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

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MA1104 Multivariable Calculus Lecture 9 Dr KU Cheng Yeaw Thursday Feb 12, 2009

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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.

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Lagrange Multipliers - Two Variables We have seen that if f ( x, y ) attains an extremum value at ( x 0 , y 0 ) subject to the constraint g ( x, y ) = k , then their gradient vectors are parallel, i.e. O f ( x 0 , y 0 ) = λ O g ( x 0 , y 0 ) 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 0 , y 0 ) . 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 0 , y 0 ) , 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 0 , y 0 , z 0 ) , then the level surface f ( x, y, z ) = c touches the surface g ( x, y, z ) = k

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To be precise: Suppose f ( x, y, z ) attains an extremum at ( x 0 , y 0 , z 0 ) 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 0 , y 0 , z 0 ) . Assume the curve C is traced out by a vector function r ( t ) = h x ( t ) , y ( t ) , z ( t ) i where r ( t 0 ) = h x 0 , y 0 , z 0 i .
Consider the function h ( t ) = f (( x ( t ) , y ( t ) , z ( t )) . Since f ( x, y, z ) has an extremum at ( x 0 , y 0 , z 0 ) , it follows that h ( t ) has an extremum at t 0 . So h 0 ( t 0 ) = 0 .

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Using Chain Rule (we require f to be differentiable), 0 = h 0 ( t 0 ) = f x ( x 0 , y 0 , z 0 ) x 0 ( t 0 ) + f y ( x 0 , y 0 , z 0 ) y 0 ( t 0 ) + f z ( x 0 , y 0 , z 0 ) z 0 ( t 0 ) = O f ( x 0 , y 0 , z 0 ) · r 0 ( t 0 ) . Therefore, if f ( x 0 , y 0 , z 0 ) is an extremum, then the gradient of f at ( x 0 , y 0 , z 0 ) is orthogonal to the tangent vector r 0 ( t 0 ) 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 0 , y 0 , z 0 ) must be orthogonal to every curve lying on the level surface S passing through ( x 0 , y 0 , z 0 ) . That is, O f ( x 0 , y 0 , z 0 ) is normal to the tangent plane at ( x 0 , y 0 , z 0 ) .

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On the other hand, recall from our study of the significance of the gradient vector (Lecture Slide 7 page 10) - here we require g ( x, y, z ) to be differentiable , O g ( x 0 , y 0 , r 0 ) · r 0 ( t 0 ) = 0 , provided O g ( x 0 , y 0 , r 0 ) 6 = 0 , for any curve r ( t ) through ( x 0 , y 0 , z 0 ) .

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