TASK of MATHEMATICS for ECONOMIC APPLICATIONS 04/06/2016I M 1) If 5Á#3œ3#3"35#3, relation can be written as##

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TASK of MATHEMATICS for ECONOMIC APPLICATIONS 04/06/2016

I M 1) If 5 Á  # 3 #  3  "  3 œ 3 5  # 3

, relation  ^#  ^# can be written as

#  3^# "  3 ^# œ 3 5  # 3  equivalent to

#  3  "  3    #  3  "  3  œ 35  # Ê $ "  #3 œ 35  # Ê  5 œ " † &  '3 œ  3 † &  '3 œ '  &3

3     .

I M 2) An element B ß B ß B ß B" # $ % belongs to the Kernel of if:0 0 B ß B ß B ß B " # $ % œ !ß !ß ! that is, in system form:

 

 

B  B œ ! B œ  B   

B  B œ ! B œ  B œ B

œ ! œ  œ  B

Ê 0 œ "

" # # "

# $ $ # "

"

B  B_$ _% B_% B_$

. So Dim Ker and every element of Ker 0 is of the form —œ B "ß  B B ß  B"ß " "œ 5†"ß  ß" "ß " and a basis for Ker 0 is UKer 0 œ"ß  ß" "ß ".

From Sylvester Theorem: Dim Ker  0 Dim Imm  0 œDim ‘^% œ %.

Since Dim Ker  0 œ " we have Dim Imm œ %  " œ3 and so Imm 0 œ‘^$ and a basis for Imm 0 is simply given by the standard basis:

UImm 0 œ / / / "ß #ß $ œ "ß !ß! ß !ß "ß! ß !ß !ß".

I M 3) Since the vector — has coordinates "ß  #ß " in the basis –, it has coordinates

α " #ß ß  in the basis if•

α  "  #       

           

" ! ! " " "

! " ! "  " "

" ! " " "  "

  œ"  #  "

that in system form gives:

 

 

 

α α

" "

α # #

œ ! œ !

œ % œ %

œ œ

Ê

  #  #

.

So in the basis the vector has coordinates • — !ß %ß  #. I M 4) The characteristic polinomial of is

: œ  œ œ œ

$  " " #  " "

# %  # ! %  #

" 7 5  "  5  7 5 

   

   

-  -ˆ

- -

- - -

œ #  %  #  "  5  " " œ

7 5  %  #

 - -   - 

- -

œ #  - - ^# %  5 - %5  #7  "  5   -#  % -œ œ #  - - ^# %  5 - %5  #7  "  5   - -  # œ œ #  - - ^# %  5 - %5  #7  "  5 -œ

œ #  - - ^# &  5 - &5  #7  ".

So : # œ !_  and - œ # is an eigenvalue of the matrix ß a 5ß 7 .

If ; - œ -^# &  5 - &5  #7  ", - œ # is a multiple eingenvalue for iff

; # œ #7  $5  ( œ ! 7 œ $5  (

  i.e. iff # .

In such case the characteristic polynomial :_ - can be factorized as:

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:_ - œ - # ^# - $  5  and the three eingevalues for are  -_"ß# œ # and -$ œ $  5. If 5 œ  " (and so 7 œ  &) it follows that -"ß#ß$ œ #. So we have:

if 7 œ $5  ( and 5 Á  " Ê 7 œ # while if 7 œ  & and 5 œ  " Ê 7 œ $.

# ^# ^#

+ +

To study the diagonalizability of when  - œ # is a multiple eingenvalue, we must check the dimension of the eingespace associated to - œ # if 7 œ $5  ( , that is:

#

Order  Rank  Rank 

 

   #ˆ œ $  œ

" " "

# # #

" 5  #

 

 ^$5(_# 

œ $  œ

" " "

! ! !

! 5  $

" 5 Á $

# 5 œ $

Rank if .

if

 

  

 

 ^$5*_# 

The matrix is diagonalizable when 7 œ " and 5 œ $, isn't diagonalizable when 7 œ $5  ( 5 Á $

# and .

II M 1) For a two times differentiable function 0 Bß C  at stationary point T! we have:

H 0 T / 0 T †  / " ! 0 T 0 T †

0 T 0 T

!

 "

/ ß/# ! " ! #

ww ww

BB ! BC !

ww ww

BC ! CC !

" #            

     

œ †‡  ^X œ †  œ

œ  " ! 0 T 0 T œ # 0 T #

0 T

        

†   œ  Ê œ 

BCww ! CCww !

ww ww

BC ! BC ! . In the same way:

H 0 T œ  " ! 0 T 0 T † 0 T

0 T 0 T

"

!

# ww

/ ß/ ! BB !

ww ww

BB ! BC !

ww ww

BC ! CC !

" "          

     

†  œ  and

H_{/ ß/}^# _" _"0 T _! œ $ implies 0_BB^ww  T_! œ $. Finally

H 0 T ! " 0 T 0 T † 0 T

0 T 0 T

!

"

# ww

/ ß/ ! CC !

ww ww

BB ! BC !

ww ww

BC ! CC !

# #          

     

œ †  œ and

H_{/ ß/}^#_# _#0 T _! œ  # implies 0_CC^ww  œ T_! #.

So the Hessian matrix of at point 0 T is 0 T $ #

# #

! ‡   œ!   

 

  .

To check the nature of T! we have:   and so T! is a maxi-

 ‡

‡

"

#

œ  $  !à  #  ! œ '  %  !

mum point.

II M 2) First step (we calculate functions and at point0 1 "ß !ß "):

  

 

0 "ß !ß " œ 1 "ß !ß " œ "

"  "  ! œ !

/  " /  ! œ !^! ^! , so the conditions are satisfied.

Second step (we calculate the Jacobian matrix of and 0 1):

N œ ` œ Þ

`

 

 0 1ß     

ß Cß B B

B D

"  "

"  " 

  "  C / 

/  D /  #C /  D /

  ^C

DB BD DB BD

Third step (we calculate the Jacobian matrix of and at point 0 1 "ß !ß "):

N"ß !ß " œ  "  "  "Þ

 " ! "

Since N"ß !ß "   

 BßD œ "  œ !

 "

"

" , while

N"ß !ß "   

 BßC œ "  œ 

 !

"

" " and

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N"ß !ß "   

 CßD œ  "  œ 

! "

" !

"

, with conditions    !

 

0 "ß !ß " œ

1 "ß !ß " œ we can de- fine a function D È B D ß C D     or a function B È C B ß D B    , both in a neighbou- rhood of point ."

Last step (we calculate the derivatives of the implicit function).

For function D È B D ß C D    :

B "   

C "

w

w BßC D

   œ  N"ß !ß " †N"ß !ß "  œ

   

"

= "    " œ !   " œ  " .

 ! "

" "

"

† "  † "   ! 

For function B È C B ß D B     :

C "   

D "

w

w CßD B

   œ  N"ß !ß " †N"ß !ß "  œ

   

"

œ   "   " œ "   " œ ! Þ

! "   "

" "

!

"

† " † "   "

II M 3) From 0 Bß Cß D œ  B ^$ )C  D^$ ^$ $B^# #%CD we get

f0 œ $ B  'B ^# , #%C  #%D  $D  #%C^# , ^# , ‡0 . B  '

 

œ

! !

!

 

'

%)C #%

 #%  'D First order conditions: f0 œ 

Ê Ê Ê Ê

œ ! œ ! œ

œ ! ” œ  # œ 

œ

  

  

  

$

#%C  #%D

 $D  #%C

$B

#% C  D

 $ D  )C

B B

D C

C  )C B  'B



B  #



#

# #

# # #

! ! % !

 

Ê Ê

œ ! ” œ  # œ 

œ

œ ! ” œ  # œ 

œ ! ” œ  #

 

 

 

B B

D C

C C  )

B B

D C

C C

#

$

#

  ! .

The function 0 has four stationary points:

T" œ!ß !ß !, , T# œ  # ! ! ß ß  T$ œ!ß  #  %ß  and T% œ  #  #  % ß ß . Second order conditions:

‡0 T ‡0 T

' '

!  ! 

 " œ #%  # œ #%

 #% !  #% !

   

! !  ! !

! !

, , both having a

second order principal minor ‡_# œ #% with a negative determinant

 !#% 

!



equal to  &('  !, and so T_" and T_# are saddle points.

‡0 T T

'

*' 

 $ œ #% $

 #% #%

 

! !

!

. At point a sequence of principal minors is:

 ‡_" œ '  ‡_# œ' ! 

! *'

œ ' !, ,œ &(' !

 ‡$ œ

! !

!

 

'

*' #% T

 #% #%

œ "!$')  ! , so we conclude that $ is a local minimum point for .0

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‡0 T

'

*' 

 % œ #%

 #% #%

 

 ! !

!

. In this case we have a second order principal mi-

nor ‡# œ  ' !  ith a negative determinant equal to , and so

! *' w  &('  ! T% is

a saddle point.

II M 4) Problem





 

 Max min

u c is equivalent to

Î 0 Bß C œ C  B Þ Þ B  C Ÿ "

C Ÿ "  B

# #

#





 

 Max min

u c .

Î 0 Bß C œ C  B Þ Þ B  C  " Ÿ !

B  C  " Ÿ !

# #

#

The admissible region is the red-drawed one in the figure:

The objective function is continous and the admissible region is a bounded and closed set, so by Weierstrass Theorem the problem admits absolute maximum and minimum.

The Lagrange function of the problem is:

ABß Cß ß- .œC  B-B  C  "^# ^# .B  C  "^#  and its gradient is:

f œ  "  # B  # Bß "  # C  ß A  - . - . B  C  " ß^# ^#  B  C  "^# . KUHN-TUCKER CONDITIONS

First case (free optimization): , the system is impossible.













- œ.œ !

 " œ !

" œ !

Ÿ ! Ÿ ! B  C  "

B  C  "

# #

#

Second case (first constraint is active):

 

  

  

  

 

  

   





- . . 

- -

Á !ß œ ! œ ! .

 "  # B œ ! B œ  "Î #

"  # C œ ! C œ "Î #

œ ! "Î %  "Î % œ "

Ÿ ! Ÿ !

Ê Ê

œ ! B œ … B  C  "

B  C  " B  C  "

# #

#Î#

C œ „ #Î#

œ „ #Î#

"Î# „ #Î# Ÿ "



 - 

;

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if - œ#Î# condition "Î# #Î# Ÿ " is not satisfied, the system is impossible;

if - œ #Î# condition "Î# #Î# Ÿ " is satisfied and point #Î#ß #Î# may be a minimum point (-  !).

Third case (second constrai t is active8 ):

  

  

  

  

  

- . -  

. .

- .

œ !ß Á ! œ !

 "  # B œ ! B œ  "Î #

"  œ ! œ "

Ÿ ! Ÿ " "Î% *Î"' Ÿ "

œ ! œ " 

Ê Ê

œ ! B œ  "Î#

œ "

œ $Î%

B  C  " B  C 

B  C  " C B C

# # # #

# #

,

condition "Î%*Î"' Ÿ " is satisfied and point  "Î#ß $Î% may be a maximum point (.  !).

Fourth case (both constrai ts are active8 ):









 "  # B  # B œ !

"  # C  œ ! œ ! œ !

Ê

- .

B  C  "

# #

#

  

  

  

  

  

# B  # B œ  "

# C  œ "

"  œ ! œ " 

Ê Ê Ê

# B  # B œ  "

# C  œ "

 œ !

œ " 

# B  # B œ  "

# C  œ "

 œ !

œ " 

- .

B  B  "

C B

B B

C B

B B "

C B

# # #

#

% #

#

# #

   #

Ê ∪ ∪ Ê

# B  # B œ  " # B  # B œ  " # B  # B œ  "

# C  œ " # C  œ " # C  œ "

œ ! œ " œ  "

œ " œ ! œ !

  

  

  

  

  

- . - . - .

B B B

C C C

Ê ∪ ∪

! œ  " œ  Î# œ  "Î#

œ ! œ " œ  "

œ " œ ! œ !

œ " œ "

  

  

  

  

  

impossible

3

B B B .

C C C

- -

. .

For both points „ "ß ! the product - .† is negative and so the points aren't max or min.

So we have Q E\0 œ 0  ß" $ œ &; 7380 œ 0 #ß  # œ  #.

# % % # #

     

For constraints qualification we consider their JacobianN œ #B #C .

#B "

 

In points  ß^{" $}_{# %} and ^_#^#ß  ^_#^# only one constraint is active and trivialy qualified.

In the figure representing the feasible region there are drawn zero level curve (yellow), positive level curves (blue) and negative level curves (pink).