Math Lagrange Multipliers

Transcription

1 Math Lagrange Multipliers Peter A. Perr Universit of Kentuck October 12, 2018

2 Homework Re-read section 14.8 Begin practice homework on section 14.8, problems 3-11 (odd), 15, 21, 23 Begin (or continue!) webwork B5 on Begin (or continue!) reviewing for Eam II Remember that there s a review session for Eam II on Monda, October 15, 6-8 PM in BS 116 Remember that Eam II is net Wednesda, October 17 at 5 PM

3 Unit II: Differential Calculus of Several Variables Lecture 13 Functions of Several Variables Lecture 14 Limits and Continuit Lecture 15 Partial Derivatives Lecture 16 Tangent Planes and Linear Approimation, I Lecture 17 Tangent Planes and Linear Approimation, II Lecture 18 The Chain Rule Lecture 19 Directional Derivatives and the Gradient Lecture 20 Maimum and Minimum Values, I Lecture 21 Maimum and Minimum Values, II Lecture 22 Lagrange Multipliers Lecture 23 Review for Eam 2

4 Goals of the Da Understand the geometrical idea behind Lagrange s Multiplier Method Use the Lagrange Multiplier Method to solve ma/min problems with one constraint Use the Lagrange Multiplier Method to solve ma/min problems with two constraints

5 A Word from Our Sponsor Pierre-Louis Lagrange ( ) was born in Ital but lived and worked for much of his life in France. Working in the generation following Newton ( ), he made fundamental contributions in the calculus of variations, in celestial mechanics, in the solution of polnomial equations, and in power series representation of functions. Lagrange lived through the French revolution, during which time the chemist Lavoisier was beheaded. Of Lavoisier s death, Lagrange remarked: It took the mob onl a moment to remove his head; a centur will not suffice to reproduce it. Image credit:

6 Reminder: Level curves and the Gradient Remember that the gradient is normal to the level curves of a function At left are the level curves of: f (, ) = 2 2 g(, ) = 2 + 2

7 Reminder: Level curves and the Gradient Remember that the gradient is normal to the level curves of a function At left are the level curves of: f (, ) = 2 2 g(, ) = Which wa does the gradient point for each picture?

8 Reminder: Level curves and the Gradient Remember that the gradient is normal to the level curves of a function At left are the level curves of: f (, ) = 2 2 g(, ) = Which wa does the gradient point for each picture? Are there an points where f and g are parallel?

9 Reminder: Level curves and the Gradient Remember that the gradient is normal to the level curves of a function At left are the level curves of: f (, ) = 2 2 g(, ) = Which wa does the gradient point for each picture? Are there an points where f and g are parallel? What about the respective tangent lines at an such points?

10 Problem Find the absolute minimum and maimum value of f (, ) = 2 2 if = 1 In this problem: f () = 2 2 the function f (, ) is the objective function 4 the equation = 1 is the constraint Where do etreme values occur?

11 Problem Find the absolute minimum and maimum value of f (, ) = 2 2 if = 1 In this problem: f () = 2 2 the function f (, ) is the objective function 4 the equation = 1 is the constraint Where do etreme values occur? Maima: f (1, 0) = f ( 1, 0) = 1 Minima: f (0, 1) = f (0, 1) = 1

12 Lagrange s Condition Problem Find the absolute minimum and maimum value of f (, ) = 2 2 if g(, ) = = 1 f () = 2 2 Etreme values occur where f and g are parallel, i.e., where 4 f = λ g Wh does this work?

13 Wh Is f = λ g at an Etremum? Problem Find the absolute minimum and maimum of f (, ) subject to a constraint g(, ) = c. The constraint restricts (, ) to a level curve ((t), (t)) of g So we want to minimize φ(t) = f ((t), g(t)), a function of one variable B the chain rule, the condition φ (t) = 0 is the same as ( f )((t), (t)) ( (t), (t)) = 0 That is, f is perpendicular to the tangent line to the level curve That is, f is parallel to g The number λ is called a Lagrange Multiplier

14 Lagrange Multipliers: One Constraint, Two Variables To find the maimum and minimum values of f (, ) subject to the constraint g(, ) = k: (a) Find all (,, λ) so that f (, ) = λ g(, ) g(, ) = k (b) Test the solutions (, ) to find the maimum and minimum values 1. Find the maimum and minimum values of f (, ) = 2 2 subject to the constraint = Find the maimum and minimum values of f (, ) = 3 + subject to the constraint = 10

15 Lagrange Multipliers: One Constraint, Three Variables To find the maimum and minimum values of f (,, z) subject to the constraint g(,, z) = k: (a) Find all (,, z, λ) so that f (,, z) = λ g(,, z) g(,, z) = k (b) Test the solutions (,, z) to find the maimum and minimum values 1. Find the maimum and minimum values of f (,, z) = e z subject to the constraint z 2 = 24

16 Lagrange Multipliers: Two Constraints, Three Variables To find the maimum and minimum values of f (,, z) subject to the constraint g(,, z) = k, h(,, z) = l: (a) Find all (,, z, λ, µ) so that f (,, z) = λ g(,, z) + µ h(,, z) g(,, z) = k h(,, z) = l (b) Test the solutions (,, z) to find the maimum and minimum values Find the etreme values of f (,, z) = + + z subject to the constraints 2 + z 2 = 2 + = 1

17 Wh Does the Two-Constraint Method Work? Find the maimum and minimum values of f (,, z) subject to the constraints g(,, z) = k h(,, z) = l The surfaces S 1 = {(,, z) : g(,, z) = k} and S 2 = {(,, z) : h(,, z) = l} intersect in a curve C We know that f (,, z) is orthogonal to C if f has an etremum at (,, z) We know that g(,, z) and h(,, z) are also orthogonal to C Hence, there are numbers λ and µ so that f (,, z) = λ g(,, z) + µ h(,, z)