14.8 - 14.8: LAGRANGE MULTIPLIERS KIAM HEONG KWA 1. Method...

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14.8: LAGRANGE MULTIPLIERS KIAM HEONG KWA 1. Method of Lagrange Multipliers Let f ( x,y,z ) and g ( x,y,z ) be continuously differentiable in the sense that all their first partial derivatives exist and are continuous. If f ( x 0 ,y 0 ,z 0 ) is a local extreme value of f on the level surface S de- scribed by g ( x,y,z ) = k and (1.1) g ( x 0 ,y 0 ,z 0 ) 6 = 0 , then (1.2) f ( x 0 ,y 0 ,z 0 ) = λ g ( x 0 ,y 0 ,z 0 ) for a real number λ that we refer to as a Lagrange multiplier . In other words, at the point ( x 0 ,y 0 ,z 0 ), the vectors f ( x 0 ,y 0 ,z 0 ) and g ( x 0 ,y 0 ,z 0 ) are parallel. The idea of a proof is this: if f ( x 0 ,y 0 ,z 0 ) is a local extreme value of f on the surface S , then it is also a local extreme value of f when f is evaluated along any smooth curve on S that passes through the point ( x 0 ,y 0 ,z 0 ). To make this precise, let r ( t ) = h x ( t ) ,y ( t ) ,z ( t ) i be a smooth curve that lies on S and passes through ( x 0 ,y 0 ,z 0 ) when t = t 0 . Then the function F ( t ) = f ( r ( t )) = f ( x ( t ) ,y ( t ) ,z ( t )) gives the value of the function f along the curve r ( t ) and has a local extreme value when t = t 0 . Furthermore, F is differentiable since so are f and r ( t ). Thanks to this, we can apply the chain rule to obtain F 0 ( t ) = f x ( x ( t ) ,y ( t ) ,z ( t )) x 0 ( t ) + f y ( x ( t ) ,y ( t ) ,z ( t )) y 0 ( t ) + f z ( x ( t ) ,y ( t ) ,z ( t )) z 0 ( t ) = f ( r ( t )) · r 0 ( t ) . Date : October 2, 2010. 1
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2 KIAM HEONG KWA In particular, since F ( t 0 ) is a local extreme value, we have F 0 ( t 0 ) = f ( r ( t 0 )) · r 0 ( t 0 ) = 0 or, equivalently, f ( x 0 ,y 0 ,z 0 ) · r 0 ( t 0 ) = 0 . This shows that f ( x 0 ,y 0 ,z 0 ) is orthogonal to the tangent plane at the point ( x 0 ,y 0 ,z 0 ). On the other hand, recall from section 14.6 that g ( x 0 ,y 0 ,z 0 ) is also orthogonal to the tangent plane at the same point. Therefore the vectors f ( x 0 ,y 0 ,z 0 ) and g ( x 0 ,y 0 ,z 0 ) must be parallel to each other, justifying (1.2). The procedure based on (1.2) to find the extreme values of f ( x,y,z ) subject to the constraint or side condition g ( x,y,z ) = k , assuming that these extreme values exist and g ( x,y,z ) 6 = 0 on the surface S described by
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This note was uploaded on 11/11/2011 for the course MATH 254.01 taught by Professor Kwa during the Fall '10 term at Ohio State.

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14.8 - 14.8: LAGRANGE MULTIPLIERS KIAM HEONG KWA 1. Method...

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