TASK MATHEMATICS for ECONOMIC APPLICATIONS 16/03/2017I M 1) Using the definition of a complex exponential we get:Dœ/œ/†/œ/œ/œ

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 16/03/2017 I M 1) Using the definition of a complex exponential we get:

D œ /^"'3 œ / † /^'3 œ /cos '  3 sin 'œ /cos'  3sin'œ œ /cos'  3/sin' , the algebraic form of D.

Since # z 'ß #)1 the number , in the complex plane, has this position:D

I M 2) To achieve the requested matrix firstly we calculate its characteristic polynomial:

: œ  œ " $  $ œ " $  $ œ

#  # $ "  !  "

" # %  " # % 

   

   

-  -ˆ - -

- - -

- -

œ "  $  $   " " $  œ

# %  " #

 - -  -  -

-

œ "  - - ^# (  '  " -  -œ "  - -  "- (. The three eigenvalues of are  -_" œ -_# œ ", -_$ œ (.

Since  " †ˆœ " # $  " †ˆ

" # $

 

  it follows 7 œ $ _"¹ Rank œ # and so the matrix is a diagonalizable one.

The second step is to find for every eigenvalue the corresponding eigenvectors.

For -_"ß# œ " we solve the system:

 

   





- ˆ †—œÊ † œ Ê Ê

" # $ B ! B  #B  $B œ !

"

" " # $

# " # $

$ " # $

 

Ê B œ  #B  $B Ê_" _# _$ —œ  #B  $B ß B ß B Þ _# _$ _# _$ Two linearly independent eigenvectors corresponding to - œ " are —_" œ  #ß "ß !  and —_# œ  $ß !ß " . For -_$ œ ( we solve the system:

 

   

- ˆ †—œÊ † œ Ê

 & # $ B !

"  % $ B !

" #  $ B !

$

"

#

$

 

Ê Ê Ê

 &B  #B  $B œ ! "#B  "#B œ !

B  %B  $B œ !  'B  'B œ !

 #B  $B œ ! œ  #B  $B

œ B œ B

 

 

 ^" ^" ^# ^# ^$ ^$  ^# ^# ^$^$ 

" # $ " # $

" $

# $

B B

B

B and so the ei- genvector corresponding to - œ ( is —_$ œ Bß Bß B Ê  —_$ œ "ß "ß " Þ 

So a matrix satisfying  † œ ƒ† is œ .

 #  $ "

" ! "

! " "

 

I M 3) First method, by the transpose of the adjoint matrix divided by the determinant.

 

   

       

 œ œ œ # †  " † œ "#  & œ (

# # $ # # $

" $ $ ! "  "

" # % " # %

"  " # $

# % "  " .

(note that   œ " † " † (, the product of the eingenvalues of the matrix). And so:

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E.4 œ   œ

$ $ " $ " $

# %  " % " #

# $ # $ # #

# % " % " #

# $ # $ # #

$ $  " $ " $

 

     



 

 

 

'  "  "

 # &  #

 $  $ %

.

 



" X

œ " E.4 œ " œ

(

'  #  $ 'Î(  #Î(  $Î(

 " &  $  "Î( &Î(  $Î(

 "  # %  "Î(  #Î( %Î(

   

   

    .

Second method, by elementary operations on >2/ 638/= 90 >2/matrix.

   

# # $ l " ! ! # # $ l " ! !

" $ $ l ! " ! ! # $Î# l  "Î# " !

" # % l ! ! " ! " &Î# l  "Î# ! "

´ ´

V  V

V  V V  V

# " "

#

$ " "

# $ " #

#

 

# # $ l " ! !

! # $Î# l  "Î# " !

! ! (Î% l  "Î%  "Î# "

´

"

# "

"

# #

%

( $

V V V

 

" " $Î# l "Î# ! !

! " $Î% l  "Î% "Î# !

! ! " l  "Î(  #Î( %Î(

´

V  V V  V

" #

# $ $

%

 

" ! $Î% l $Î%  "Î# !

! " ! l  "Î( &Î(  $Î(

! ! " l  "Î(  #Î( %Î(

V  V"´$ $

%

 

" ! ! l 'Î(  #Î(  $Î(

! " ! l  "Î( &Î(  $Î(

! ! " l  "Î(  #Î( %Î(

.

In the right block we have the requested inverse matrix.

I M 4) If œ — —

+ + +

 

"" "# "$

#" ## #$

$" $# $$

" #

, since and belong to the Kernel, we have:

† œÊ œ Ê Ê

! !  œ !

" !  œ !

—_"

"" "# "$ "# "$

#" ## #$ ## #$

$" $# $$ $# $$

  

 

+ + + + +

  





Ê Ê

œ  





+ + + + +

œ

"$ "# "" "# "#

#$ ## #" ## ##

$$ $# $" $# $#



 

  ;

† œÊ œ Ê Ê

 " !  œ !

 ! !  œ !

—#

"" "# "# "" "#

#" ## ## #" ##

$" $# $# $" $#

  

 

+ + + + +

  





Ê

œ   





+ + + + +

œ

"" "# "# "# "#

#" ## ## ## ##

$" $# $# $# $#

, 

 

  .

Finally, from 0 "ß "ß " œ #ß "ß  "    , must satisfy:

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  

 

  " # œ # œ #

  " "  œ " œ  "

  "  "  œ  "

œ Ê Ê

+ + + + +

+ + + +

"# "# "# "# "#

## ## ## ## ##

$# $# $# $#

 

 

 

 

 

+_$# œ "

.

So  œ

 #

 "

" " "

 

# #

" "

 

.

To find a basis for the Kernel and a basis for the Image, since Rank  œ " we have simply to note that —_" and —_# are linear independent, so U_{O/< 0}_{ } œ — —_"ß _# and

U_{M77 0}_{ } œ#ß "ß  ".

II M 1) The equation 0 Bß Cß D œ B/  ^CD C/^BD D/^BC œ ! verifies 0 "ß "ß ! œ !  . f0 œ / ^CD  C/^BD  D/^BCà B/^CD  /^BD D/^BCà B/^CD  C/^BD /^BC from which we get f0 "ß "ß ! œ !à !à /   ^#. Since 0 T Á !D^w  , the equation defines an impli- cit function D œ D Bß C  with:

fD "ß " œ 0 T à œ !à ! œ !ß !

0 T 0 T / /

     0 T     

   

     

Bw

w w # #

D D

Cw .

II M 2) The problem Max min  

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

Þ Þ %B  C Ÿ %

# #

Max min  

u c Î 0 Bß C œ B  BC ; the feasible region is drawn in red in the figure Þ Þ %B  C  % Ÿ !

# #

# # on

the next page; the objective function is continous and the feasible region is a bounded and closed set, so by Weiertrass Theorem the problem has absolute maximum and mini- mum. The Lagrangian function of the problem is:

ABß Cß-œB  BC^# ^#-%B  C  %^# ^# 

whose gradient vector is f œ #A  B  C^# )-B #BCà  #-Cà %B  C  %^# ^# . KUNH-TUCKER CONDITIONS

First case (free optimization): , the unique solution

 

 

 

 

 

- œ ! -

# œ ! œ !

œ ! œ !

Ÿ ! Ê

œ !

Ÿ

B  C B

#BC C

%B  C  % ! %

#

# #

point is  !ß ! . #  !ß ! !  !ß !

# !

‡ 0 # C ‡ # ‡ 

C #B !

œ , œ  with œ !; second order conditions are useless to give us informations on the nature of the point  !ß ! , but from the figure on the next page it is easy to see that  !ß ! is a saddle point for the objective function of the problem.

Second case (active constraint):

 

 

 

 

 

- -

Á ! Á !

#  ) œ ! #  ) œ !

 # œ !  œ !

œ ! œ

B  C B Ê B  C B

#BC C #C B

%B  C  % %B  C %

# #

# # # # from which we get:

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 

 

 

 

 

-

- - Á !

#  % œ !

œ ! œ

Ê

Á ! œ "Î%

œ ! œ B "

C

%B %

C B „"

 

#

and

 

 

 

 

    

-

- -

-

- -

-

- -

-

- -

Á ! œ ) œ

"# œ Ê

Á ! œ ) œ

#  " $  # œ

C  #

B

 # %

C  #

B

!

# #

#

# #

from which we get:

   

   

   

   

   

- -

Á ! Á !

œ $ œ #!Î*

œ  "Î# œ #Î$

Ê Ê

Á ! Á !

œ $ œ #!Î$

œ  "Î# œ #Î$

C C

B B

C „ C „

B B

#  and #  .

So we get six constrained critical points:

T_"ß# œ„"ß ! T, _$ß% œ #Î$ß „#!Î$, Q +B ? ; T_&ß' œ "Î#ß„$, 738 ? . 0T_"ß#œ ", 0T_$ß%œ %%Î*, 0T_&ß'œ  &Î%.

Q +B 0 œ %%Î* at points T à 738 œ  &Î%_$ß% 0 at points T_&ß'.

Note also that by figure below point is a local maximum; is not a maximum point.T_# T_"

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

II M 3) For 0 Bß Cß D œ  logB  C  / ^DBB C  CD^# ^$, differentiable function, we have: H 0 T@  ! œf0 T! † @, where is the unit vector of @ "ß "ß " .  T œ "ß !ß "!  

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f0 Bß Cß D  œ "  /  #BC à  "  B  D à  /  $CD

B  C B  C

 ^DB ^# ^$ ^DB ^#

f0"ß !ß " œ "  "  ! à  "  "  " à  "  ! œ #à  "à  " Þ     Since

@ œ " à " à " H 0 T œ #à  "à  " † " à " à " œ !

$ $ $ $ $ $

    we get _@   _!      .

II M 4) The Mac Laurin polynomial of degree 2 at point  !ß ! for the function is0

T Bß C œ 0 !ß !  f0 !ß !  " !ß ! 0 !ß ! œ !

#      † Bß C #Bß C † † B  

‡0  C ; ,

f0 œ / ^^{B C}^ ^ B  C /  ^^{B C}^ ^{ }à /^{B C}^ ^ B  C /  ^^{B C}^ ^ œ

f0 œ"  B  C / ^^{B C}^ ^à "  B  C /  ^^{B C}^ ^à f0 !ß ! œ "ß "   ;

‡0 Bß C  œ /  "  B  C / /  "  B  C / œ

 /  "  B  C /  /  "  B  C /

       

B C B C B C B C

   

   

‡0 Bß C  œ     ‡0 œ 

   

#  B  C /  B  C / # !

 B  C /  #  B  C / !ß !

!  #

   

B C B C

 

  , .

So T_#Bß C œ B  C " # # œB  C

# B  C^# ^# B  C^# ^#.