# The Method of Lagrange Multipliers

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1 The Method of Lagrange Multipliers S. Sawyer October 25, Lagrange s Theorem. Suppose that we want to maximize (or imize a function of n variables f(x = f(x 1, x 2,..., x n for x = (x 1, x 2,..., x n (1.1a p constraints g 1 (x = c 1, g 2 (x = c 2,..., and g p (x = c p (1.1b As an example for p = 1, find { n x 2 i : or x 1,...,x 5 x 1,...,x n 5 x 2 i } x i = 1 A first guess for (1.1 (with f(x = n solutions of the n equations { x1 + 2x 2 + x 3 = 1 and x 3 2x 4 + x 5 = 6 x2 i (1.2a (1.2b in (1.2 might be to look for x i f(x = 0, 1 i n (1.3 However, this leads to x i = 0 in (1.2, which does not satisfy the constraint. Lagrange s solution is to introduce p new parameters (called Lagrange Multipliers and then solve a more complicated problem: Theorem (Lagrange Assug appropriate smoothness conditions, imum or maximum of f(x the constraints (1.1b that is not on the boundary of the region where f(x and g j (x are defined can be found by introducing p new parameters λ 1, λ 2,..., λ p and solving the system p f(x + λ j g j (x = 0, 1 i n (1.4a x i j=1 g j (x = c j, 1 j p (1.4b This amounts to solving n+p equations for the n+p real variables in x and λ. In contrast, (1.3 has n equations for the n unknowns in x. Fortunately, the system (1.4 is often easy to solve, and is usually much easier than using the constraints to substitute for some of the x i.

2 The Method of Lagrange Multipliers Examples. (1 There are p = 1 constraints in (1.2a, so that (1.4a becomes x 2 k + λ x k = 2x i + λ = 0, 1 i n x i Thus x i = λ/2 for 1 i n. The constraint n x i = 1 in (1.4b implies n x i = nλ/2 = 1 or λ = 2/n, from which x i = 1/n for 1 i n. For x i = 1/n, f(x = n/n 2 = 1/n. One can check that this is a imum as opposed to a maximum or saddle point by noting that f(x = 1 if x 1 = 1, x i = 0 for 2 i n. (2 A System with Two Constraints: There are p = 2 constraints in (1.2b, which is to find 5 x 2 i { x1 + 2x 2 + x 3 = 1 and x 3 2x 4 + x 5 = 6 (2.1 The method of Lagrange multipliers says to look for solutions of ( 5 x 2 k + λ(x 1 + 2x 2 + x 3 + µ(x 3 2x 4 + x 5 = 0 (2.2 x i where we write λ, µ for the two Lagrange multipliers λ 1, λ 2. The equations (2.2 imply 2x 1 + λ = 0, 2x 2 + 2λ = 0, 2x 3 + λ + µ = 0, 2x 4 2µ = 0, and 2x 5 + µ = 0. Combining the first three equations with the first constraint in (2.1 implies 2 + 6λ + µ = 0. Combining the last three equations in (2.2 with the second constraint in (2.1 implies 12+λ+6µ = 0. Thus 6λ + µ = 2 λ + 6µ = 12 Adding these two equations implies 7(λ + µ = 14 or λ + µ = 2. Subtracting the equations implies 5(λ µ = 10 or λ µ = 2. Thus (λ + µ + (λ µ = 2λ = 0 and λ = 0, µ = 2. This implies x 1 = x 2 = 0, x 3 = x 5 = 1, and x 4 = 2. The imum value in (2.1 is 6. (3 A BLUE problem: Let X 1,..., X n be independent random variables with E(X i = µ and Var(X i = σ 2 i. Find the coefficients a i that imize Var a i X i E a i X i = µ (2.3

3 The Method of Lagrange Multipliers This asks us to find the Best Linear Unbiased Estimator n a ix i (abbreviated BLUE for µ for given values of σi 2. Since Var(aX = a 2 Var(X and Var(X + Y = Var(X + Var(Y for independent random variables X and Y, we have Var( n a ix i = n a2 i Var(X i = n a2 i σ2 i. Thus (2.3 is equivalent to finding a 2 i σi 2 a i = 1 Using one Lagrange multiplier λ for the constraint leads to the equations 2a i σi 2 + λ = 0 or a i = λ/(2σi 2. The constraint n a i = 1 then implies that the BLUE for µ is / n a i X i where a i = c/σi 2 for c = 1 (1/σk 2 (2.4 If σi 2 = σ2 for all i, then a i = 1/n and n a ix i = (1/n n X i = X is the BLUE for µ. Conversely, if Var(X i = σi 2 is variable, then the BLUE n a ix i for µ puts relatively less weight on the noisier (higher-variance observations (that is, the weight a i is smaller, but still uses the information in the noiser observations. Formulas like (2.4 are often used in survey sampling. 3. A Short Proof of Lagrange s Theorem. The extremal condition (1.3 (without any constraints can be written in vector form as f(x = Now by Taylor s Theorem ( x 1 f(x, x 2 f(x,..., f(x x n = 0 (3.1 f(x + hy = f(x + hy f(x + O(h 2 (3.2 where h is a scalar, O(h 2 denotes terms that are bounded by h 2, and x y is the dot product. Thus (3.1 gives the vector direction in which f(x changes the most per unit change in x, where unit change in measured in terms of the length of the vector x. In particular, if y = f(x 0, then f(x hy < f(x < f(x + hy

4 The Method of Lagrange Multipliers for sufficiently small values of h, and the only way that x can be a local imum or maximum would be if x were on the boundary of the set of points where f(x is defined. This implies that f(x = 0 at non-boundary imum and maximum values of f(x. Now consider the problem of finding max f(x g(x = c (3.3 for one constraint. If x = x 1 (t is a path in the surface defined by g(x = c, then by the chain rule d dt g( x 1 (0 = d dt x 1(0 g ( x 1 (0 = 0 (3.4 This implies that g ( x 1 (0 is orthogonal to the tangent vector (d/dtx 1 (0 for any path x 1 (t in the surface defined by g(x = c. Conversely, if x is any point in the surface g(x = c and y is any vector such that y g(x = 0, then it follows from the Implicit Function Theorem there exists a path x 1 (t in the surface g(x = c such that x 1 (0 = x and (d/dtx 1 (0 = y. This result and (3.4 imply that the gradient vector g(x is always orthogonal to the surface defined by g(x = c at x. Now let x be a solution of (3.3. I claim that f(x = λ g(x for some scalar λ. First, we can always write f(x = c g(x+y where y g(x = 0. If x(t is a path in the surface with x(0 = x and (d/dtx(0 f(x = 0, it follows from (3.2 with y = (d/dtx(0 that there are values for f(x for x = x(t in the surface that both larger and smaller than f(x. Thus, if x is a maximum of imum of f(x in the surface and f(x = c g(x + y for y g(x = 0, then y f(x = y g(x + y y = y y = 0 and y = 0. This means that f(x = c g(x, which completes the proof of Lagrange s Theorem for one constraint (p = 1. Next, suppose that we want to solve max f(x g 1 (x = c 1,..., g p (x = c p (3.5 for p constraints. Let x be a solution of (3.5. Recall that the each vector g j (x is orthogonal to the surface g j (x = c j at x. Let L be the linear space L = span{ g j (x : 1 j p } I claim that f(x L. This would imply f(x = p λ j g j (x j=1

5 The Method of Lagrange Multipliers for some choice of scalar values λ j, which would prove Lagrange s Theorem. To prove that f(x L, first note that, in general, we can write f(x = w + y where w L and y is perpendicular to L, which means that y z = 0 for any z L. In particular, y g j (x = 0 for 1 j p. Now find a path x 1 (t through x in the intersection of the surfaces g j (x = c j such that x 1 (0 = x and (d/dtx 1 (0 = y. (The existence of such a path for sufficiently small t follows from a stronger form of the Implicit Function Theorem. It then follows from (3.2 and (3.5 that y f(x = 0. Since f(x = w + y where y w = 0, it follows that y f(x = y w + y y = y y = 0 and y = 0, This implies that f(x = w L, which completes the proof of Lagrange s Theorem. 4. Warnings. The same warnings apply here as for most methods for finding a maximum or imum: The system (1.4 does not look for a maximum (or imum of f(x constraints g j (x = c j, but only a point x on the set of values detered by g j (x = c j whose first-order changes in x are zero. This is satisfied by a value x = x 0 that provides a imum or maximum typical for f(x in a neighborhood of x 0, but may only be a local imum or maximum. There may be several local ima or maxima, each yielding a solution of (1.4. The criterion (1.4 also holds for saddle points of f(x that are local maxima in some directions or coordinates and local ima in others. In these cases, the different values f(x at the solutions of (1.4 have to be evaluated individually to find the global maximum. A particular situation to avoid is to look for a maximum value of f(x by solving (1.4 or (1.3 when f(x takes arbitrarily large values when any of the components of x are large (as is the case for f(x in (1.2 and (1.4 has a unique solution x 0. In that case, x 0 is probably the global imum of f(x the constraints, and not a maximum. In that case, rather than find the best possible value of f(x, one may end up with the worst possible value. After solving (1.3 or (1.4, one often has to look at the problem more carefully to see if it is a global maximum, a global imum, or neither. Another situation to avoid is when the maximum or imum is on the boundary of the values for which f(x is defined. In that case, the maximum or imum is not an interior value, and the first-order changes in f(x (that is, the partial derivatives of f(x may not be zero at that point. An example is f(x = x on the unit interval 0 x 1. The imum value of f(x = x on the interval is x = 0 and the maximum is x = 1, but neither are solutions of f (x = 0.

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