Maxima and Minima of Functions of Several Variables Critical Points and Local Extrema Solutions

Maxima and Minima of Functions of Several Variables ( from Stewart , Calculus, Chapter 14) Critical Points and Local Extrema JA Critical Point: A point ( a , 6) in the domain of / where fx ( a , b ) = 0 and fv ( a, b ) = 0 or where one of these partial derivatives does not exist. local maximum y Thm. If / has a local maximum or local minimum at ( a, 6) then ( a , b ) is a critical point of /. Is ** T Sv Second Derivative Test Suppose the second partial derivatives of / are continuous near ( a, b ) and suppose that fx ( a , b ) = 0 and fy ( a , b ) = 0 ( that is, (a , b ) is a critical point of / ) Let f {** (/«)(/ ) -(A )2 D( x, y ) Jyy x y J ‘ be the determinant of the Hessian matrix of second order partial derivatives of /. j (x - w "? 0 =*> foorave 'A f v (a) If D( a , b ) > Q and fXx ( a , b ) > 0, then / ( a , b ) is a local minimum. | ( b) If D ( a , b ) > 0 and fxx (.a , b ) < 0, then / (a , b ) is a local maximum. P/x 6 ? CcMayf (c) If D( a , 6) < 0, then fa , b ) is a saddle point. (d ) If D = 0, the test gives no information - / could have a local maximum or local minimum at (a, b ) , or ( a , b ) could be a saddle point. -\ - ] . - , „ ^ - ' ^ local minimum absolute \ minimum . , W absolute maximum . Ex. 1 Let f { x , y ) a:4 + 2 y 2 - 4 xy. Find all critical points of /, and classify each as a local maximum , local minimum , or saddle point. = f * ( x . jl - “lx 1 - 1l r L " IS ( i; y- X l k > vt W -v t S \ i 0* c ts - X), U • r X* r\ & \ Q C tw.M = o 3> fc, dV o - f-'fi 12 s, Jr (0 .4 ^ A A = “i / ^ l^ r-i -C\ - 2.7 0 * 75 X Ces i tffx’-A "0 Cx-iiCx ( r r - W * «\ fi . 1 \.yji r . ; i r\ C\ 0 a.+ " ( ro 70 min I Co, o^) . + * '"+ -i > local I ! ;>V X* U .o f a. 4 • , fi ' f - *9 -' A Method of Lagrange Multipliers To find the maximum and minimum values of a function / subject to the constraint g = k for some constant k ( assuming these extreme values exist and Vg 0 when g = k ): ^ 1. Find all points where g=k and V / = A Vs for some A 6 R . These values of A are called the Lagrange multipliers. 2. Evaluate / at each of the points from Step 1. The largest is the maximum value of the minimum value of / , subject to the constraint . / and the smallest is Ex. 2. Find the absolute maximum and minimum values of f ( x , y ) = x2 + 2y2 on the circle x 2 + y2 * 1. zA z - x2 + 2 y2 i - \ c liihn vmsjc o f HVR. r - C>nilL ^ .rJ + ff 00 +1«. W of r £ y y= l * AM J 'v X l * ' -f ±' i HjX^ atl JW- fc [\ +0 (f \1fdLV /. vH. ^ i ^*^ ( i Vs'Ji , -v aW '1 ( Oj ^ A if'fs ' ^ Q *h ' Vft \ . * 01 A 5 ? tj^ Aie- V^ h ocA , ' A fidhct 4Wj l u f'eA'MA^rrt- - 5- 7 X *0 iI (i b &V : ^ *4 o f f 0« -VM -to* Ui . imO , YA^ ‘ -5 or =7 1' 0 f w w r* '* ’ o ± rt = l_ — x . CM 2 x. - J 1 JUJVX ^ ^ ^ StMJt -VVn x 1 + 2v* = 1 4 WvjV < 2*, / " c^n/ e TU 2 0 i/ T 2 of f i » Kitl« \i occurs of OCLVI ^ \ (L faUY / ex . Absolute Maxima and Minima If f ( x , y ) is continuous on a closed , Extreme Value Theorem for Functions of Two Variables 2 bounded set D in R , then / attains an absolute maximum value / (11, 2/1 ) and an absolute minimum value / jx2, 2/2 ) at some points (21, 1/1) and (12, 2/2) in D. /These extreme values occur either at the critical points of / or on the boundary of D . . . To find ( 1) (2) (3) ~ the absolute maximum and minimum values of a continuous function / on a closed, bounded set D : Find the values of / at each critical point in D. Find the maximum and minimum of f ( x , y ) on the boundary of D. The largest of these values is the absolute maximum and the smallest is the absolute minimum. — Ex. 3. Find the points at which the absolute maximum arid minimum values of the function f { x , y ) 2x2 + 3i/2 - A x - 5 on the disk x2 + y2 < 16 occur. State all points where the extrema occur as well as the maximum and minimum values. Cyvh u,\ tiV -u > ) =r> \H e < s - ' ^ 2. It - S * - “ il , 2 2 Hb - S |S ' £ (- 2, i(n) - $ *UUK «\> S ^ min H )M ~ s . ~ >x X - ~2 4 Vsfe 41 -0 occurs cuf <> f ^ X( lAo . \ V * rw V. 0 a f Evi*Uk. £ ( V ) ° ' i* u' SX 3) 3 ) — f ii ~ 7 a-V ( Zj ' Applications . Ex. 4 Find the shortest distance from the point (1, 0, -2) to the plane x + 2 y + z = 4. frrrt 1 -v * < -AVj = > • 2« 2 ) ^^ So . -^ kt + 2-ftx-ii + * -T =4 ' H/6 ’ 4 - 1 -- ^a. Or + +2 = o-M O 1 ) 1! TW A + ZJ + 2 = 2 j) = t )v ? S A A 2 +2 If 2 +2I = X ** ~7A l = “ i *A 3 1 £ + * " % - 7 /b ' t> - > + £ wfc . . 4b + «/v. ^ -+ v ft ^ , » Ex. 5 A rectangular box without a lid is to be made from 12 m 2 of cardboard Find the maximum volume of such a box. * . Vfo/ S O'Xj -fc Mjsd f f) X ® © y X 2r = * - f§) V /) /Voie So VC d «.cK K +v > ^ XOr A= ^ -Hv4 TXrj , & ^0 ^ - ^ " . o fy\< \-{- H« S JwiwtVr TW +KSL ^ ^ ) ^ ^ or X ^ => " -Hs X= = 2 Pyfc * *3 ^22^ 2: X * So - TU - - * 2* ^ =1 Z ICZ ^ ^ 12 ^ -- I ^ =± l 1^ * <} i + 1( 2* )* + 2» & 2r 2 VC 2 \) 0> * (f > ^ |fY| \ >0SM bit b.L. +Kin >tf 3, ar t rvwuv t«. 0 VflW /ne u ’3 V\t + Zji + A 3 = 2 = X + 2. * 1 Ll 0 + Xj \C 2? *;j x 9 , l U i ( 2At ZhXfr * 2K$ }\( Zi + X ) ” 3 Xx, =ZX sulci. -zr O 2 X ( 2x+ 2.0 ° 2 Can 1 " 1 in , \i w(V =\ , X =3 = 2 - sO 1 fc