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by: Kavon Feest

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# ANAL GEO & CALCULUS MATH 016B

Kavon Feest

GPA 3.93

Staff

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COURSE
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Staff
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Class Notes
PAGES
2
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KARMA
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## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 016B at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 67 views. For similar materials see /class/226603/math-016b-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15
Math 16B 7 F05 Supplementary Notes 3 The Lagrange Multiplier Method The method of Lagrange multipliers is often effective in nding solutions of constrained ex tremum problems In the two variable version of such a problem one is given a function fy and one wishes to maximize it or minimize it under the constraint that another function gy vanishes ie one wishes to nd a maximum or minimum of f on the level curve gy 0 As explained in our textbook where you will also nd examples Lagrange7s method proceeds as follows One introduces a third variable A traditionally called a Lagrange multiplier and one de nes a function Fzy A of three variables by FwyV ay MM The basic theorem underlying the method states that if fxy attains a maximum or a minimum at the point 11 under the constraint gxy 0 then there is a value c of A such that 11 0 is a critical point of F 6 F 6F lt1 6 we 0 ewe o 7 6y ab c 0 9F 5 Thus in principle one can nd the candidates for the desired constrained extremum of f by solving the three simultaneous equations 1 for abc In the nicest situations there will be only one solution which gives immediately the sought for extremum ab of f The aim here is to explain the geometric underpinning of the method So assume fxy does have a maximum or a minimum at a b under the constraint gy 0 We shall assume further that ab is a critical point of neither f nor 9 the most common case Note rst that the partial derivatives of F are given by 6F76f Ady 6F76f A69 dFi ez ax ax ey ey ey 6A 9 The third equality in 1 therefore just says that gab 0 ie that 17 b satis es the constraint The other two equalities in 1 can be written as 2 ab icgm ab 70370119 What do these mean To shorten the notation let7s de ne a 2th n gmm a 2796mm 3 gum Rewritten in the new notation 2 becomes 3 a 7007 B 703 Suppose for de niteness that ab is a maximum of fy under the constraint gy 0 and let m fab Consider the level curve fy m see Figure 31 It separates the region where f is larger than m from the region where f is smaller than m On the level curve gxy 0 the function f takes no value larger than m so that curve although it touches the level curve fy m at a7b7 cannot pass through the latter curve it must stay in the region where fy m From this it follows that the two curves fy m and gy 0 share a common tangent line at the point a7b see the gure The tangent lines at 17 b to the curves fx7 y m and gx7 y 0 have the respective equations 4 aa yb07 ampa5yb0 see Supplementary Notes 1 Now simple algebraic reasoning left to the reader shows that the two equations 4 de ne the same line if and only if the coef cients og are proportional to the coef cients 0737 ie7 there is a number 7 such that 04 yo and B 73 This gives 3 with c 77 To summarize7 the rst two equalities in 1 just say that the level curves fy fab and gxy 0 have a common tangent line at the point 11

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