MAXIMUM AND MINIMUM VALUES
ELE CT RON IC V ER SI ON O F LE CT UR E
HoChiMinh City Universityof Technology
Faculty of Applied Science, Department of Applied Mathematics
(HCMUT-OISP) MAXIMUMAND MINIMUM VALUES 1/ 51
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Local and Absolute Extrema of Functions in Two Variables, Schemes and Mind Maps of Mathematics
The concepts of local and absolute extrema of functions in two variables, including definitions, necessary and sufficient conditions, methods to find local extrema, and the use of lagrange multipliers for constrained optimization. Examples and exercises are provided.
Typology: Schemes and Mind Maps
2021/2022
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MAXIMUM AND MINIMUM VALUES
ELECTRONIC VERSION OF LECTURE
HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics
OUTLINE
(^1) LOCAL MAXIMA AND MINIMA
(^2) CONSTRAINED OPTIMIZATION AND LAGRANGE MULTIPLIERS
(^3) ABSOLUTE MAXIMUM AND MINIMUM VALUES
Local Maxima and Minima Definition
DEFINITION 1.
1
A function f ( x , y ) has a local maximum at ( x 0 , y 0 )
if f ( x , y ) É f ( x 0 , y 0 ), when ( x , y ) is near ( x 0 , y 0 ). The
number f ( x 0 , y 0 ) is called a local maximum value.
2 A function f ( x , y ) has a local minimum at ( x
0 ,^ y 0 )
if f ( x , y ) Ê f ( x 0 , y 0 ), when ( x , y ) is near ( x 0 , y 0 ). The
number f ( x 0 , y 0 ) is called a local minimum value.
Local Maxima and Minima Definition
REMARK.
1 If f ( x , y ) É f ( x
0 ,^ y 0 ),^ for all^ ( x ,^ y )^ ∈^ D^ f then^ f^ has an
absolute maximum at ( x 0 , y 0 ).
2 If f ( x , y ) Ê f ( x
0 ,^ y 0 ),^ for all^ ( x ,^ y )^ ∈^ D^ f then^ f^ has an
absolute minimum at ( x 0 , y 0 ).
Local Maxima and Minima Necessary Condition for Local Maximum and Local Minimum
PROOF.
Let g ( x ) = f ( x , y 0 ). If f has a local maximum (or local
minimum) at ( x 0 , y 0 ), then
g ( x ) = f ( x , y 0 ) É f ( x 0 , y 0 ),
when x is near x 0. By Fermat’s Theorem for function
of one variable g ( x ), we have g
′ ( x 0 ) = 0
⇒ g
′ ( x 0 ) = f
′ x ( x^0 ,^ y^0 )^ =^ 0.
Similarly, by applying Fermat’s Theorem to the
function h ( y ) = f ( x 0 , y ), we obtain f
′ y ( x 0 , y 0 ) = 0 ■
Local Maxima and Minima Sufficient Condition for Local Maximum and Local Minimum
THEOREM 1.
Suppose the second partial derivatives of z = f ( x , y )
are continuous on a disk with center ( x 0 , y 0 ), and
suppose that f
′ x
( x 0 , y 0 ) = 0 and f
′ y
( x 0 , y 0 ) = 0. Let
A = f
′′ xx ( x 0 , y 0 ), B = f
′′ x y ( x 0 , y 0 ), C = f
′′ y y ( x 0 , y 0 ),
A B
B C
= AC − B
2 .
1
If
A > 0
then f ( x 0 , y 0 ) is a local minimum.
2 If
A < 0
then f ( x 0 , y 0 ) is a local maximum.
3 If ∆ < 0, then f ( x
0 ,^ y 0 )^ is^ NOT a local maximum or
minimum (the point ( x 0 , y 0 ) is a saddle point).
Local Maxima and Minima How to Find Local Maximum and Local Minimum
HOW TO FIND LOCAL MAXIMUM AND LOCAL
MINIMUM II
If ∆ > 0, A > 0, then ( xi , yi ) is minimum point.
If ∆ > 0, A < 0 then ( xi , yi ) is maximum point.
If ∆ < 0, then ( xi , yi ) is a saddle point.
If ∆ = 0, then anything is possible. More advanced
methods are required. For example, we consider
∆ f = f ( x , y ) − f ( xi , yi )
Local Maxima and Minima How to Find Local Maximum and Local Minimum
EXAMPLE 1.
Find the local maximum and minimum values and
saddle points of f ( x , y ) = x
3
- 2 y
3 − 3 x
2 − 6 y
Step 1. Finding Stationary Points
f
′ x =^3 x
2 − 6 x = 0
f
′ y = 6 y
2 − 6 = 0
⇒ The 4 stationary points are
P 1 (0, −1), P 2 (0, 1), P 3 (2, −1), P 4 (2, 1).
Local Maxima and Minima How to Find Local Maximum and Local Minimum
P 2 (0, 1), A = f
′′ xx (0, 1) = −6, B = f
′′ x y
C = f
′′ y y
∆ = AC − B
2 = (−6).(12) − (0)
2 < 0.
⇒ P 2 is a saddle point.
P 3 (2, −1), A = f
′′ xx (2,^ −1)^ =^ 6,^ B^ =^ f^
′′ x y (2,^ −1)^ =^ 0,
C = f
′′ y y (2,^ −1)^ = −12,
∆ = AC − B
2 = (6).(−12) − (0)
2 < 0.
⇒ P 3 is a saddle point.
Local Maxima and Minima How to Find Local Maximum and Local Minimum
P 4 (2, 1), A = f
′′ xx (2, 1) = 6, B = f
′′ x y
C = f
′′ y y
∆ = AC − B
2 = (6).(12) − (0)
2
A > 0
⇒ P 4 is local minimum point,
f
LocMin
= f (2, 1) = −8.
Local Maxima and Minima How to Find Local Maximum and Local Minimum
EXAMPLE 1.
A rectangular box without a lid is to be made from
12m
2
of cardboard. Find the maximum volume of
such a box.
SOLUTION Let the length, width, and height of the
box be x , y , and z. Then the volume of the box is
V = x y z.
We can express V as a function of just two variables x
and y by using the fact that the area of the four sides
and the bottom of the box is
2 xz + 2 y z + x y = 12
Local Maxima and Minima How to Find Local Maximum and Local Minimum
Solving the equation for z, we get
z = (12 − x y )/[2( x + y )]
so the expression for V becomes
V = x y
12 − x y
2( x + y )
12 x y = x
2 y
2
2( x + y )
We find the stationary points ( x , y ) by solving the
system of equations
∂V
∂x
y
2 (12 − 2 x y − x
2 )
2( x + y ) 2
∂V
∂y
x
2 (12 − 2 x y − y
2 )
2( x + y )
2
Constrained Optimization and Lagrange Multipliers
CONSTRAINED OPTIMIZATION AND LAGRANGE
MULTIPLIERS
In many applications, the goal is not to identify
theoretical maximum or minimum values, but to
achieve the absolute best possible product given
a large set of constraints such as limited
resources or technology.
We consider the following equality constrained
optimization problem
Maximize (or Minimize) the function f ( x , y )
subject to the condition ϕ ( x , y ) = k
Constrained Optimization and Lagrange Multipliers Necessary Condition for equality constrained problem
THEOREM 2.
If f ( x , y ) and ϕ ( x , y ) have continuous first partial
derivatives and f ( x 0 , y 0 ) is an extremum of f , subject
to the constraint ϕ ( x , y ) = k and
ϕ
′ x ( x 0 , y 0 ) ̸= 0, ϕ
′ y
( x 0 , y 0 ) ̸= 0 then exist some constant λ
such that
f
′ x ( x^0 ,^ y^0 )^ −^ λϕ
′ x ( x^0 ,^ y^0 )^ =^0
f
′ y ( x 0 , y 0 ) − λϕ
′ y ( x 0 , y 0 ) = 0
ϕ ( x 0 , y 0 ) = k