Lagrange multiplier

"Lagrangian function" redirects here. Not to be confused withLagrangian (physics).

Inmathematical optimization,themethod of Lagrange multipliersis a strategy for finding the localmaxima and minimaof afunctionsubject toequation constraints(i.e., subject to the condition that one or moreequationshave to be satisfied exactly by the chosen values of thevariables).It is named after the mathematicianJoseph-Louis Lagrange.

Summary and rationale

The basic idea is to convert a constrained problem into a form such that thederivative testof 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 theLagrangian functionor Lagrangian.In the general case, the Lagrangian is defined as $${\mathcal {L}}(x,\lambda )\equiv f(x)+\langle \lambda,g(x)\rangle$$ for functions$f,g$;$\lambda$is called the Lagrange multiplier.

In simple cases, where theinner productis defined as thedot product,the Lagrangian is $${\mathcal {L}}(x,\lambda )\equiv f(x)+\lambda \cdot g(x)$$

The method can be summarized as follows: in order to find the maximum or minimum of a function$f$subjected to the equality constraint$g(x)=0$,find thestationary pointsof${\mathcal {L}}$considered as a function of$x$and the Lagrange multiplier$\lambda ~$.This means that allpartial derivativesshould be zero, including the partial derivative with respect to$\lambda ~$.

${\frac {\partial {\mathcal {L}}}{\partial x}}=0$and${\frac {\ \partial {\mathcal {L}}\ }{\partial \lambda }}=0\;$

or equivalently

${\frac {\partial f(x)}{\partial x}}+\lambda \cdot {\frac {\partial g(x)}{\partial x}}=0$and$g(x)=0~.$

The solution corresponding to the original constrained optimization is always asaddle pointof the Lagrangian function,which can be identified among the stationary points from thedefinitenessof thebordered Hessian matrix.

The great advantage of this method is that it allows the optimization to be solved without explicitparameterizationin 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 theKarush–Kuhn–Tucker conditions,which can also take into account inequality constraints of the form$h(\mathbf {x} )\leq c$for a given constant$c$.

Statement

The following is known as the Lagrange multiplier theorem.

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

$${\begin{aligned}&{\text{maximize }}f(x)\\&{\text{subject to: }}g(x)=0\end{aligned}}$$

Then there exists a unique Lagrange multiplier$\lambda _{\star }\in \mathbb {R} ^{c}$such that$\operatorname {D} f(x_{\star })=\lambda _{\star }^{\mathsf {T}}\operatorname {D} g(x_{\star })~.$(Note that this is a somewhat conventional thing where$\lambda _{\star }$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).

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 thegradientof the function (at that point) can be expressed as alinear combinationof the gradients of the constraints (at that point), with the Lagrange multipliers acting ascoefficients.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 thedirectional derivativeof the function is0in every feasible direction.

Single constraint

Figure 1: The red curve shows the constraintg(x,y) =c.The blue curves are contours off(x,y).The point where the red constraint tangentially touches a blue contour is the maximum off(x,y)along the constraint, sinced₁>d₂.

For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider theoptimization problem $${\begin{aligned}{\underset {x,y}{\text{maximize}}}\quad &f(x,y)\\{\text{subject to}}\quad &g(x,y)=0.\end{aligned}}$$ (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 firstpartial derivatives.We introduce a new variable ($\lambda$) called aLagrange multiplier(orLagrange undetermined multiplier) and study theLagrange function(orLagrangianorLagrangian expression) defined by $${\mathcal {L}}(x,y,\lambda )=f(x,y)+\lambda \cdot g(x,y),$$ where the$\lambda$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$\nabla g(x_{0},y_{0})\neq 0,$then there exists$\lambda _{0}$such that ($x_{0},y_{0},\lambda _{0}$) is astationary pointfor the Lagrange function (stationary points are those points where the first partial derivatives of${\mathcal {L}}$are zero). The assumption$\nabla g\neq 0$is called constraint qualification. However, not all stationary points yield a solution of the original problem, as the method of Lagrange multipliers yields only anecessary conditionfor optimality in constrained problems.Sufficient conditions for a minimum or maximumalso exist,but if a particularcandidate solutionsatisfies the sufficient conditions, it is only guaranteed that that solution is the best onelocally– that is, it is better than any permissible nearby points. Theglobaloptimum 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 hasg= 0.If it were, we could walk alongg= 0to 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 is0,that is, we are verifying that the directional derivative is 0 in any relevant (viable) direction.

We can visualizecontoursoffgiven byf(x,y) =dfor various values ofd,and the contour ofggiven byg(x,y) =c.

Suppose we walk along the contour line withg=c.We are interested in finding points wherefalmost 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 off,since by definitionfdoes not change as we walk along its contour lines. This would mean that the tangents to the contour lines offandgare parallel here.
2. We have reached a "level" part off,meaning thatfdoes not change in any direction.

To check the first possibility (we touch a contour line off), notice that since thegradientof a function is perpendicular to the contour lines, the tangents to the contour lines offandgare parallel if and only if the gradients offandgare parallel. Thus we want points(x,y)whereg(x,y) =cand $$\nabla _{x,y}f=\lambda \,\nabla _{x,y}g,$$ for some$\lambda$

where $$\nabla _{x,y}f=\left({\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right),\qquad \nabla _{x,y}g=\left({\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right)$$ are the respective gradients. The constant$\lambda$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$\lambda$is preceded by a minus sign).

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

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

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

Note that$\ \nabla _{\lambda }{\mathcal {L}}(x,y,\lambda )=0\$implies$\ g(x,y)=0\,$as the partial derivative of${\mathcal {L}}$with respect to$\lambda$is$\ 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 $$\nabla _{x_{1},\dots,x_{n},\lambda }{\mathcal {L}}(x_{1},\dots,x_{n},\lambda )=0$$ which amounts to solvingn+ 1equations inn+ 1unknowns.

The constrained extrema offarecritical pointsof the Lagrangian${\mathcal {L}}$,but they are not necessarilylocal extremaof${\mathcal {L}}$(see§ Example 2below).

One mayreformulate the Lagrangianas aHamiltonian,in which case the solutions are local minima for the Hamiltonian. This is done inoptimal controltheory, in the form ofPontryagin'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 themagnitudeof the gradient of the Lagrangian, as these minima are the same as the zeros of the magnitude, as illustrated inExample 5: Numerical optimization.

Multiple constraints

Figure 2: A paraboloid constrained along two intersecting lines.

Figure 3: Contour map of Figure 2.

The method of Lagrange multipliers can be extended to solve problems with multiple constraints using a similar argument. Consider aparaboloidsubject to two line constraints that intersect at a single point. As the only feasible solution, this point is obviously a constrained extremum. However, thelevel setof$f$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 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$g_{i}(\mathbf {x} )=0,i=1,\dots,M\,.$Every point$\mathbf {x}$on the contour of a given constraint function$g_{i}$has a space of allowable directions: the space of vectors perpendicular to$\nabla g_{i}(\mathbf {x} )\,.$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$\ A\$and denote the span of the constraints' gradients by$S\,.$Then$A=S^{\perp }\,,$the space of vectors perpendicular to every element of$S\,.$

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$\mathbf {x}$such that any allowable direction of movement away from$\mathbf {x}$is perpendicular to$\nabla f(\mathbf {x} )$(otherwise we could increase$f$by moving along that allowable direction). In other words,$\nabla f(\mathbf {x} )\in A^{\perp }=S\,.$Thus there are scalars$\lambda _{1},\lambda _{2},\ \dots,\lambda _{M}$such that $$\nabla f(\mathbf {x} )=\sum _{k=1}^{M}\lambda _{k}\,\nabla g_{k}(\mathbf {x} )\quad \iff \quad \nabla f(\mathbf {x} )-\sum _{k=1}^{M}{\lambda _{k}\nabla g_{k}(\mathbf {x} )}=0~.$$

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

As before, we introduce an auxiliary function $${\mathcal {L}}\left(x_{1},\ldots,x_{n},\lambda _{1},\ldots,\lambda _{M}\right)=f\left(x_{1},\ldots,x_{n}\right)-\sum \limits _{k=1}^{M}{\lambda _{k}g_{k}\left(x_{1},\ldots,x_{n}\right)}\$$ and solve $$\nabla _{x_{1},\ldots,x_{n},\lambda _{1},\ldots,\lambda _{M}}{\mathcal {L}}(x_{1},\ldots,x_{n},\lambda _{1},\ldots,\lambda _{M})=0\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$\ n+M\$unknowns.

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

Modern formulation via differentiable manifolds

The problem of finding the local maxima and minima subject to constraints can be generalized to finding local maxima and minima on adifferentiable manifold$\ M~.$In what follows, it is not necessary that$M$be a Euclidean space, or even a Riemannian manifold. All appearances of the gradient$\ \nabla \$(which depends on a choice of Riemannian metric) can be replaced with theexterior derivative$\ \operatorname {d} ~.$

Single constraint

Let$\ M\$be asmooth manifoldof dimension$\ m~.$Suppose that we wish to find the stationary points$\ x\$of a smooth function$\ f:M\to \mathbb {R} \$when restricted to the submanifold$\ N\$defined by$\ g(x)=0\,$where$\ g:M\to \mathbb {R} \$is a smooth function for which0is aregular value.

Let$\ \operatorname {d} f\$and$\ \operatorname {d} g\$be theexterior derivativesof$\ f\$and$\ g\$.Stationarity for the restriction$\ f|_{N}\$at$\ x\in N\$means$\ \operatorname {d} (f|_{N})_{x}=0~.$Equivalently, the kernel$\ \ker(\operatorname {d} f_{x})\$contains$\ T_{x}N=\ker(\operatorname {d} g_{x})~.$In other words,$\ \operatorname {d} f_{x}\$and$\ \operatorname {d} g_{x}\$are proportional 1-forms. For this it is necessary and sufficient that the following system of$\ {\tfrac {1}{2}}m(m-1)\$equations holds: $$\operatorname {d} f_{x}\wedge \operatorname {d} g_{x}=0\in \Lambda ^{2}(T_{\star x}M)$$ where$\ \wedge \$denotes theexterior product.The stationary points$\ x\$are the solutions of the above system of equations plus the constraint$\ g(x)=0~.$Note that the$\ {\tfrac {1}{2}}m(m-1)\$equations are not independent, since the left-hand side of the equation belongs to the subvariety of$\ \Lambda ^{2}(T_{\star x}M)\$consisting ofdecomposable elements.

In this formulation, it is not necessary to explicitly find the Lagrange multiplier, a number$\ \lambda \$such that$\ \operatorname {d} f_{x}=\lambda \cdot \operatorname {d} g_{x}~.$

Multiple constraints

Let$\ M\$and$\ f\$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$\ G:M\to \mathbb {R} ^{p}(p>1)\,$with component functions$\ g_{i}:M\to \mathbb {R} \,$for which$0\in \mathbb {R} ^{p}$is aregular value.Let$N$be the submanifold of$\ M\$defined by$\ G(x)=0~.$

$\ x\$is a stationary point of$f|_{N}$if and only if$\ \ker(\operatorname {d} f_{x})\$contains$\ \ker(\operatorname {d} G_{x})~.$For convenience let$\ L_{x}=\operatorname {d} f_{x}\$and$\ K_{x}=\operatorname {d} G_{x}\,$where$\ \operatorname {d} G$denotes the tangent map or Jacobian$\ TM\to T\mathbb {R} ^{p}~.$The subspace$\ker(K_{x})$has dimension smaller than that of$\ker(L_{x})$,namely$\ \dim(\ker(L_{x}))=n-1\$and$\ \dim(\ker(K_{x}))=n-p~.$$\ker(K_{x})$belongs to$\ \ker(L_{x})\$if and only if$L_{x}\in T_{\star x}M$belongs to the image of$\ K_{\star x}:\mathbb {R} _{\star }^{p}\to T_{\star x}M~.$Computationally speaking, the condition is that$L_{x}$belongs to the row space of the matrix of$\ K_{x}\,$or equivalently the column space of the matrix of$K_{\star x}$(the transpose). If$\ \omega _{x}\in \Lambda ^{p}(T_{\star x}M)\$denotes the exterior product of the columns of the matrix of$\ K_{\star x}\,$the stationary condition for$\ f|_{N}\$at$\ x\$becomes $$L_{x}\wedge \omega _{x}=0\in \Lambda ^{p+1}\left(T_{\star x}M\right)$$ Once again, in this formulation it is not necessary to explicitly find the Lagrange multipliers, the numbers$\ \lambda _{1},\ldots,\lambda _{p}\$such that $$\ \operatorname {d} f_{x}=\sum _{i=1}^{p}\lambda _{i}\operatorname {d} (g_{i})_{x}~.$$

Interpretation of the Lagrange multipliers

In this section, we modify the constraint equations from the form$g_{i}({\bf {x}})=0$to the form$\ g_{i}({\bf {x}})=c_{i}\,$where the$\ c_{i}\$aremreal constants that are considered to be additional arguments of the Lagrangian expression${\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 $${\begin{aligned}&{\mathcal {L}}(x_{1},x_{2},\ldots;\lambda _{1},\lambda _{2},\ldots;c_{1},c_{2},\ldots )\\[4pt]={}&f(x_{1},x_{2},\ldots )+\lambda _{1}(c_{1}-g_{1}(x_{1},x_{2},\ldots ))+\lambda _{2}(c_{2}-g_{2}(x_{1},x_{2},\dots ))+\cdots \end{aligned}}$$ then $$\ {\frac {\partial {\mathcal {L}}}{\partial c_{k}}}=\lambda _{k}~.$$

So,λ_kis the rate of change of the quantity being optimized as a function of the constraint parameter. As examples, inLagrangian mechanicsthe equations of motion are derived by finding stationary points of theaction,the time integral of the difference between kinetic and potential energy. Thus, the force on a particle due to a scalar potential,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 ascostate equations.

Moreover, by theenvelope theoremthe 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 ($\star$), then it can be shown that $${\frac {\ \operatorname {d} f\left(\ x_{1\star }(c_{1},c_{2},\dots ),\ x_{2\star }(c_{1},c_{2},\dots ),\ \dots \ \right)\ }{\operatorname {d} c_{k}}}=\lambda _{\star k}~.$$

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$\ \lambda _{\star k}\$is themarginal costof the constraint, and is referred to as theshadow price.

Sufficient conditions

Main article: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 borderedHessian matrixof second derivatives of the Lagrangian expression.

Examples

Example 1

Illustration of the constrained optimization problem1

Suppose we wish to maximize$\ f(x,y)=x+y\$subject to the constraint$\ x^{2}+y^{2}=1~.$Thefeasible setis the unit circle, and thelevel setsoffare diagonal lines (with slope −1), so we can see graphically that the maximum occurs at$\ \left({\tfrac {1}{\sqrt {2}}},{\tfrac {1}{\sqrt {2}}}\right)\,$and that the minimum occurs at$\ \left(-{\tfrac {1}{\sqrt {2}}},-{\tfrac {1}{\sqrt {2}}}\right)~.$

For the method of Lagrange multipliers, the constraint is $$g(x,y)=x^{2}+y^{2}-1=0\,$$ hence the Lagrangian function, $${\begin{aligned}{\mathcal {L}}(x,y,\lambda )&=f(x,y)+\lambda \cdot g(x,y)\\[4pt]&=x+y+\lambda (x^{2}+y^{2}-1)\,\end{aligned}}$$ is a function that is equivalent to$\ f(x,y)\$when$\ g(x,y)\$is set to0.

Now we can calculate the gradient: $${\begin{aligned}\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )&=\left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial y}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right)\\[4pt]&=\left(1+2\lambda x,1+2\lambda y,x^{2}+y^{2}-1\right)\ \color {gray}{,}\end{aligned}}$$ and therefore: $$\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\quad \Leftrightarrow \quad {\begin{cases}1+2\lambda x=0\\1+2\lambda y=0\\x^{2}+y^{2}-1=0\end{cases}}$$

Notice that the last equation is the original constraint.

The first two equations yield $$x=y=-{\frac {1}{2\lambda }},\qquad \lambda \neq 0~.$$ By substituting into the last equation we have: $${\frac {1}{4\lambda ^{2}}}+{\frac {1}{4\lambda ^{2}}}-1=0\,$$ so $$\lambda =\pm {\frac {1}{\sqrt {2\ }}}\,$$ which implies that the stationary points of${\mathcal {L}}$are $$\left({\tfrac {\sqrt {2\ }}{2}},{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {1}{\sqrt {2\ }}}\right),\qquad \left(-{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {\sqrt {2\ }}{2}},{\tfrac {1}{\sqrt {2\ }}}\right)~.$$

Evaluating the objective functionfat these points yields $$f\left({\tfrac {\sqrt {2\ }}{2}},{\tfrac {\sqrt {2\ }}{2}}\right)={\sqrt {2\ }}\,\qquad f\left(-{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {\sqrt {2\ }}{2}}\right)=-{\sqrt {2\ }}~.$$

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

Example 2

Illustration of the constrained optimization problem2

Now we modify the objective function of Example1so that we minimize$\ f(x,y)=(x+y)^{2}\$instead of$\ f(x,y)=x+y\,$again along the circle$\ g(x,y)=x^{2}+y^{2}-1=0~.$Now the level sets of$f$are still lines of slope −1, and the points on the circle tangent to these level sets are again$\ ({\sqrt {2}}/2,{\sqrt {2}}/2)\$and$\ (-{\sqrt {2}}/2,-{\sqrt {2}}/2)~.$These tangency points are maxima of$\ f~.$

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

Example 3

Illustration of constrained optimization problem3.

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$\ x\$- and$\ y\$-coordinates lie on the circle around the origin with radius$\ {\sqrt {3\ }}~.$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$\ \lambda ~.$

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

Applying the ordinary Lagrange multiplier method yields $${\begin{aligned}{\mathcal {L}}(x,y,\lambda )&=f(x,y)+\lambda \cdot g(x,y)\\&=x^{2}y+\lambda (x^{2}+y^{2}-3)\,\end{aligned}}$$ from which the gradient can be calculated: $${\begin{aligned}\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )&=\left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial y}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right)\\&=\left(2xy+2\lambda x,x^{2}+2\lambda y,x^{2}+y^{2}-3\right)~.\end{aligned}}$$ And therefore: $$\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\quad \iff \quad {\begin{cases}2xy+2\lambda x=0\\x^{2}+2\lambda y=0\\x^{2}+y^{2}-3=0\end{cases}}\quad \iff \quad {\begin{cases}x(y+\lambda )=0&{\text{(i)}}\\x^{2}=-2\lambda y&{\text{(ii)}}\\x^{2}+y^{2}=3&{\text{(iii)}}\end{cases}}$$ (iii) is just the original constraint. (i) implies$\ x=0\$or$\ \lambda =-y~.$If$x=0$then$\ y=\pm {\sqrt {3\ }}\$by (iii) and consequently$\ \lambda =0\$from (ii). If$\ \lambda =-y\,$substituting this into (ii) yields$\ x^{2}=2y^{2}~.$Substituting this into (iii) and solving for$\ y\$gives$\ y=\pm 1~.$Thus there are six critical points of$\ {\mathcal {L}}\:$ $$({\sqrt {2\ }},1,-1);\quad (-{\sqrt {2\ }},1,-1);\quad ({\sqrt {2\ }},-1,1);\quad (-{\sqrt {2\ }},-1,1);\quad (0,{\sqrt {3\ }},0);\quad (0,-{\sqrt {3\ }},0)~.$$

Evaluating the objective at these points, one finds that $$f(\pm {\sqrt {2\ }},1)=2;\quad f(\pm {\sqrt {2\ }},-1)=-2;\quad f(0,\pm {\sqrt {3\ }})=0~.$$

Therefore, the objective function attains theglobal maximum(subject to the constraints) at$\ (\pm {\sqrt {2\ }},1\ )$and theglobal minimumat$\ (\pm {\sqrt {2\ }},-1)~.$The point$\ (0,{\sqrt {3\ }})\$is alocal minimumof$\ f\$and$\ (0,-{\sqrt {3\ }})\$is alocal maximumof$\ f\,$as may be determined by consideration of theHessian matrixof$\ {\mathcal {L}}(x,y,0)~.$

Note that while$\ ({\sqrt {2\ }},1,-1)\$is a critical point of$\ {\mathcal {L}}\,$it is not a local extremum of$\ {\mathcal {L}}~.$We have $${\mathcal {L}}\left({\sqrt {2\ }}+\varepsilon,1,-1+\delta \right)=2+\delta \left(\varepsilon ^{2}+\left(2{\sqrt {2\ }}\right)\varepsilon \right)~.$$

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

Example 4 – Entropy

Suppose we wish to find thediscrete probability distributionon the points$\ \{p_{1},p_{2},\ldots,p_{n}\}\$with maximalinformation entropy.This is the same as saying that we wish to find theleast structuredprobability distribution on the points$\ \{p_{1},p_{2},\cdots,p_{n}\}~.$In other words, we wish to maximize theShannon entropyequation: $$f(p_{1},p_{2},\ldots,p_{n})=-\sum _{j=1}^{n}p_{j}\log _{2}p_{j}~.$$

For this to be a probability distribution the sum of the probabilities$\ p_{i}\$at each point$\ x_{i}\$must equal 1, so our constraint is: $$g(p_{1},p_{2},\ldots,p_{n})=\sum _{j=1}^{n}p_{j}=1~.$$

We use Lagrange multipliers to find the point of maximum entropy,$\ {\vec {p}}^{\,*}\,$across all discrete probability distributions$\ {\vec {p}}\$on$\ \{x_{1},x_{2},\ldots,x_{n}\}~.$We require that: $$\left.{\frac {\partial }{\partial {\vec {p}}}}(f+\lambda (g-1))\right|_{{\vec {p}}={\vec {p}}^{\,*}}=0\,$$ which gives a system ofnequations,$\ k=1,\ \ldots,n\,$such that: $$\left.{\frac {\partial }{\partial p_{k}}}\left\{-\left(\sum _{j=1}^{n}p_{j}\log _{2}p_{j}\right)+\lambda \left(\sum _{j=1}^{n}p_{j}-1\right)\right\}\right|_{p_{k}=p_{\star k}}=0~.$$

Carrying out the differentiation of thesenequations, we get $$-\left({\frac {1}{\ln 2}}+\log _{2}p_{\star k}\right)+\lambda =0~.$$

This shows that all$\ p_{\star k}\$are equal (because they depend onλonly). By using the constraint $$\sum _{j}p_{j}=1\,$$ we find $$p_{\star k}={\frac {1}{n}}~.$$

Hence, the uniform distribution is the distribution with the greatest entropy, among distributions onnpoints.

Example 5 – Numerical optimization

Lagrange multipliers cause the critical points to occur at saddle points (Example5).

The magnitude of the gradient can be used to force the critical points to occur at local minima (Example5).

The critical points of Lagrangians occur atsaddle points,rather than at local maxima (or minima).Unfortunately, many numerical optimization techniques, such ashill climbing,gradient descent,some of thequasi-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 thegradientof the Lagrangian as below), or else use an optimization technique that findsstationary points(such asNewton's methodwithout an extremum seekingline search) and not necessarily extrema.

As a simple example, consider the problem of finding the value ofxthat minimizes$\ f(x)=x^{2}\,$constrained such that$\ x^{2}=1~.$(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: $${\mathcal {L}}(x,\lambda )=x^{2}+\lambda (x^{2}-1)~.$$

The two critical points occur at saddle points wherex= 1andx= −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: $${\begin{aligned}&{\frac {\partial {\mathcal {L}}}{\partial x}}=2x+2x\lambda \\[5pt]&{\frac {\partial {\mathcal {L}}}{\partial \lambda }}=x^{2}-1~.\end{aligned}}$$

If the target function is not easily differentiable, the differential with respect to each variable can be approximated as $${\begin{aligned}{\frac {\ \partial {\mathcal {L}}\ }{\partial x}}\approx {\frac {{\mathcal {L}}(x+\varepsilon,\lambda )-{\mathcal {L}}(x,\lambda )}{\varepsilon }},\\[5pt]{\frac {\ \partial {\mathcal {L}}\ }{\partial \lambda }}\approx {\frac {{\mathcal {L}}(x,\lambda +\varepsilon )-{\mathcal {L}}(x,\lambda )}{\varepsilon }},\end{aligned}}$$ where$\varepsilon$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: $${\begin{aligned}h(x,\lambda )&={\sqrt {(2x+2x\lambda )^{2}+(x^{2}-1)^{2}\ }}\\[4pt]&\approx {\sqrt {\left({\frac {\ {\mathcal {L}}(x+\varepsilon,\lambda )-{\mathcal {L}}(x,\lambda )\ }{\varepsilon }}\right)^{2}+\left({\frac {\ {\mathcal {L}}(x,\lambda +\varepsilon )-{\mathcal {L}}(x,\lambda )\ }{\varepsilon }}\right)^{2}\ }}~.\end{aligned}}$$

(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 ofhoccur atx= 1andx= −1,just as in${\mathcal {L}}~.$Unlike the critical points in${\mathcal {L}}\,,$however, the critical points inhoccur at local minima, so numerical optimization techniques can be used to find them.

Applications

Control theory

Inoptimal controltheory, the Lagrange multipliers are interpreted ascostatevariables, and Lagrange multipliers are reformulated as the minimization of theHamiltonian,inPontryagin's minimum principle.

Nonlinear programming

The Lagrange multiplier method has several generalizations. Innonlinear programmingthere are several multiplier rules, e.g. the Carathéodory–John Multiplier Rule and the Convex Multiplier Rule, for inequality constraints.

Power systems

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

Safe Reinforcement Learning

The method of Lagrange multipliers applies toconstrained Markov decision processes. It naturally produces gradient-based primal-dual algorithms in safe reinforcement learning.

See also

• 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 testin maximum likelihood estimation
• Lagrangian relaxation

Further reading

• Beavis, Brian; Dobbs, Ian M. (1990)."Static Optimization".Optimization and Stability Theory for Economic Analysis.New York, NY: Cambridge University Press. pp. 32–72.ISBN0-521-33605-8.
• Bertsekas, Dimitri P.(1982).Constrained optimization and Lagrange multiplier methods.New York, NY: Academic Press.ISBN0-12-093480-9.
• Beveridge, Gordon S.G.; Schechter, Robert S. (1970)."Lagrangian multipliers".Optimization: Theory and Practice.New York, NY: McGraw-Hill. pp. 244–259.ISBN0-07-005128-3.
• Binger, Brian R.; Hoffman, Elizabeth (1998). "Constrained optimization".Microeconomics with Calculus(2nd ed.). Reading: Addison-Wesley. pp. 56–91.ISBN0-321-01225-9.
• Carter, Michael (2001)."Equality constraints".Foundations of Mathematical Economics.Cambridge, MA: MIT Press. pp. 516–549.ISBN0-262-53192-5.
• Hestenes, Magnus R.(1966). "Minima of functions subject to equality constraints".Calculus of Variations and Optimal Control Theory.New York, NY: Wiley. pp. 29–34.
• Wylie, C. Ray; Barrett, Louis C. (1995). "The extrema of integrals under constraint".Advanced Engineering Mathematics(Sixth ed.). New York, NY: McGraw-Hill. pp. 1096–1103.ISBN0-07-072206-4.

External links

Exposition

• Steuard."Conceptual introduction".slimy.com.— plus a brief discussion of Lagrange multipliers in thecalculus of variationsas used in physics.
• Carpenter, Kenneth H."Lagrange multipliers for quadratic forms with linear constraints"(PDF).Kansas State University.

Additional text and interactive applets

• Resnik."Simple explanation with an example of governments using taxes as Lagrange multipliers".umiacs.umd.edu.University of Maryland.
• Klein, Dan."Lagrange multipliers without permanent scarring] Explanation with focus on the intuition"(PDF).nlp.cs.berkeley.edu.University of California, Berkeley.
• Sathyanarayana, Shashi."Geometric representation of method of Lagrange multipliers".wolfram.com(Mathematicademonstration).Wolfram Research.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.
• "Lagrange multipliers – two variables".MIT Open Courseware (ocw.mit.edu)(Applet).Massachusetts Institute of Technology.
• "Lagrange multipliers".MIT Open Courseware (ocw.mit.edu)(video lecture). Mathematics 18-02: Multivariable calculus.Massachusetts Institute of Technology.Fall 2007.
• Bertsekas."Details on Lagrange multipliers"(PDF).athenasc.com(slides / course lecture). Non-Linear Programming.— Course slides accompanying text on nonlinear optimization
• Wyatt, John (7 April 2004) [19 November 2002]."Legrange multipliers, constrained optimization, and the maximum entropy principle"(PDF).www-mtl.mit.edu.Elec E & C S / Mech E 6.050– Information, entropy, and computation.— Geometric idea behind Lagrange multipliers
• "Using Lagrange multipliers in optimization".matlab.cheme.cmu.edu(MATLAB example). Pittsburgh, PA: Carnegie Mellon University. 24 December 2011.