TASK MATHEMATICS for ECONOMIC APPLICATIONS 06/09/2017

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 06/09/2017

I M 1) Since D œ  #(3 œ #(3^$ , to find the three cubic roots of we remember thatD

3 œ  3 œ œ $ œ

# #

cos 1 sin 1 . So _$ $ $

D #(3 3

œ $  5  3  5 5 œ !ß "ß #

' $ ' $

# #

cos1  sin1  with . The three roots are:

1 1

D_" œ $  3 œ $  3 œ $  3

' ' # # #

$ $ $

cos 1 sin 1    ,

D# œ $ &  3 & œ  $ $  3 œ$ $  $  3

' ' # # #

cos 1 sin 1      and

D$ œ $ $  3 $ œ  $3

# #

cos 1 sin 1  .

I M 2) The characteristic polynomial of is

: œ  œ œ œ

 

   

   

- -ˆ

- -

- - -



$ " " # " "

# % # ! % #

"  *  $ %   *  $

œ #    %  œ

 

 - -   - 

- -

% # " "

 *  $ % #

œ #  - - ^#- '  %   - -  # œ #   - - ^# #  #- .

Since -^# #  # œ !- for - œ "„3, the matrix  admits the three eigenvalues -" œ #, -# œ "  3 and -$ œ "  3 and it is diagonalizable because it admits three distinct eingenvalues.

I M 3) By conditions proposed on basis – and vector it follows:—

— œ † œ # "ß  "ß #  #ß "ß "  "ß #ß 5 œ "ß  "ß $  5 Þ

" # " #

 " " #  "

# " 5 "

   

           

It is easy to find 5 œ  " . So — œ "ß  "ß #  and   — œ"  "  % œ'. I M 4) Since the non-singular matrices and have the eigenvector corresponding  — to the eigenvalue , it follows -  —† œ- — and  —† œ- —, with -Á !.

Remember that for the inverse matrix  we have  — —. And so -

" "† œ "

Œ —† œ ^#†^" $ † ^"††—œ^#†^"†— $ † ^"† —† œ œ  † † †— $ † † —† œ † † — $ † †- —œ

-

" " " "

œ " † †  $ † † œ " †  $ † " œ -   — -  ^" — -  - — -  - — œ —†  $ —† œ- — $- —œ %- — .

The eigenvalue of matrix that corresponds to eigenvector is .Œ — %- II M 1) Since  Bß C œ   B C œ B/  B C

B^# /^C

C # , the problem is equivalent to find Max minÎ 0 Bß C œ  B/  B C^C ^# .

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M SG f œ œ ! Ê :  Bß C  /  #BC B/  B . /  #BCœ !

B/  B

 ^C ß ^C ^#  ^C_C _#

Ê œ !

œ !

œ ! œ !

œ Ê œ Ê œ "Î#

œ

 

  

/  #BC B /  B

/ B

/  #/ C / "  #C

B / B /

C

B /

C C

C

C C C

C C

   

ß à

; system impossible.

.

 We get only one stationary point T œ/ "Î# .ß 

‡0Bß Cœ  #C /  #Bà ‡0Bß Cœ  #BC/ /  #B /  #B B/

C

C C C C ^#.

MM SG: ‡0 T œ  //^# œ  #/  !. So is a saddle point. There areT no values Bß C which maximise or minimize the determinant of the matrix . 

II M 2) Problem Max min  

u c Î 0 Bß C œ is equivalent to the problem:

Þ Þ

B C B  C Ÿ "

#

# #

Max min  

u c Î 0 Bß C œ ; the a in

Þ Þ B  C  " Ÿ ! B C^#

# # dmissible region is drawed in red in the figure the next page; the objective function is continuous, the admissible region is bounded and closed, so by Weierstrass Theorem the problem admits the absolute maximum and minimum values.

The Lagrangian function of the problem is: ABß Cß-œ B C ^# -B  C  "^# ^#  whose gradient is: fABß Cß- œ #BC  # Bß B  # Cß - ^# - B  C  "^# ^# . KUHN-TUCKER CONDITIONS

First case (free optimization): , the system is satisfied

 

 

 

 

 

- œ ! -

#BC œ !

B œ ! Ê

œ !

! œ ! B œ !

#

B  C  " Ÿ !^# ^# C Ÿ "^# for every point T œ !ß C  with  " Ÿ C Ÿ ".

‡  0 œ #C #B ‡0 œ # ! ‡0  œ !

#B ! T !C ! T

,   and     ; we can't conclude anything about the nature of the points T by the hessian of .0

If we consider the difference 0 Bß C 0!ß Cœ B C^# in a neighbourhood of T, we have that 0 Bß C 0!ß C

C C C

is    

   

 !  ! B Á !

œ ! œ ! B œ !

 !  ! B Á !

if so we conclude that

and and and

if  " Ÿ C  ! T, is a local maximum point, if !  C Ÿ " T, is a local minimum point, while  !ß ! is a saddle point.

Second case (constrai t is active8 ):







 -

- - Á !

#BC  # B œ ! B  # C œ !_# Ê B  C  " œ !^# ^#

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







 









 







 -

- -

- - -

- - - Á !

#B C  œ ! B  # C œ !

Á ! B œ !

 # C œ ! Ê

Á ! B œ ! œ ! Á !

C œ B œ #

$

#

# #

#

B  C œ "

C œ " C œ „ "

œ "

# #

ß

#

à

; the system is impossible.

Ê

 

 

 

 





  -

-

- Á !

C œ B œ #

Á ! C œ B œ

„ $Î$

Î$

œ „ $Î$

„ $Î$

„ 'Î$

œ „ $Î$





# Ê

. We

get four constrained critical points: T_"ß# œ„'Î$ß$Î$ß$Î$, possible maximum points since -  ! and T_$ß% œ„'Î$ß $Î$ß $Î$ possible minimum points since -  !. Furthermore 0T_"ß#œ#$Î* 0, T_$ß%œ #$Î*. So:

Max 0 œ 0„ 'ß $œ # $ ; min 0 œ 0„ 'ß  $ œ  # $.

$ $ * $ $ *

     

For constraints qualification consider its Jacobian: ‰ 1 œ#Bß #C and  ‰ 1 œ !ß ! only at point  !ß ! , point not belonging to the constraint. So the constrait is qualified.

On the figure below there are drawn zero level curves (yellow), positive level curves (blue) and negative level curves (pink).

II M 3) Since  —_"  œ"  " œ^# ^# # and  —_# œ"   "^#  ^# œ#, thus

? œ " † œ #ß # @ œ " † œ #ß  #

# # # #

         

— — — —

" " and # # .

0 Bß C  is a twice differentiable function and so, from f0 œ # B  #C "  #Bß  it follows:

W?0 T œ † ? œ B  #C "  #B † #ß # œ # "  #C

# # #

     

f0 # ß    and

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W_@0 T œ † @ œ B  #C "  #B † #ß  # œ # B  #C  "

# # #

     

f0 # ß  % .

From the given conditions we get:

   

  



 

W W

?

@

#

0 T œ !

0 T œ #

"  #C œ !

B  #C  " œ #

"  #C œ ! B  #C  " œ #

Ê Ê Ê



  

%  %

ÊB œ "    #  

C œ "Î# T œ "ß"  # ! 0 T ? † † @

, # . ‡ 0 œ #  œ ‡ 0 œ

, W^#_?ß@ ^X

œ #ß # † † #ß # † œ "

# # # #

 # # #

 # !   #

       

 

 



# ^ 



#

œ .

II M 4) Let us calculate the gradiente of 0 Bß C :

f0 œlog"   ß log"   

"  " 

B  C  B  C † #B B  C  B  C † #C

B  C B  C

# # # #

    ,

f0 œ ß œ  0 œ  œ 

    0  

T C "  T "

log log  T log .

$ $ log$

; ^w ^B^w_w $

C