Extreme Values and Saddle Points

De…nitions Let f (x; y) be de…ned on a region D containing the point (a; b).
Then
1. f (a; b) is a local maximum value of f if f (a; b) f (x; y) for all domain
points (x; y) in an open disk centered at (a; b).
2. f (a; b) is a local minimum value of f if f (a; b) f (x; y) for all domain
points (x; y) in an open disk centered at (a; b).
An open disk centered at (a; b) is a set of points (x; y) for which
p
(x a)2 + (y b)2 < c (1)
for some may be small enough c.
Test for the maximum, minimum.
If f (x; y) is di¤erentiable at the point (a; b), and the function f (x; y) has a
maximum or minium at the point (a; b) then the partial derivatives at this point
is 0.
A di¤erentiable function f (x; y) has a saddle point at a critical point (a; b)
if in every open disk centered at (a; b) there are domain points (x; y) where
f (x; y) > f (a; b) and domain points (x; y) where f (x; y) < f (a; b). The corre-
sponding point (a; b; f (a; b)) on the surface z = f (x; y) is called a saddle point
of the surface.
Theorem. Second Derivative Test for Local Extreme Values
Suppose that f (x; y) and its …rst and second partial derivatives are con-
tinuous throughout a disk centered at (a; b) and that fx (a; b) = fy (a; b) = 0.
Then
2
i) f has a local maximum at (a; b) if fxx < 0 and fxx fyy fxy > 0 at (a; b).
2
ii) f has a local minimum at (a; b) if fxx > 0 and fxx fyy fxy > 0 at (a; b).
2
iii) f has a saddle point at (a; b) if fxx fyy fxy < 0 at (a; b).
2
iv) the test is inconclusive at (a; b) if fxx fyy fxy = 0 at (a; b). In this case,
we must …nd some other way to determine the behavior of f at (a; b).
Example 1. f (x; y) = x2 + y 2 xy + 2x:

fx = 2x y+2
fy = 2y x

That is to …nd the candidate point for max, min and saddle points we need to
solve the system

2x y+2 = 0
2y x = 0

Solution is: x = 34 ; y = 23 . Thus the point ( 43 ; 32 ) is the candidate point.

Now we calculate the second derivatives at this point

1
 fxx = 2
fyy = 2
fxy = 1
2
fxx fyy fxy = 2 2 ( 1)2 = 4 1 = 3 > 0
fxx > 0; that is the point ( 43 ; 23 ) is a point of minimum.
x2 + y 2 xy + 2x

80

60

z 40
-4 -4
20
-2 -2
0
0 0

y2 2x
4 4

Example 2.Find the local extreme values of f (x; y) = xy x2 y2 2x 2y + 4

Partial derivatives:

fx = y 2x 2
fy = x 2y 2

Thus to …nd the candidates for max, min.. we need to solve the system

y 2x 2 = 0
x 2y 2 = 0

Solving we get a unique solution [x = 2; y = 2] : That is candidate point is

(-2,-2).
Now calculate the 2nd derivatives

2
 fxx = 2; fyy = 2; fxy = 1
2
and the expression fxx fyy fxy = 4 1 > 0 is positive (always). On the other
hand fxx = 2 < 0: Test says that this point is a point of maximum and the
value of the function is

f ( 2; 2) = 4 4 4+4+4+4=8
Example 3.Find the local extreme values of f (x; y) = 6xy 2y 3 + 3y 2 3x2
Partial derivatives:

fx = 6y 6x
fy = 6x 6y 2 + 6y

Thus to …nd the candidates for max, min.. we need to solve the system

6y 6x = 0
2
6x 6y + 6y = 0

From …rst one we obtain y=x and so roots are (0,0) and (2,2): That is candidate
points:(0,0) and (2,2).
Now calculate the 2nd derivatives

fxx = 6; fyy = 12y + 6; fxy = 6

2
1) …rst candidate (0,0): The expression fxx fyy fxy = ( 6)( 12y + 6) 36:
replace x=0,y=0:
2
fxx fyy fxy = 36 36 < 0
that is (0,0) is saddle point
2
2) second candidate (2,2):In the expression fxx fyy fxy = ( 6)( 12y + 6)
36: replace x=2,y=2:
2
fxx fyy fxy = ( 6)( 24 + 6) 36 = 72 > 0;
We have fxx = 6 < 0 and therefore this point is a point of maximum.
Example 4**.Find the local extreme values of f (x; y) = x3 +y 3 +3y 2 3x2 +1
Partial derivatives:

fx = 3x2 6x
2
fy = 3y + 6y

Thus to …nd the candidates for max, min.. we need to solve the system

3x2 6x = 0
2
3y + 6y = 0

3
DI·KKAT EDI·N, BU EN FAZLA YANLIŞ YAPILAN ÖRNEK TÜRLERI·NDEN
BI·RI·DI·R, OKULDA BUNLARI I·YI· ÖGRENM ¼ I·YORUZ GALI·BA.
This system has 4 solutions!!! [x = 0; y = 2] ; [x = 2; y = 2] ; [x = 0; y = 0] ;
[x = 2; y = 0]
Then we calculate second derivatives:

fxx = 6x 6; fyy = 6y + 6; fxy = 0

and calculate this expression separately for all candidates:
1) (0,-2):
fxx = 6; fyy = 6; fxy = 0

2
fxx fyy fxy = 36;
fxx < 0

maximum point
2) (2,-2)
fxx = 6; fyy = 6; fxy = 0
2
fxx fyy fxy = 36;
saddle point
3) (0,0)

fxx = 6; fyy = 6; fxy = 0

2
fxx fyy fxy = 36

saddle point
4) (2.0)

fxx = 6; fyy = 6; fxy = 0

2
fxx fyy fxy = 36
fxx > 0

minimum point
Example 5.Find the local extreme values of f (x; y) = x4 + y 4 4xy
Partial derivatives:

fx = 4x3 4y = 0
3
fy = 4y 4x = 0

x3 y = 0
3
y x = 0

birinciden y-yi bul ikincide yaz

4
 y = x3
9
x x = 0

Roots: DI·KKAT EDI·NI·Z, burada da hatalar çok fazla oluyor

x(x8 1) = 0

x1 = 0; x2 = 1; x3 = 1
In total ve have 3 roots (y=x3 oldu¼
gundan)
(0,0),(1,1)(-1,-1)
Bunlar¬birer-birer denemek gerekiyor arkas¬kolay....

Absolute maxima and minima on closed

bounded region
If we are interested in absolute max, min problems on some closed bounded
region, we proceed like in the case of a single variable functions:
THE EXTREME VALUES OF f (x; y) OCCUR ONLY at
1) boundary points
2) interior ponts where fx = fy = 0 (or derivatives fails to exist)
Example. Find the absolute max and min values of f (x; y) = 2 + 2x + 4y
x2 y 2 on the region in the …rst quadrant enclosed by the lines x=0, y=0 and
y=9-x.
First we …nd …rst derivative to …nd candidates in interior points:

fx = 2 2x = 0
fy = 4 2y = 0

x=1,y=2 this candidate point (1,2) belongs to the region under consideration.
Calculate the value of the function at this point

f (1; 2) = 2 + 2 + 8 1 4=7
Now we consider boundary. şimdi s¬n¬r hatlar¬nda max,min ar¬yoruz
1) line x=0: replace x=0 in the function f (x; y)x=0 = 2 + 4y y 2 = g(y):
That is we need to …nd a max,min of a "normal" function g(y) = 2 + 4y y 2
over the interval [0,9]. Burada bir de¼gişkenli fonksiyona geçiyoruz yani birinci
dönem:
derivative =g 0 (y) =4-2y=0, that is y=2. g(2)=f (0; 2)=2+8-4=6. Endpoints
y=0 and y=9. g(0) = f (0; 0) = 4; g(9) = f (0; 9) = 2 + 4 9 81 = 43
Yani x=0 hatt¬üzere f (x; y) nin en küçük de¼ geri -43 en büyük de¼
geri 6-d¬r

5
 2) line y=0:replace y=0 in the function f (x; y)y=0 = 2+2x x2 = h(x):That
is we need to …nd a max,min of a "normal" function h(x) = 2 + 2x x2
over the interval [0,9].h0 (x) = 2 2x = 0; x = 1: h(1)=f (1,0)=2+2-1=3,
h(0)=f (0; 0)=2,h(9)=f (9; 0)=2+2*9-81= 61
Yani y=0 hatt¬üzere f (x; y) nin en küçük de¼ geri -61 en büyük de¼geri 3-dür
3) line y=9-x:replace y=9-x in the function f (x; y)y=9 x = 2 + 2x + 4(9
x) x2 (9 x)2 = 2x2 + 16x 43 = v(x)
yine bir de¼gişkene geldik. v 0 (x) = 4x + 16 = 0; x = 4; v(4) = f (4; 5) =
2 16 + 16 4 43 = 11: Endpoints 0 and 9:
v(0) = f (0; 9) = 43; v(9) = f (9; 0) = 2 81 + 16 9 43 = 61
We list all candidate values: tüm adaylardaki de¼ gerlere bak¬yoruz

7; 6; 4; 43; 3; 2 61; 11; 43; 61

The smallest one is the min value and highest one is the max value. En
küçüyümin en büyüyü max olacak
Cevap: f (9; 0) = 61 min value and f (1; 2) = 7 max value
Exercise: Find the absolute max and min values of f (x; y) = 2x2 4x +
y 2 4y + 1 on the closed triangular region bounded by lines x=0, y=2, y=2x.