,$ as may be determined by consideration of the Hessian matrix of $\backslash \; \backslash mathcal\{L\}(x,y,0)\; ~.$

Note that while $\backslash \; (\backslash sqrt\{2\backslash \; \},\; 1,\; -1)\backslash$ is a critical point of $\backslash \; \backslash mathcal\{L\}\backslash \; ,$ it is not a local extremum of $\backslash \; \backslash mathcal\{L\}\; ~.$ We have

$$\backslash mathcal\{L\}\; \backslash left\; (\backslash sqrt\{2\backslash \; \}\; +\; \backslash varepsilon,\; 1,\; -1\; +\; \backslash delta\; \backslash right\; )\; =\; 2\; +\; \backslash delta\; \backslash left(\; \backslash varepsilon^2\; +\; \backslash left\; (2\backslash sqrt\{2\backslash \; \}\; \backslash right)\backslash varepsilon\; \backslash right)\; ~.$$

Given any neighbourhood of $\backslash \; (\backslash sqrt\{2\backslash \; \},\; 1,\; -1)\backslash \; ,$ one can choose a small positive $\backslash \; \backslash varepsilon\backslash$ and a small $\backslash \; \backslash delta\backslash$ of either sign to get $\backslash \; \backslash mathcal\{L\}$ values both greater and less than $\backslash \; 2\; ~.$ This can also be seen from the Hessian matrix of $\backslash \; \backslash mathcal\{L\}\backslash$ evaluated at this point (or indeed at any of the critical points) which is an indefinite matrix. Each of the critical points of $\backslash \; \backslash mathcal\{L\}\backslash$ is a saddle point of $\backslash \; \backslash mathcal\{L\}\; ~.$

Suppose we wish to find the discrete probability distribution on the points $\backslash \; \backslash \{p\_1,\; p\_2,\; \backslash ldots,\; p\_n\backslash \}\backslash$ with maximal information entropy. This is the same as saying that we wish to find the least structured probability distribution on the points $\backslash \; \backslash \{p\_1,\; p\_2,\; \backslash cdots,\; p\_n\backslash \}\; ~.$ In other words, we wish to maximize the Shannon entropy equation:

$$f(p\_1,p\_2,\backslash ldots,p\_n)\; =\; -\backslash sum\_\{j=1\}^n\; p\_j\; \backslash log\_2\; p\_j\; ~.$$

For this to be a probability distribution the sum of the probabilities $\backslash \; p\_i\backslash$ at each point $\backslash \; x\_i\backslash$ must equal 1, so our constraint is:

$$g(p\_1,p\_2,\backslash ldots,p\_n)=\backslash sum\_\{j=1\}^n\; p\_j\; =\; 1\; ~.$$

We use Lagrange multipliers to find the point of maximum entropy, $\backslash \; \backslash vec\{p\}^\{\backslash ,*\}\backslash \; ,$ across all discrete probability distributions $\backslash \; \backslash vec\{p\}\backslash$ on $\backslash \; \backslash \{x\_1,x\_2,\; \backslash ldots,\; x\_n\backslash \}\; ~.$ We require that:

$$\backslash left.\; \backslash frac\{\backslash partial\}\{\backslash partial\; \backslash vec\{p\}\}(f+\backslash lambda\; (g-1))\; \backslash right|\_\{\backslash vec\{p\}=\backslash vec\{p\}^\{\backslash ,*\}\}\; =\; 0\backslash \; ,$$

which gives a system of {{mvar|n}} equations, $\backslash \; k\; =\; 1,\backslash \; \backslash ldots,\; n\backslash \; ,$ such that:

$$\backslash left.\; \backslash frac\{\backslash partial\}\{\backslash partial\; p\_k\}\backslash left\backslash \{\; -\backslash left\; (\backslash sum\_\{j=1\}^n\; p\_j\; \backslash log\_2\; p\_j\; \backslash right\; )\; +\; \backslash lambda\; \backslash left(\backslash sum\_\{j=1\}^n\; p\_j\; -\; 1\; \backslash right)\; \backslash right\backslash \}\; \backslash right|\_\{\; p\_k\; =\; p\_\{\backslash star\; k\}\; \}\; =\; 0\; ~.$$

Carrying out the differentiation of these {{mvar|n}} equations, we get

$$-\backslash left(\backslash frac\{1\}\{\backslash ln\; 2\}+\backslash log\_2\; p\_\{\backslash star\; k\}\; \backslash right)\; +\; \backslash lambda\; =\; 0\; ~.$$

This shows that all $\backslash \; p\_\{\backslash star\; k\}\backslash$ are equal (because they depend on {{mvar|λ}} only). By using the constraint

$$\backslash sum\_j\; p\_j\; =1\backslash \; ,$$

we find

$$p\_\{\backslash star\; k\}\; =\; \backslash frac\{1\}\{n\}\; ~.$$

Hence, the uniform distribution is the distribution with the greatest entropy, among distributions on {{mvar|n}} points.

Image:lagnum1.png

Image:lagnum2.png

The critical points of Lagrangians occur at saddle points, rather than at local maxima (or minima).{{cite book |first=Michael T. |last= Heath |author-link=Michael Heath (computer scientist) |year=2005 |title=Scientific Computing: An introductory survey |page=203 |publisher=McGraw-Hill |isbn=978-0-07-124489-3 |url=https://books.google.com/books?id=gwBrMAEACAAJ}} Unfortunately, many numerical optimization techniques, such as hill climbing, gradient descent, some of the quasi-Newton methods, among others, are designed to find local maxima (or minima) and not saddle points. For this reason, one must either modify the formulation to ensure that it's a minimization problem (for example, by extremizing the square of the gradient of the Lagrangian as below), or else use an optimization technique that finds stationary points (such as Newton's method without an extremum seeking line search) and not necessarily extrema.

As a simple example, consider the problem of finding the value of {{mvar|x}} that minimizes $\backslash \; f(x)\; =\; x^2\backslash \; ,$ constrained such that $\backslash \; x^2\; =\; 1\; ~.$ (This problem is somewhat untypical because there are only two values that satisfy this constraint, but it is useful for illustration purposes because the corresponding unconstrained function can be visualized in three dimensions.)

Using Lagrange multipliers, this problem can be converted into an unconstrained optimization problem:

$$\backslash mathcal\{L\}(\; x,\; \backslash lambda\; )\; =\; x^2\; +\; \backslash lambda(x^2-1)\; ~.$$

The two critical points occur at saddle points where {{math|x {{=}} 1}} and {{math|x {{=}} −1}}.

In order to solve this problem with a numerical optimization technique, we must first transform this problem such that the critical points occur at local minima. This is done by computing the magnitude of the gradient of the unconstrained optimization problem.

First, we compute the partial derivative of the unconstrained problem with respect to each variable:

$$\backslash begin\{align\}$$

& \frac{\partial \mathcal{L} }{ \partial x } = 2x + 2x \lambda \\[5pt]

& \frac{\partial \mathcal{L} }{ \partial \lambda }=x^2-1 ~.

\end{align}

If the target function is not easily differentiable, the differential with respect to each variable can be approximated as

$$\backslash begin\{align\}$$

\frac{\ \partial \mathcal{L}\ }{ \partial x } \approx \frac{\mathcal{L}(x + \varepsilon,\lambda) - \mathcal{L}(x,\lambda)}{\varepsilon}, \\[5pt]

\frac{\ \partial \mathcal{L}\ }{ \partial \lambda } \approx \frac{\mathcal{L}(x, \lambda + \varepsilon) - \mathcal{L}(x,\lambda)}{\varepsilon},

\end{align}

where $\backslash varepsilon$

is a small value.

Next, we compute the magnitude of the gradient, which is the square root of the sum of the squares of the partial derivatives:

$$\backslash begin\{align\}$$

h(x,\lambda) & = \sqrt{ (2x+2x\lambda)^2 + (x^2-1)^2\ } \\[4pt]

& \approx \sqrt{ \left(\frac{\ \mathcal{L}(x+\varepsilon,\lambda)-\mathcal{L}(x,\lambda)\ }{\varepsilon}\right)^2 + \left(\frac{\ \mathcal{L}(x,\lambda+\varepsilon) - \mathcal{L}(x,\lambda)\ }{\varepsilon}\right)^2\ }~ .

\end{align}

(Since magnitude is always non-negative, optimizing over the squared-magnitude is equivalent to optimizing over the magnitude. Thus, the "square root" may be omitted from these equations with no expected difference in the results of optimization.)

The critical points of {{mvar|h}} occur at {{math|1=x = 1}} and {{math|1=x = −1}}, just as in $\backslash mathcal\{L\}\; ~.$ Unlike the critical points in $\backslash mathcal\{L\}\backslash ,\; ,$ however, the critical points in {{mvar|h}} occur at local minima, so numerical optimization techniques can be used to find them.

In optimal control theory, the Lagrange multipliers are interpreted as costate variables, and Lagrange multipliers are reformulated as the minimization of the Hamiltonian, in Pontryagin's maximum principle.

The Lagrange multiplier method has several generalizations. In nonlinear programming there are several multiplier rules, e.g. the Carathéodory–John Multiplier Rule and the Convex Multiplier Rule, for inequality constraints.{{Cite journal |last=Pourciau |first=Bruce H. |date=1980 |title=Modern multiplier rules |journal=American Mathematical Monthly |volume=87 |issue=6 |pages=433–452 |doi=10.2307/2320250 |jstor=2320250 |url=http://www.maa.org/programs/maa-awards/writing-awards/modern-multiplier-rules|url-access=subscription }}

In many models in mathematical economics such as general equilibrium models, consumer behavior is implemented as utility maximization and firm behavior as profit maximization, both entities being subject to constraints such as budget constraints and production constraints. The usual way to determine an optimal solution is achieved by maximizing some function, where the constraints are enforced using Lagrangian multipliers.{{cite book |first1=M. I. |last1=Kamien |author-link=Morton Kamien |first2=N. L. |last2=Schwartz |author-link2=Nancy Schwartz |title=Dynamic Optimization: The Calculus of Variations and Optimal Control in Economics and Management |location=New York |publisher=Elsevier |edition=Second |year=1991 |isbn=0-444-01609-0 |url=https://books.google.com/books?id=0IoGUn8wjDQC }}{{cite journal |last1=Glötzl |first1=Erhard |last2=Glötzl |first2=Florentin |last3=Richters |first3=Oliver |title=From constrained optimization to constrained dynamics: extending analogies between economics and mechanics |journal=Journal of Economic Interaction and Coordination |volume=14 |pages=623–642 |date=2019 |issue=3 |doi=10.1007/s11403-019-00252-7|hdl=10419/171974 |hdl-access=free }}{{cite journal |first1=John V. |last1=Baxley |first2=John C. |last2=Moorhouse |title=Lagrange Multiplier Problems in Economics |journal=The American Mathematical Monthly |volume=91 |issue = 7 |date=1984 |pages=404–412|doi=10.1080/00029890.1984.11971446 }}.{{cite journal |first1=Jitka |last1=Janová |date=2011 |title=Applications of a constrained mechanics methodology in economics |journal=European Journal of Physics |volume=32 |issue=6 |pages=1443–1463 |doi=10.1088/0143-0807/32/6/001|arxiv=1106.3455 |bibcode=2011EJPh...32.1443J }}

Methods based on Lagrange multipliers have applications in power systems, e.g. in distributed-energy-resources (DER) placement and load shedding.

{{cite conference

| last1 = Gautam | first1 = Mukesh

| last2 = Bhusal | first2 = Narayan

| last3 = Benidris | first3 = Mohammed

| year = 2020

| title = A sensitivity-based approach to adaptive under-frequency load shedding

| conference = 2020 IEEE Texas Power and Energy Conference (TPEC)

| publisher = Institute of Electronic and Electrical Engineers

| pages=1–5

| doi=10.1109/TPEC48276.2020.9042569

}}

The method of Lagrange multipliers applies to constrained Markov decision processes.

{{cite book

| last1 = Altman | first1 = Eitan

| year = 2021

| title = Constrained Markov Decision Processes

| publisher = Routledge

}}

It naturally produces gradient-based primal-dual algorithms in safe reinforcement learning.

{{cite conference

| last1 = Ding | first1 =Dongsheng

| last2 = Zhang | first2 = Kaiqing

| last3 = Jovanovic | first3 = Mihailo

| last4 = Basar | first4 = Tamer

| year = 2020

| title = Natural policy gradient primal-dual method for constrained Markov decision processes

| conference = Advances in Neural Information Processing Systems

}}

Considering the PDE problems with constraints, i.e., the study of the properties of the normalized solutions, Lagrange multipliers play an important role.

{{div col}}

• Adjustment of observations
• Duality
• Gittins index
• Karush–Kuhn–Tucker conditions: generalization of the method of Lagrange multipliers
• Lagrange multipliers on Banach spaces: another generalization of the method of Lagrange multipliers
• Lagrange multiplier test in maximum likelihood estimation
• Lagrangian relaxation

{{colend}}

{{Reflist|25em}}

{{refbegin|25em|small=y}}

• {{cite book

|first1=Brian |last1=Beavis

|first2=Ian M. |last2=Dobbs

|year=1990

|chapter=Static Optimization

|title=Optimization and Stability Theory for Economic Analysis

|place=New York, NY

|publisher=Cambridge University Press

|isbn=0-521-33605-8

|pages=32–72

|chapter-url=https://books.google.com/books?id=L7HMACFgnXMC&pg=PA32

}}

• {{cite book

|first=Dimitri P. |last=Bertsekas |author-link=Dimitri Bertsekas

|year=1982

|title=Constrained optimization and Lagrange multiplier methods

|place=New York, NY

|publisher=Academic Press

|isbn=0-12-093480-9

}}

• {{cite book

|first1=Gordon S.G. |last1=Beveridge

|first2=Robert S. |last2=Schechter

|year=1970

|chapter=Lagrangian multipliers

|title=Optimization: Theory and Practice

|place=New York, NY

|publisher=McGraw-Hill

|pages=244–259

|isbn=0-07-005128-3

|chapter-url=https://books.google.com/books?id=TfhVXlWtOPQC&pg=PA244

}}

• {{cite book

|first1=Brian R. |last1=Binger

|first2=Elizabeth |last2=Hoffman

|year=1998

|chapter=Constrained optimization

|title=Microeconomics with Calculus |edition=2nd

|location=Reading

|publisher=Addison-Wesley

|pages=56–91

|isbn=0-321-01225-9

}}

• {{cite book

|first=Michael |last=Carter

|year=2001

|chapter=Equality constraints

|title=Foundations of Mathematical Economics

|location=Cambridge, MA

|publisher=MIT Press

|pages=516–549

|isbn=0-262-53192-5

|chapter-url=https://books.google.com/books?id=KysvrGGfzq0C&pg=PA516

}}

• {{cite book

|first=Magnus R. |last=Hestenes |author-link=Magnus Hestenes

|year=1966

|chapter=Minima of functions subject to equality constraints

|title=Calculus of Variations and Optimal Control Theory

|place=New York, NY

|publisher=Wiley

|pages=29–34

}}

• {{cite book

|first1=C. Ray |last1=Wylie

|first2=Louis C. |last2=Barrett

|year=1995

|chapter=The extrema of integrals under constraint

|title=Advanced Engineering Mathematics |edition=Sixth

|place=New York, NY

|publisher=McGraw-Hill

|pages=1096–1103

|isbn=0-07-072206-4

}}

{{refend}}

{{wikibooks|Calculus optimization methods|Lagrange multipliers}}

• {{cite web

|author=Steuard

|title=Conceptual introduction

|website=slimy.com

|url=http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html

}} — plus a brief discussion of Lagrange multipliers in the calculus of variations as used in physics.

• {{cite web

|author=Carpenter, Kenneth H.

|title=Lagrange multipliers for quadratic forms with linear constraints

|publisher=Kansas State University

|url=http://ece.k-state.edu/people/faculty/carpenter/documents/lagrange.pdf

}}

• {{cite web

|author=Resnik

|title=Simple explanation with an example of governments using taxes as Lagrange multipliers

|website=umiacs.umd.edu

|publisher=University of Maryland

|url=http://www.umiacs.umd.edu/~resnik/ling848_fa2004/lagrange.html

}}

• {{cite web

|author=Klein, Dan

|title=Lagrange multipliers without permanent scarring] Explanation with focus on the intuition

|website=nlp.cs.berkeley.edu

|publisher=University of California, Berkeley

|url=http://nlp.cs.berkeley.edu/tutorials/lagrange-multipliers.pdf

}}

• {{cite web

|author=Sathyanarayana, Shashi

|title=Geometric representation of method of Lagrange multipliers

|type=Mathematica demonstration

|website=wolfram.com

|publisher=Wolfram Research

|url=http://demonstrations.wolfram.com/GeometricRepresentationOfMethodOfLagrangeMultipliers

|quote=Needs Internet Explorer / Firefox / Safari.

}} — Provides compelling insight in 2 dimensions that at a minimizing point, the direction of steepest descent must be perpendicular to the tangent of the constraint curve at that point.

• {{cite web

|title=Lagrange multipliers – two variables

|type=Applet

|website=MIT Open Courseware (ocw.mit.edu)

|publisher=Massachusetts Institute of Technology

|url=http://ocw.mit.edu/ans7870/18/18.02/f07/tools/LagrangeMultipliersTwoVariables.html

}}

• {{cite web

|title=Lagrange multipliers

|date=Fall 2007

|type=video lecture

|series=Mathematics 18-02: Multivariable calculus

|volume=Lecture 13

|website=MIT Open Courseware (ocw.mit.edu)

|publisher=Massachusetts Institute of Technology

|url=http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-13-lagrange-multipliers/

}}

• {{cite web

|author=Bertsekas

|title=Details on Lagrange multipliers

|type=slides / course lecture

|series=Non-Linear Programming

|volume=Lectures 11 and 12

|website=athenasc.com

|url=http://www.athenasc.com/NLP_Slides.pdf

}} — Course slides accompanying text on nonlinear optimization

• {{cite web

|author=Wyatt, John

|date=7 April 2004 |orig-date=19 November 2002

|title=Legrange multipliers, constrained optimization, and the maximum entropy principle

|website=www-mtl.mit.edu

|series={{nobr|Elec E & C S / Mech E 6.050}} – Information, entropy, and computation

|volume=Unit 9

|url=http://www-mtl.mit.edu/Courses/6.050/2004/unit9/wyatt.apr.7.pdf

}} — Geometric idea behind Lagrange multipliers

• {{cite web

|title=Using Lagrange multipliers in optimization

|date=2011-12-24

|type=MATLAB example

|website=matlab.cheme.cmu.edu

|publisher=Carnegie Mellon University

|place=Pittsburgh, PA

|url=http://matlab.cheme.cmu.edu/2011/12/24/using-lagrange-multipliers-in-optimization/

}}

{{Calculus topics}}

{{Joseph-Louis Lagrange}}

{{authority control}}

Category:Multivariable calculus

Category:Mathematical optimization

Category:Mathematical and quantitative methods (economics)