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people:grads:brown:brown_teaching:math323:chapter_14:section_8

Chapter 14: Partial Derivatives
Calculus III Discussion Links
Chapter 14 - Section 7 - Maxima and Minima

Chapter 14 - Section 8 - Lagrange Multipliers

Section Overview

In this section we learn how to use the Lagrange Technique to locate extreme (maximum or minimum) values of a multivariable function subject to some constraints. As we shall see, the technique outlined is extremely similar to the method we used in Calculus I to locate extreme values of a single variable function on an interval. We will describe the technique three times, once for a three-variable function with one constraint, once for a three-variable function with two constraints, and finally for the case of an n-variable function with m constraints. We will assume throughout that our objective function is differentiable and our constraints have non-zero gradient, except for a couple examples in which we will discuss how and why the technique may fail when a constraint gradient is the zero vector.

Three variable function, single constraint

Let w=f(x,y,z) denote our objective function and g(x,y,z)=k our constraint; that is, we seek for the extreme values attained by the function f on the level surface g=k. Suppose f does have an extreme value at a point P=(x0,y0,z0) on the surface g=k. Let C be a curve with differentiable parametrization r(t) that lies on the surface g=k and passes through the point P. Let t0 denote the parameter value corresponding to the point P, so r(t0)=x0,y0,z0, and let h(t)=fr(t). Now h has an extreme value at t0, so it has a critical point there. Since f is differentiable at P and r is differentiable, it follows that the composition h is differentiable at t0, and so the critical point t0 is a root of the derivative. Hence, by the chain rule we have

0=h(t0)=fx(x0,y0,z0)dxdt(t0)+fy(x0,y0,z0)dydt(t0)+fz(x0,y0,z0)dzdt(t0)=f(x0,y0,z0)r(t0)

Thus the gradient of f at the point P is orthogonal to the line tangent to the curve C at the point P. Since this holds for all curves in the surface g=k through the point P, it follows that the gradient of f at P is parallel to the direction of the plane tangent to the surface g=k at the point P; in other words, at the point P, if g0 then f=λg for some number λ.

Let us recall the technique used in Calculus I to locate extreme values of a differentiable function y=f(x) on an open interval I. If f has an extreme value at a point x0I, then x0 is a critical point of f, and since we are assuming that f is differentiable on I we must have f(x0)=0. We use this result to develop our technique as follows. Assuming that f has extreme values on I, we locate them in two steps:

  1. Solve f=0
  2. Evaluate f at each solution c found in step 1.; largest value is the maximum, smallest is the minimum.

The Lagrange technique works in a very similar way; assuming extreme values exist, we first find critical points, then evaluate at each to determine the extreme values. There is a slight difference in our Lagrange technique though, in that it is not the critical points of f which we seek for but rather one which involves f together with our constraint g=k. Let λ denote a fourth variable and set F=fλ(gk). Critical points of F are found by solving the vector equation 0=F, which yields the following system of four equations with four unknowns:

0=Fx=fxλgx0=Fy=fyλgy0=Fz=fzλgz0=Fλ=gk

We see that the first three equations represent the vector equation f=λg and the fourth equation is our constraint g=k. From our discussion above, we can conclude that if f has an extreme value at a point P=(x0,y0,z0) on the surface g=k and if g0 at P then there exists a number λ0 such that f=λ0g at P and so (x0,y0,z0,λ0) is a critical point of F. Therefore, assuming that f has extreme values on the surface g=k and that the gradient of g is non-zero, we can find the extreme values in two steps:

  1. Solve F=0.
  2. Evaluate f at the projections in R3 of each solution found in step 1.; largest is maximum, smallest is minimum

Note that in step 1. above, we don't actually need the λ part of the solution. You may find it's values in the course of finding x,y, and z, but more often it will just be used to help relate the important variables. The points found in step 1. are in four-space, but we need points in three space to evaluate f; specifically, we need the x,y,z part. The projection referred to above works as follows:

(x,y,z,λ)(x,y,z,0)(x,y,z)

In words, we replace our fourth coordinate with zero, then identify the point (x,y,z,0) in R4 with the point (x,y,z) in R3.

Example Set One

One

Two

Three variable function, two constraints

Let f(x,y,z) be our objective function and g(x,y,z)=k and h(x,y,z)= our two constraints. Let F=fλ(gk)μ(h)

If f has an extreme value at a point (x0,y0,z0) on the intersection of the level surfaces g=k and h= and both g and h are non-zero, then there exist numbers λ0 and μ0 such that (x0,y0,z0,λ0,μ0) is a critical point of F. Thus, to find extreme values of f subject to the given constraints, we solve 0=F and check each point.

Example Set Two

One

n variable function, m constraints

As usual we form an auxiliary function F with of objective and all constraints and seek for solutions to the vector equation 0=F. Let x1,x2,,xn denote the n-variables and g1=k1,g2=k2,,gm=km the m constraints. Then we introduce m variables λ1,λ2,,λm and set F=fmi=1λi(giki)

Example Set Three

One


Chapter 15: Multiple Integrals

Discussion

Section 14.8 Discussion

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people/grads/brown/brown_teaching/math323/chapter_14/section_8.txt · Last modified: 2020/03/19 16:15 by brown