Extrema

I.  Homework

II.  Definition of Relative Max and Min

A function f(x,y) has a relative maximum at (a,b) if there is a delta neighborhood centered at (a,b) such that

f(a,b) > f(x,y)

similarly there is a relative minimum at (x,y) if

f(a,b) < f(x,y)

for all (x,y) in the delta neighborhood.  

Theorem:  If f(x,y) has a relative maximum or minimum then f_x  =  f_y = 0 there.

A visual proof will be given in class.

Example   Let f(x,y) = x² + xy - 2y + x - 1

then f_x  = 2x + y + 1 and  f_y = x - 2

so that x - 2 = 0 or x = 2

Hence 2(2) + y + 1 = 0  so y = 5

A possible extrema is (2,-5)

Theorem (Second Derivative Test For Functions of Two Variables)

We call the matrix:  

fxx	fxy
fyx	fyy

the hessian then the determinant is  

D = f_xxf_yy - f_xy² 

then if f_x  =  f_y = 0 we have the following

D	fxx	Type
>0	>0	Rel Min
>0	<0	Rel Max
<0	any	Saddle
= 0	any	Test Fails

We will discus saddle points in class

Example:

Let f(x,y) = -x² - 5y² + 8x - 10y - 13   

then f_x  =   -2x + 8 and  f_y =  -10y - 10 these are zero at (4,-1)

we have f_xx = -2,  f_yy = -10, and f_xy = 0 Hence

D = (-2)(-10) - 0 > 0

so f has a relative maximum at (4,-1)

Example:  Let

f(x,y) = 4xy - x⁴ - y⁴

The f_x  =  4y - 4x³ and  f_y = 4x - 4y³

We solve:

4y - 4x³ = 0 so that y = x³  

Hence 4x - 4(x³)³ = 0

x - x⁹ = 0 so that x = 1 or x = 0 or x = -1

this gives us the points (1,1), (0,0) and (-1,-1)

We have f_xx = -12x²,  f_yy = -12y², and f_xy =  4

Hence D = 144x²y² - 16

We write the table:

Point	D	fxx	Type
(1,1)	126	-12	Max
(0,0)	-16	0	Saddle
(-1,-1)	126	-12	Max