14
Functions of Several Variables
Book: Calculus and Analytic Geometry (11th Edition) by
George B. Thomas, Jr. and Ross L. Finney
Book: Calculus Early Transcendentals (6th Edition) by
James Stewert.
14.7
Extreme Values and Saddle
Points
Objective: In this section, we will learn how to use partial
derivatives to locate maxima, minima and
saddle points of functions of two variables.
Review (Functions of single variable)
Absolute Extremas
Max: f (c) f (x) x [a, b] y f (x)
Min: f (c) f (x) x [a, b]
f (b)
f (a)
Local Extremas
a b x
Max: f (c) f (x) x [c h, c h]
Min: f (c) f (x) x [c h, c h]
Review (Functions of single variable)
Absolute Extremas
Max: f (c) f (x) x [a, b] y f (x)
Min: f (c) f (x) x [a, b]
f (b)
f (a)
Local Extremas
a b x
Max: f 0 f 0 Geometrically
Tangent line is parallel to x-axis means
Min: f 0 f 0 slope of the tangent is zero.
f0
Multivariable Functions:
Absolute Extremas
Max: f (a, b) f (x, y) (x, y) D
Entire Domain
Min: f (a, b) f (x, y) (x, y) D
Multivariable Functions:
Local Extremas
Max: f (a, b) f (x, y) (x, y) D
An open disk
Min: f (a, b) f (x, y) (x, y) D
Geometrically
Tangent lines are parallel to x and y axes means
slope of both tangent lines is zero, therefore
fx 0 f y
Example:
Example:
Solution:
Therefore the critical points are:
(0,0) and (1,1)
Example:
Solution:
Example:
Solution:
• This example illustrates the fact that a function
need not have a maximum or minimum value at
a critical point.
• This fact leads to the definition of a Saddle
Point.
We can see that f(0, 0) = 0 is:
A maximum in the direction of the x-axis.
A minimum in the direction of the y-axis.
Near the origin, the graph has the shape of a saddle.
So, (0, 0) is called
a saddle point of f.
Multivariable Functions:
Saddle Point: A critical point at which f(x,y) does not have a relative
extrema is called a saddle point.
Max: f (a, b) f (x, y) (x, y) D
Both conditions hold !!!
Min: f (a, b) f (x, y) (x, y) D
Geometrically
Even in this case, the slopes of both
tangents is zero
fx 0 f y
Neither maxima nor minima.
EXTREME VALUE AT CRITICAL POINT
• We need to be able to determine whether or not a
function has an extreme value at a critical point.
• The following test is analogous to the Second
Derivative Test for functions of one variable.
Theorem: Second Derivative Test
If f (x,y) and its first and second derivatives are continuous near
a critical point (a,b) then
f can have a local maximum or local minimum at (a, b), or (a, b)
can be a saddle point of f.
Theorem: Second Derivative Test
The expression , is called the discriminant or
Hessian of ƒ. It is sometimes easier to remember it in determinant
form,
f xx f xy
D f xx f yy f xy 2
f xy f yy
Example:
Solution:
Step I: Find the critical points
f x 2x , f y y 2 1
f x -2x 0 x0
f y y 2 1 0 y 1
Critical points are (0,1) and (0,-1)
Step II: Find the Second derivatives
f xx 2 f xy 0 f yy 2y
D f xx f yy f xy2 2(2y) 0 4y
Geometrically, the surface can be drawn on any computer algebra
system which verifies that the there is a local maxima and one saddle
point.
Example: Locate and identify the critical points of the following
function
f (x, y) 4xy x 4 y 4
Solution:
Step I: Find the critical points f x 4 y 4x 3 , f y 4x 4 y 3
f x 4 y 4x 3 0 y x3
f y 4x 4 y 3 0 x x 9 0
x(1 x 8 ) 0 x 0,1,1
Critical points are (0,0), (1,1) and (-1,-1)
Step II: Find the Second derivatives
f xx 12x 2 f xy 4 f yy 12 y 2
D f xx f yy f xy2 12x 2 (12 y 2 ) 16 144x 2 y 2 16
Step III: Test the critical points
At (0,0) D 144x 2 y 2 16 -16 0
f (0,0) 0
0,0,0 Is a saddle point
At (1,1) D 144x 2 y 2 16 128 0
f xx 12x 2 12 0
f (1,1) 2
1,1,2 Is a maximum point
At (-1,-1) D 144x 2 y 2 16 128 0
f xx 12x 2 12 0
f (1,1) 2
-1,-1,2 Is a maximum point
Example: Locate and identify the critical points of the following
function
1 1
f (x, y) xy ,
x 0, y 0
x y
Solution:
Step I: Find the critical points
f x 1/ x 2 y , f y x 1/ y 2
f x 1/ x 2 y 0 y 1/ x 2
f y x 1/ y 2 0 x x 4 0
x(1 x 3 ) 0 x 0,1
The only critical point is: (1,1)
Step II: Find the Second derivatives
f xx 2 / x 3 f xy 1 f yy 2 / y 3
D f xx f yy f xy2 (2 / x 3 )(2 / y 3 ) 1 4 / x 3 y 3 1
Step III: Test the critical points
At (1,1) D 4 / x 3 y 3 1 3 0
f xx 2 / x 3 2 0
f (1,1) 2
1,1,2 Is a minimum point
Example: Find the absolute extremum of the function
f (x, y) x 2 xy y 2 6x 2
defined in the region R: R (x, y) : 0 x 5,3 y 0
Solution:
Solution: f (x, y) x 2 xy y 2 6x 2
defined in the region R: R (x, y) : 0 x 5,3 y 0
Interior points y
f x 2x y 6 , fy 2 y x A B x
f y 2 y x 0 x 2 y
f x 2x y 6 0 3y - 6 0 D
C
Critical point is (4,-2)
f (4,2) (4) 2 4(2) (2) 2 6(4) 2 10
Boundary points
1: Along AB: y=0
f (x,0) x 2 6x 2
df
2x 6 0 x 3
dx
Critical point on AB is (3,0)
f (3,0) (3) 2 6(3) 2 7
End points of AB
f (0,0) (0) 2 6(0) 2 2
f (5,0) (5) 2 6(5) 2 3
2: Along BC; x=5
f (5, y) y 2 5y 3
df
2 y 5 0 y 5 / 2
dy
Critical point on BC is (5,-5/2)
f (5,5 / 2) (5 / 2) 2 5(5 / 2) 3 37 / 4
End points of BC
f (5,0) (0) 2 5(0) 3 3
f (5,3) (3) 2 5(3) 3 9
3: Along CD; y=-3
f (x,3) x 2 9x 11
df
2x 9 0 x 9 / 2
dx
Critical point on AC is (9/2,-3)
f (9 / 2,3) (9 / 2) 2 9(9 / 2) 11 37 / 4
End points of CD
f (5,3) (5) 2 9(5) 11 9
f (0,3) (0) 2 9(0) 11 11
4: Along AD; x=0
f (0, y) y 2 2
df
2y 0 y 0
dy
Critical point on AD is (0,0)
f (0,0) 0 2 2 2 y
End points of AD
f (0,3) (3) 2 2 11 A B x
f (0,0) 0 2 2 2 D
C
Absolute Maximum f (0,3) 11
Absolute Minimum f (4,2) 10
y
A B x
Absolute Maximum f (0,3) 11 D
C
Absolute Minimum f (4,2) 10
Example:
d ( x 1) y ( z 2)
2 2 2
d ( x 1) y (6 x 2 y)
2 2 2
We can minimize d by minimizing the simpler expression
d f ( x, y )
2
( x 1) y (6 x 2 y)
2 2 2
By solving the equations
f x 2( x 1) 2(6 x 2 y ) 4 x 4 y 14 0
f y 2 y 4(6 x 2 y ) 4 x 10 y 24 0
we find that the only critical point is ( 11 , 5 ) .
6 3
( 116 , 53 )
d ( x 1) y (6 x 2 y )
2 2 2
5 2
6 5 2
3 6
5 2
5
6 6
5
6 6
Example:
A rectangular box without a lid is to be made from 12 m2
of cardboard. Find the maximum volume of such a box.
Solution:
12 xy 12 xy x 2 y 2
V xy
2( x y) 2( x y)
We compute the partial derivatives:
V y 2 (12 2 xy x 2 )
x 2( x y ) 2
V x 2 (12 2 xy y 2 )
y 2( x y ) 2
