TASK MATHEMATICS for ECONOMIC APPLICATIONS 15/1/2018

(1)

TASK MATHEMATICS for ECONOMIC APPLICATIONS 15/1/2018

I M 1) D œ /691 "'31 œ /691 "' 3/ 1 œ "'cos1 3sin1œ

œ "'cos1 3sin1œ  "'. So ^% D œ^%  "'œ ^% "' ^% cos1 3sin1œ

œ #  5  3  5 5 œ !ß "ß #ß $

% % % %

# #

cos1 1 sin1 1

with .

The four roots are:

for 5 œ ! À #  3 œ #  3 # à

% %

cos1 sin1  

for 5 œ " À # $  3 $ œ  #  3 # à

% %

cos 1 sin 1  

for 5 œ # À # &  3 & œ  #  3 # à

% %

cos 1 sin 1  

for 5 œ $ À # (  3 ( œ #  3 #.

% %

cos 1 sin 1  

I M 2) To find the eigenvalues of the matrix , we need determinant  -ˆœ !:

 

 ! 5

! 5  !

5 ! 

5  5 

-

- - - - -

œ  ^# 5^# œ   5  5 œ ! . So the three eigenvalues are -_" œ-_# œ 5ß-_$ œ  5.

For 5 Á ! it is 7 œ #ß 7 œ ". Since

5

+ +

5 5  5ˆœ

 ! 5

! ! !

5 ! 

 

  we get

7 œ $ ¹₅ Rank 5ˆœ $  " œ # (obviously, since the matrix is symmetric).

Since -_$ œ  5 is a simple eigenvalue, 7¹_5 œ 7⁺_5 œ ". If 5 œ ! we easily get -" œ-# œ-$ œ ! and 7 œ 7 œ $1 .

! +

For 5 Á !, to find eigenvectors corresponding to -!" œ-# œ 5 we solve the system:

 5ˆ†—œ † œ Ê

 ! 5 B !

! ! ! C !

5 !  D !

B  D œ ! a C

     

      

5

5 so eigenvectors are: • œ Bß Cß B Þ 

For the basis we need two linearly independent eigenvectors, and we choose:

•_" œ "ß !ß "  and •_# œ !ß "ß ! Þ 

For 5 Á !, to find eigenvectors corresponding to -$ œ  5 we solve the system:

 5ˆ†—œ † œ Ê

! 5 B !

! #5 ! C !

5 ! D !

B  D œ ! C œ !

     

      

5

so the unique eigenvector is • œ Bß !ß  B Þ  For the basis we choose •$ œ "ß !ß  " Þ 

If 5 œ ! Ê œ −

! ! !

 • ‘

 

 , the null matrix, and its eigenvectors are all ^$.

(2)

I M 3) Since      we get:

   

0 "ß "ß " œ $ß $ß "

0 "ß "ß  " œ  "ß &ß  "

     

 

 

 

 

" B C " $ "  B  C œ $ B  C œ #

# B C " $ #  B  C œ $ B  C œ "

 " B C " "  "  B  C œ " B  C œ

† œ Ê Ê

" " " " " "

# # # # # #

$ $ $ $ $ $ #

and

     

 

 

 

 

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

# B C " & #  B  C œ & B  C œ $

 " B C  "  "  "  B  C œ  "

† œ Ê Ê

" " " " " "

# # # # # #

$ $ $ $ B  C œ !_$ _$

and so:

B  C œ # B B  C œ " B B  C œ  #^" ^" Ê C^" B  C œ $^# ^# Ê C^#

" " " # # #

œ ! œ #

œ # à œ  " à

B  C œ # B

B  C œ ! Ê C œ

" ! #

# #  "

 " " "

$ $ $

œ "

œ " . So 

 

  . Using elementary operations on the rows: V Ã V  #V à V Ã V  #V and then V Ã V  V" we get:

# # " $ $ " $ $ # #

     

 " ! #  " ! #       

# #  " ! #  &

 " " " ! " $

Ê Ê Á !

" ! #

! #  &

! ! ^""_#

. So  and Rank  œ $ . From this we get Dim Imm   œ $ and Dim Ker   œ $  $ œ !.

I M 4) If the vector has coordinates — #ß  "ß # in the basis it means that:–

—œ † œ •

" " " # #

! " "  " !

" ! " " $

     

 

  . If we use the basis we get:

— œ † œ Ê Ê

" " # B # B  B  #B œ # B œ "

! " " B ! ! B  B  B œ ! B œ  "

" ! # B $ B  ! B  #B œ $ B œ "

     

 

 

 

 

" " # $ "

# " # $ #

$ " # $ $

.

In fact .

     

 

" " # " #

! " "  " !

" ! # " $

† œ œ —

II M 1) From the equation 0 Bß C œ C  logB  B  " /  C œ !  ^C we get 0 "ß ! œ !  . f0 Bß C œ C† "  / Êf0 "ß ! œ "ß  "

   B ^CßlogB  B  " /  "  ^C     .

Since 0_C^w "ß ! œ  " Á !, we can define an implicit function C œ C B , for which:

C " œ  "ß ! œ  œ "

"ß !

w   

 

0 "

0  "

Bw Cw

.

Furthermore ‡Bß C  †   / Ê‡ "ß !  

 /

#

œ  œ #

B  " /

!

C _B^" _B^" ^C

"

B C

#

C and so:

C " œ  0  #0 C  0 C œ  œ %

0  "

!  # † # † "  ! † "

ww ww ww w ww w

BB BC CC #

Cw

   

. For the expression of the Taylor polynomial of the second degree at B œ " we get:

T Bß " œ !  C " B  "  C " B  "" œ B  "  # B  " œ #B  $B  "

#  w   # ww  #  # # .

(3)

II M 2) Firstly we write the problem in the form:

Max/min u.c.





 



0 Bß C œ B  C B  %C Ÿ %

#C Ÿ #  B

# #





 

 Max/min

u.c. .

0 Bß C œ B  C B  %C  % Ÿ ! B  #C  # Ÿ !

# #

# # The objective function of the problem is a continuous function, the feasible region is a compact set, and so maximum and minimum values surelyX exist. The constraints are qualified.

The Lagrangian function of the problem is:

ABß Cß- -_"ß _#œB  C^# ^#-_"B  %C  %^# ^# -_#B  #C  #. 1) case -_" œ !ß-_# œ ! À

 

 

 

 

 

     

A A

wB wC

œ #B œ ! œ  #C œ !

Ê

œ ! œ ! B  %C Ÿ %

#C Ÿ #  B

!  ! Ÿ %  #

! Ÿ #  !

#

# #

B

C !

Bß C œ !ß ! œ

. ‡ ‡ ! .

Since 0_{B B}^ww †0_{C C}^ww  ! the point  !ß ! is a saddle point.

2) case -_" Á !ß-_# œ ! À

 

 

 

 

 

A - -

wB " "

wC " "

œ #B  # B œ #B "  œ ! œ  #C  ) C œ  #C "  % œ !

Ê

œ ! œ !

 

B  %C œ %

#C Ÿ #  B

!  ! œ %

! Ÿ #  !

# #

B

C impossible

∪ À !ß " !ß  "

B B

„#

 

 

 

 

 

   

œ !

œ  Ê

œ !

œ   !

-" " -

% " "

%

%C œ %

#C Ÿ #  B

C œ „"

Ÿ #  !

# and may be minimum points;

∪ À #ß !  #ß !

B

!

 

 

 

 

 

   

-

"

œ "

C œ !

Ê

œ C œ !

œ "  !

# œ %

#C Ÿ #  B

„#

Ÿ #„#

and may be maximum points;

∪







 - -

"

" "

%

œ "

œ  B  %C œ %

#C Ÿ #  B

# # impossible . 3) case -_" œ !ß-_# Á! À

(4)

 

  

  

  

 





A -

- -

- - -

wB #

wC #

"

# #

#

# " #

# #

œ #B  œ ! œ  #C  # œ !

Ê Ê

œ œ 

 #

œ œ

#C œ #  B B  %C Ÿ %

 œ #

B  %C Ÿ %

 œ 

 Ÿ % À

# # # #

#

% $

$ %

$

% '%

* *

B C

not satisfied .

The point  ß# % .

$ $ Â X 4) case -" Á!ß-# Á! À









  

A - -

wB " #

wC " #

œ #B  # B  œ ! œ  #C  ) C  # œ !

Ê œ ! œ # Ê

œ œ !

B  %C œ %

#C œ #

B  %C œ %

#C œ #

B B

C C

# #

B 

B  ∪

. From 1

 

 

 

 

 

B œ ! C œ "

 œ !

 #C  )  # œ ! Ê

B œ ! C œ "

œ  œ ! -

- -

#

" #

" "

%

#

just seen, it may be a minimum point;

 

 

 

 

 

B œ # B œ #

C œ ! C œ !

%  %  œ ! œ "

 # œ ! œ !

- - Ê -

- -

" # "

# #

just seen, it may be a maximum point.

Now we study the function in the boundary points.

For B #C œ # Ê B œ #  #C we get 0 C œ  #  #C^#  C œ $C  )C  %^# ^# from which 0 C œ '^w  C  ) ! C %. So for  " Ÿ C Ÿ " the function is increasing.

for $

For B  %C œ % we put B œ # >

C œ >

# #

 cos

sen to get 0 > œ %  cos^#> sen^#> œ &cos^#>  "

from which 0 > œ  "!^w  cos sen> > !Þ

Since we get:





 "!  ! cos sen

>

! ! Ÿ > Ÿ Ÿ > Ÿ # ! ! Ÿ > Ÿ

for or

for

1 1

# $# 1

1

So  #ß ! and  #ß ! are maximum points with 0 #ß ! œ 0  #ß ! œ %    ,

 !ß " and !ß  " are minimum points with 0 !ß " œ 0 !ß  " œ  "    .

(5)

II M 3) For the function 0 Bß C œ  C^#B  #BC  #(B^$ ^# firstly we find its stationary points. f œ  $0  B  #C  #( #C  %BC^# ^# ß .

I SG:  $ œ ! Ê $ œ ! Ê

œ ! œ !

 B  #C  #(  B  #C  #(

#C  %BC #C "  #B

# # # #

 

Ê œ Ê ∪ Ê

œ !

œ $ œ  $

œ ! œ !

œ œ

œ œ 

   

 

 

 

B *

C

B B

C C

C B

B B

C C

# #

and "!& and

" )

#

" "

# #

"!& "!&

# # # #

 

There are four stationary points:   "ß ! ß  "ß ! ß ^"_#ß "!& ß ^"_#ß  "!&Þ

# # # #

 

II SG: œ  % .

‡Bß C   'B % C  C #  %B

‡ ‡

     ‡  

$ß !  ")   $ß !

"!

œ ! À  ! À

!   !

"

# is a Maximum point;

‡ ‡

     ‡  

 œ ! À    ! À

! "%  !

$ß ! ") _"  $ß !

# is a minimum point;

‡     ‡

    

" "!&

#ß ß

# #

#"!



 œ  À  ! À



 3 is a Saddle point.

0 ^#

"

#

"!&

# #



‡     ‡

    

" "!&

#ß  ß 

# #

#"!



 œ  3 À  ! À is a Saddle point.

0 ^#

"

#

"!&

# #



II M 4) From 0 Bß C œ  αB  C  #B  $C^# " ^$ ^# we get:

0 Bß C œ # B  # ß 0_B^w  α _{B B}^ww Bß C œ # ß 0 Bß C œ $ C  'C ß 0 α _C^w  " ^# _{C C}^ww Bß C œ ' C  ' " . From the given conditions:

0 "ß " œ #  # œ 0_B^w  α _{C C}^ww  "ß " œ '  '" Ê#  # œ '  'α " Ê#α' œ  ) à"

0_C^w "ß  " œ $  ' œ 0 " _{B B}^ww  "ß  " œ # αÊ$  ' œ #" αÊ#α$ œ ' Þ"

# ' œ  )  œ  "% 

# $ œ ' # œ $  '

œ œ "!

α

α α α

  $



" "

"

Ê Ê

"%

$ .