11/22-1. Read 10.1 and 10.2

HW 9.513 due 12/5

HW 9.6 due 1217

Final : 12/17 3:30PM- 5:30PM

Optimization z = fix, y )

zndperiuatiue.ie# Suppose ( x 'T y't ) is a cryptic. I

D)exit,y* ,

= [ fxx . fyy - ffxy) ) /c×*,y*,= #

( i ) D > 0 and fxx y* ) 2 o ⇒ LOCAL MIN

Iii) D 70 and fxxfxtyx ) 20 ⇒ LOCAL MAX

iii) D LO ⇒ SADDLE POINT

Civ D -

- o ⇒ Test Fails

Ey : fix , =x3 Ex : fix = X"

ft' .¥¥

ftxl ' 3×2=0fix ,=

4×3=0⇒ x*=o

⇒x fix ) =

Rx4×*

f' 'm -

- GX= 12.02=0

Test Test Failsf'

'

67=0 Fails

EI : Z = fix, y ) = X2ty2 paraboloid

crit.pt

¥::X ,

⇒ * " 99.

.

..

.

.

I'

o

2nd D test o

-

:

fxx = ( fx ) x= ( 2x )×= 2

fyy = # y= (Lyly = 2

AY = fyx = O

Plan = fxx.fyy-ffx.DZ ⇒

Dl 70

= 2.2 - o = 4 > OFxx y 'T so

fxxko,

. ,= 270 LOCAL MIN

II : Z = fix, yl = Xa - ya

cat pt-

'

.

f× -- 2X = O

⇒ = 0,0 ) •

Fy =- 2y = O

-

2nd D test a

-

:

Fxx = 2,

fyy = - 2,

fxy =fy×=O

Dla,a

= [ 2. C- 2) - 023=-4 LO

⇒ Cgo ) is aSADDLE POINT

Ey : Find and identify the critical points of

fix , yl-- X 't y

'- 9 x y

E. as * * of'#¥÷IJans :

f ,= 3×2 to -9 Y =Ogz ya - q×=3C¥X2%

⇒ I x4 - 9 ×= o ⇒ I x ( x

'- 277=0

so x' to ⇒y¥÷c¥EEi critical ptsor

4*7-27=1, yes,

#5=365--3 ;yf3,D⇒ X

't=3

Apply2tt :

Fxx = ×= (3×2-91)

x= GX fxy = ( 35 - g) y

= - 9

fyy = ffyly = Gf -9×1,= Gy

⇒ D)ex , ,= Fxx . fyy - ftp.I36xy-fgf-36xy - 81

QI : D1 egg= 360.0 - 81 =

- 8120, D LO

ISADDLEPOINTatco.co#! D Icq ,

= 36-3.3-81=9.36- 8170

Fxxkzp)= 6.3=1820 ⇒

D > o,

Fxx > O/L0CACMINatG##

Ex : For fix, y ) = ye 't x y

'

the critical pt is CE,

- e" )

Classify this point.

⇒ ¥¥:¥¥I¥¥'

Fxx = y ex, fyy = 2x

,fxy = e×t2yFix = ex t ay

DIE,

. em) = fxx . fyy - ¥5lez ,-

I' ) a

= f-e'' 7.em. 2. ta - Ce" '

- ad73-e-c-e- e = - 2e so ⇒

Ck ,- e

" I= - e ! SADDLE POINT

Exe : For what value of A- does fix ,yl= - x 't xytyt Ayhave a LOCAL MAX at C - 2

,-4 ) ?

Constrained Optimization-

Contour Plot of fix,y )

Minimize local 10M¥or Z= FIX

, y ) min

mini"su¥ty÷±g!• B. •

gcx ,y)=C

If C x't, y

't ) is a local mo'TMax

of fix ,yl subject to the constraint X

⇒ There exists ① fxcxtyg-xgxcx.tt )}If⇒Fg② fycxtyy-xgylx.FI

a > such that③ gaily ) -

- C

Exe : Find the MIN and MAX of

=L = fix ,yl=7xt7y subject to gcx ,y)=x7yZI