Lecture 4,Extremum

8/2/2019 Lecture 4,Extremum

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Instructor Dr. Rehana Naz 1 Mathematical Economics II

Lectures 4

Section 11.4 from Fundamental methods of Mathematical Economics, McGraw Hill 2005, 4 th Edition.

by A. C. Chiang & Kevin Wainwright is covered. Please read this section from book. The brief

summary of this section and some examples are solved here for better understanding.

Two-variables case: We have already discussed the extremum of function of two choice variables in Lecture 1.

Here we will express second order conditions using leading principal minors and principal minors of a matrix.

First-order condition for extremum

Given a function of twochoice variables

= (,) A first order necessary condition for the existence of a relative maximum or a relative minimum of a function f

whose partial derivatives,exist is

= 0,

= 0

which yields the critical values ,Sufficient conditions:

Determinantal test for relative extremum: For the function = (,), the second order total differentialcan be written as

= [ ] The matrix

= is known as the Hessian matrix for function = (,).The leading principal minors of H are denoted by

|| = ||,|| = || The second order sufficient condition for an extremum of z is as follows:

1. is a maximum if|| < 0,|| > 0 ( is negative definite).2.

is a minimum if |

| > 0,|

| > 0(

is positive definite) .

3.

Neither maximum nor minimum if any of nonzero leading principal minor does not fit in above twosign patterns. ( is indefinite)In using this condition, we must evaluate all the leading principal minors at the critical value critical

values ,.


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Objective functions wih more than two variables

First-order condition for extremum

Given a function of three choice variables

= (,,) A first order necessary condition for the existence of a relative maximum or a relative minimum of a function f

whose partial derivatives,, exist is = 0, = 0, = 0

which yields the critical values

, ,.

Sufficient conditions:

Determinantal test for relative extremum: For the function = (,,), the second order totaldifferential can be written as

= [ ]

The matrix

=

is known as the Hessian matrix for function = (,,).The leading principal minors of H are denoted by

|| = | |,|| = ,|| = ||The second order sufficient condition for an extremum of z is as follows:

1.

is a maximum if|

| < 0,|

| > 0,|

| < 0 (

is negative definite).

2.

is a minimum if |

| > 0,|

| > 0,|

| > 0 (

is positive definite).

3. Neither maximum nor minimum if any of nonzero leading principal minor does not fit in abovetwo sign patterns. ( is indefinite)

In using this condition, we must evaluate all the leading principal minors at the critical value critical

values , ,.


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n-variable case: Leading Principal minors test

Condition Maximum Minimum

First-order necessary

condition

= 0, = 0, = 0, = 0 = 0, = 0, = 0, = 0Second-order sufficient

condition

|| < 0,|| > 0,|| < 0,..Even order leading principal minors>0

Odd order leading principal minors 0,|| > 0, || > 0, | | > 0 is positive definite .


values , , .Example 1: Given

= (,,) = 2 + 2 + 2 + 4 2(a) Find the extreme values of function.(b) Check whether they are maxima or minima. Find also minimum or maximum value of function.Solution:

(a) The first order partial derivatives are = 2 2 + 2

=

2

+ 4

= 2 + 8 2The first-order necessary condition for maximum or minimum is

= 2 2 + 2 = 0 = 2 + 4 = 0 = 2 + 8 2 = 0This linear system yields = 1, = , = .(b) The Hessian matrix is

= 2 2 22 4 02 0 8

The leading prinical minors of H are

|| = |2| = 2,|| = 2 22 4 = 4, || = || = 2 2 22 4 02 0 8

= 16


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Since all leading principal minors are positive, ( is positive definite), we conclude that relative minimumoccurs at the critical point and minimum value is = .Example 2: Given

=

(

,

,

) = 25

(a) Find the extreme values of function.(b) Check whether they are maxima or minima. Find also minimum or maximum value of function.

Solution:

(a) The first order partial derivatives are = 2 = 2 = 2

The first-order necessary condition for maximum or minimum is

= 2 = 0, = 2 = 0, = 2 = 0This linear system yields = 0, = 0, = 0.

(b) The Hessian matrix is = 2 0 00 2 0

0 0 2The leading prinical minors of H are

|

| = |

2| =

2,|

| =

2 0

0 2=

4,|

| = |

| =

2 0 0

0

2 0

0 0 2=

8

Since || < 0,|| > 0, || < 0, ( negative definite), we conclude that relative maximum occurs at thecritical point and maximum value is = 25.Caution: The test of leading principals minors fails if any of leading principal minors is zer and nonzero one fit

in attren for maximum or minimum. In such a case maximum and minimum can be checked by calculating all

principal minors of Hessian matrix.

n-variable case: Principal minors test

Condition Maximum Minimum

First-order necessarycondition = 0, = 0, = 0, = 0 = 0, = 0, = 0, = 0Second-order necessary

condition condition

Even order principal minors0Odd order principal minors0 is negative semidefinite.

All principal minors 0. is positive semidefinite.


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values , , .Example 3: Given

= (,,) = 2 + 6 + 2 6 + 5(a) Find the extreme values of function.(b) Check whether they are maxima or minima. Find also minimum or maximum value of function.

Solution:

(a) The first order partial derivatives are found and set to zero = 6 + 6 = 0 = 2 2 = 0 = 6 12 = 0

This linear system yields = , = 1, = and = 0, = 1, = 0.(b) The Hessian matrix is

= 12 0 60 2 06 0 12

Evaluation at = , = 1, = The leading prinical minors of H are

|| = 12(12) = 6,|| = 6 00 2 = 12,|| = || = 6 0 60 2 06 0 12 = 72

Since |

| < 0,|

| > 0,|

| < 0, (

negative definite), we conclude that relative maximum occurs at

the critical point and maximum value is = .Evaluation at = , = , = The leading prinical minors of H are

|| = |12(0)| = 0,|| = 0 00 2 = 0,|| = || = 0 0 60 2 06 0 12 = 72

The non-zero leading principal minor || > 0 suggests that may be positive semi definite. The leadingprincipal minors test fails for the point = , = , = . In this case we need to calculate all principalminors.

The first order principal minors are

0, -2, -12.

All first order principal minors 0But third order principal minor > 0 therefore is indefinite. Neither maximum nor minimum occurs at thispoint.


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Example 4: Find the extreme value for following function

= (,,) = + 3 + 2 3Check whether they are maxima or minima.

Solution:

The first order partial derivatives are found and set to zero

= 3 + 3 = 0 = 2 = 0 = 3 6 = 0This system yields two solutions:

= 0, = 1, = 0 implying = 1

=

,

= 1,

=

implying

=

The Hessian matrix is

= 6 0 30 2 03 0 6

Evaluation at = , = , = The leading prinical minors of H are

|| = |6(0)| = 0,|| = 0 00

2 = 0,|| = || = 0 0 30 2 0

3 0

6

= 18The non-zero leading principal minor || > 0 suggests that may be positive semi definite. The leadingprincipal minors test fails for the point = , = , = . In this case we need to calculate all principalminors.

The first order principal minors are

0, -2, -6.

All first order principal minors 0But third order principal minor > 0 therefore is indefinite. Neither maximum nor minimum occurs at thispoint.

Evaluation at = , = 1, = The leading prinical minors of H are

|| = 6(12) = 3,|| = 3 00 2 = 6,|| = || = 3 0 30 2 03 0 6 = 18


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Since || < 0,|| > 0,|| < 0, ( negative definite), we conclude that relative maximum occurs atthe critical point and maximum value is = .

Example 5: Given

=

(

,

,

,

) = 8

+ 4

+ 2

+ 5

+ 3.5

24

+ 20

75

(a) Find the extreme values of function.(b) Check whether they are maxima or minima.Solution:

(a) The first order partial derivatives are found and set to zero = 8 24 = 0 = 24 + 4 = 0 = 10 + 20 = 0 = 7 = 0

This system yields

= 3,

= 0,

=

2,

= 0 and

= 3,

=

,

=

2,

= 0

(b) The Hessian matrix is

= 8 0 0 00 48 + 4 0 00 0 10 00 0 0 7

Evaluation at = 3, = 0, = 2, = 0The leading prinical minors of H are

|| = |8| = 8,|| = 8 00 4 = 12,|| = 8 0 00 4 00 0 10 = 320, || = || = 8 0 0 00 4 0 00 0 10 00 0 0 7

= 2240Since || > 0,|| > 0,|| > 0, || > 0, ( positive definite), we conclude that relative minimumoccurs at the critical point.

Evaluation at = 3, = , = 2, = 0The leading prinical minors of H are

|

| = |8| = 8,|

| =

8 00

4

=

32

As || < 0 suggests that is indefinite. Neither maximum nor minimum occurs at this point.

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