20e-lecture12

# 20e-lecture12 - Lecture 12 Monday April 27th...

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Lecture 12 - Monday April 27th [email protected] Key words: Several constraints, Jacobian, Implicit function theorem, Inverse function theorem Know how to do Lagrange multipliers with several constraints, computing Jacobian determinants, statement of implicit function theorems 12.1 Several Constraints The method of Lagrange multipliers works for optimizing functions under several con- straints. If we want to ﬁnd the extreme points of a diﬀerentiable function f : U R where U is a closed set subject to constraints g 1 ( x ) = g 2 ( x ) = ··· = g r ( x ) = 0, we introduce multipliers λ 1 2 ,...,λ r and ﬁnd the critical points of φ ( x,λ ) = f ( x ) + λ 1 g 1 ( x ) + ··· + λ r g r ( x ) where λ = ( λ 1 2 ,...,λ r ). Example 1. Find the closest distance between the parabola y = 1 - x 2 and the parabola y = 2 - | x | for 0 x 2. Solution. Let ( x,y ) be a point on the curve y = 1 - x 2 and let ( X,Y ) be a point on the curve Y = 2 -| X | . We want to minimize ( x - X ) 2 +( y - Y ) 2 – the square of the distance between the two points. We can assume that all the variables are positive, by symmetry (see the ﬁgure). y = 1 - x 2 Y = 2 - | X | 1

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The constraints are g 1 ( x,y ) = y - 1 + x 2 = 0 and g 2 ( X,Y ) = Y - 2 + X = 0. By the method of Lagrange multipliers, we set φ = 0 where λ and μ are the Lagrange multipliers and φ ( x,y,X,Y,λ,μ ) = ( x - X ) 2 + ( y - Y ) 2 + λ ( y - 1 + x 2 ) + μ ( Y - 2 + X ) .
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## This note was uploaded on 02/12/2010 for the course MATH 20E taught by Professor Enright during the Spring '07 term at UCSD.

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20e-lecture12 - Lecture 12 Monday April 27th...

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