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Method to solve constrained optimization problems From Wikipedia, the free encyclopedia

In mathematical optimization, the **method of Lagrange multipliers** is a strategy for finding the local maxima and minima of a function subject to equation constraints (i.e., subject to the condition that one or more equations have to be satisfied exactly by the chosen values of the variables).^{[1]} It is named after the mathematician Joseph-Louis Lagrange.

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.^{[2]} In the general case, the Lagrangian is defined as
for functions ; is called the Lagrange multiplier.

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

The method can be summarized as follows: in order to find the maximum or minimum of a function subject to the equality constraint , find the stationary points of considered as a function of and the Lagrange multiplier . This means that all partial derivatives should be zero, including the partial derivative with respect to .^{[3]}

and

or equivalently

and

The solution corresponding to the original constrained optimization is always a saddle point of the Lagrangian function,^{[4]}^{[5]} which can be identified among the stationary points from the definiteness of the bordered Hessian matrix.^{[6]}

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 for a given constant .

The following is known as the Lagrange multiplier theorem.^{[7]}

Let be the objective function, be the constraints function, both belonging to (that is, having continuous first derivatives). Let be an optimal solution to the following optimization problem such that, for the matrix of partial derivatives , :

Then there exists a unique Lagrange multiplier such that (Note that this is a somewhat conventional thing where 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 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.^{[8]} 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 0 in every feasible direction.

For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider the optimization problem
(Sometimes an additive constant is shown separately rather than being included in , in which case the constraint is written as in Figure 1.) We assume that both and have continuous first partial derivatives. We introduce a new variable () called a **Lagrange multiplier** (or **Lagrange undetermined multiplier**) and study the **Lagrange function** (or **Lagrangian** or **Lagrangian expression**) defined by
where the term may be either added or subtracted. If is a maximum of for the original constrained problem and then there exists such that () is a *stationary point* for the Lagrange function (stationary points are those points where the first partial derivatives of are zero). The assumption 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.^{[9]}^{[10]}^{[11]}^{[12]}^{[13]} 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:

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

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
for some

where are the respective gradients. The constant 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 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 is a solution regardless of .

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

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

Note that implies as the partial derivative of with respect to is

To summarize

The method generalizes readily to functions on variables
which amounts to solving *n* + 1 equations in *n* + 1 unknowns.

The constrained extrema of f are *critical points* of the Lagrangian , but they are not necessarily *local extrema* of (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.

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 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 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 constraints and are walking along the set of points satisfying Every point on the contour of a given constraint function has a space of allowable directions: the space of vectors perpendicular to 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 and denote the span of the constraints' gradients by Then the space of vectors perpendicular to every element of

We are still interested in finding points where does not change as we walk, since these points might be (constrained) extrema. We therefore seek such that any allowable direction of movement away from is perpendicular to (otherwise we could increase by moving along that allowable direction). In other words, Thus there are scalars such that

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

As before, we introduce an auxiliary function and solve which amounts to solving equations in 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 ^{[14]} In what follows, it is not necessary that be a Euclidean space, or even a Riemannian manifold. All appearances of the gradient (which depends on a choice of Riemannian metric) can be replaced with the exterior derivative

Let be a smooth manifold of dimension Suppose that we wish to find the stationary points of a smooth function when restricted to the submanifold defined by where is a smooth function for which 0 is a regular value.

Let and be the exterior derivatives of and . Stationarity for the restriction at means Equivalently, the kernel contains In other words, and are proportional 1-forms. For this it is necessary and sufficient that the following system of equations holds: where denotes the exterior product. The stationary points are the solutions of the above system of equations plus the constraint Note that the equations are not independent, since the left-hand side of the equation belongs to the subvariety of consisting of decomposable elements.

In this formulation, it is not necessary to explicitly find the Lagrange multiplier, a number such that

Let and be as in the above section regarding the case of a single constraint. Rather than the function described there, now consider a smooth function with component functions for which is a regular value. Let be the submanifold of defined by

is a stationary point of if and only if contains For convenience let and where denotes the tangent map or Jacobian ( can be canonically identified with ). The subspace has dimension smaller than that of , namely and belongs to if and only if belongs to the image of Computationally speaking, the condition is that belongs to the row space of the matrix of or equivalently the column space of the matrix of (the transpose). If denotes the exterior product of the columns of the matrix of the stationary condition for at becomes Once again, in this formulation it is not necessary to explicitly find the Lagrange multipliers, the numbers such that

In this section, we modify the constraint equations from the form to the form where the are m real constants that are considered to be additional arguments of the Lagrangian expression .

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 then

So, *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,

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 (), then it can be shown that

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 is the marginal cost of the constraint, and is referred to as the shadow price.^{[15]}

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.^{[6]}^{[16]}

Suppose we wish to maximize subject to the constraint The feasible set is the unit circle, and the level sets of f are diagonal lines (with slope −1), so we can see graphically that the maximum occurs at and that the minimum occurs at

For the method of Lagrange multipliers, the constraint is hence the Lagrangian function, is a function that is equivalent to when is set to 0.

Now we can calculate the gradient: and therefore:

Notice that the last equation is the original constraint.

The first two equations yield By substituting into the last equation we have: so which implies that the stationary points of are

Evaluating the objective function f at these points yields

Thus the constrained maximum is and the constrained minimum is .

Now we modify the objective function of Example **1** so that we minimize instead of again along the circle Now the level sets of are still lines of slope −1, and the points on the circle tangent to these level sets are again and These tangency points are maxima