Slide_09 - MA1104 Multivariable Calculus Lecture 9 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 9 Dr KU Cheng Yeaw Thursday Feb 12, 2009 Recap Last lecture ... • we learned how to locate local and absolute extremum of functions of several variables. • we began to look at the problem of finding extremum point of f ( x,y ) subject to a given constraint g ( x,y ) = k . The method we developed is called the Method of Lagrange Multipliers. Overview In this lecture, • we prove the Method of Lagrange Multiplier for functions of three variable f ( x,y,z ) . • we consider some applications of this method. • we extend this method to optimization problems with two given constraints. Lagrange Multipliers - Two Variables We have seen that if f ( x,y ) attains an extremum value at ( x ,y ) subject to the constraint g ( x,y ) = k , then their gradient vectors are parallel, i.e. O f ( x ,y ) = λ O g ( x ,y ) for some λ ∈ R . The basic idea is to look at the problem on the xy-plane: how the curve g ( x,y ) = k and the level curves of f relate to each other when trying to locate the point ( x ,y ) . However, our argument was not precise. Lagrange Multipliers - Three Variables We can extend the our argument to the problem of finding the extreme values of f ( x,y,z ) subject to the constraint g ( x,y,z ) = k . This time, we supply a more formal proof; and we shall see what assumptions we need to make. Instead of considering level curves of f ( x,y ) and conclude that when maximum occurs at ( x ,y ) , then the level curve f ( x,y ) = c touches g ( x,y ) = k (so the maximum value is c ), we shall consider level surfaces of f ( x,y,z ) and see that when maximum occurs at ( x ,y ,z ) , then the level surface f ( x,y,z ) = c touches the surface g ( x,y,z ) = k To be precise: Suppose f ( x,y,z ) attains an extremum at ( x ,y ,z ) subject to the constraint that g ( x,y,z ) = k . Assume both f and g are differentiable (we will see why we need these assumptions) . Consider the surface S given by g ( x,y,z ) = k and let C be any curve lying on this surface that passes through ( x ,y ,z ) . Assume the curve C is traced out by a vector function r ( t ) = h x ( t ) ,y ( t ) ,z ( t ) i where r ( t ) = h x ,y ,z i . Consider the function h ( t ) = f (( x ( t ) ,y ( t ) ,z ( t )) . Since f ( x,y,z ) has an extremum at ( x ,y ,z ) , it follows that h ( t ) has an extremum at t . So h ( t ) = 0 . Using Chain Rule (we require f to be differentiable), = h ( t ) = f x ( x ,y ,z ) x ( t ) + f y ( x ,y ,z ) y ( t ) + f z ( x ,y ,z ) z ( t ) = O f ( x ,y ,z ) · r ( t ) . Therefore, if f ( x ,y ,z ) is an extremum, then the gradient of f at ( x ,y ,z ) is orthogonal to the tangent vector r ( t ) at that point. But C was an arbitrary curve on the level surface S defined by g ( x,y,z ) = k , it follows that O f ( x ,y ,z ) must be orthogonal to every curve lying on the level surface S passing through ( x ,y ,z ) ....
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This note was uploaded on 03/19/2012 for the course MATH 1104 taught by Professor Kuchengyeaw during the Spring '08 term at National University of Singapore.

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Slide_09 - MA1104 Multivariable Calculus Lecture 9 Dr KU...

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