TASK MATHEMATICS for ECONOMIC APPLICATIONS11/11/2017

TASK MATHEMATICS for ECONOMIC APPLICATIONS 11/11/2017

I M 1) D œ *  * œ * "  3  * 3  " œ ")3 œ ")3 œ  *3 3  " "  3 3  " "  3 3  "  #

   

   ^# . The

number in trigonometric form is D D œ *cos $ sen $ and

#1  3 #1 the three cubic roots of are D ^$ D œ^$  *3œ^$ *  $  3  $ œ

# #

cos 1 sen 1 

œ $ Î#  #5  3 $ Î#  #5 5 œ !ß "ß #

$ $

 ^$ * cos 1 1 sen 1 1

with .

The three roots are: D" œ  3 œ 3;

# #

^$ *cos1 sen1 ^$ *

D_# œ (  3 ( œ   3" œ   3

' ' # # #

        

 

$ $

$

* * $ * $

cos 1 sen 1 ;

D_$ œ ""  3 "" œ  3" œ  3

' ' # # #

       

 

$ $

$

* * $ * $

cos 1 sen 1 .

I M 2) If the vector "ß "ß #ß  " belongs to the Kernel of , matrix  must satisfy theJ

condition or and it easily

         

         

         

         

         

         

 

 

" #  " " ! ! !

# # " 5 ! '  5 !

" # 7 " ! #  #7 !

† œ œ

"

"

#

 "

follows 5 œ ' and 7 œ  ".

For such values we get J "ß "ß #ß #  œ † œ

" #  " " $

# # " ' ")

" #  " " $

"

"

#

#

     

     

     

     

     

     

 

 

.

The dimensions of the Image and the Kernel are both equal to , since in the matrix #  the first and the third rows are equal while the second row is not proportional to the first and so DimImm J œRank  œ # ÊDim Ker œ %  # œ #.

I M 3) If "  3 is an eigenvalue of the matrix , the determinant   "  3 ˆ must be equal to :  !  "  3ˆœ ! Þ        

$  "  3 " " # " "

# %  "  3 # # $ #

" 7  $  "  3 " 7  %

 

 

  œ œ

 3

 3

 3

œ #  3  3    3 œ

 3  3

 $  # #  # 

" "

# $

7  %  % 7

œ #  3 $ 3 % 3  #7  #    % 3  #  #7    $ 3œ œ #  3 3  3  "#  #7   )  #3  #  #7  $  3 œ  ^#   

œ #  3 3  "$  #7   "!  #3  #7  $   3 œ     œ  3  "&3  #73  #'  %7  "!  #3  #7  $  3 œ^#

œ ")3  ")  #73  #7 œ # * 3  "  7 3  "    œ # 3  " *  7  . The determinant is zero if and only if 7 œ  * .

To find the last eigenvalue of the matrix ("  3 is clearly the second eigenvalue) we calculate the characteristic polynomial of :

: œ  œ œ







   

 

 

 

 

 

 

- -ˆ

-

-

-



$ " "

# % #

"  *  $

œ      œ

 

$ % #     % 

 *  $  $  *

- - -

- -

# # #

" "

œ$- - ^#- '   #  )   -  - ## œ

œ -^$ %-^# *  ")  #  ) - - - ## œ -^$ %-^# '  %- that can be easily factorized as # - - ^# #  #- .

The three eigenvalues of  are -_" œ # and -_#ß$ œ " „ 3.

I M 4) If the vector has coordinates — #ß "ß # in the basis – – –_"à _#à _$ it results:

#–"–# #–$ œ— –$ œ " — #–"–# œ

from which we get # 

œ "

#

 

 

  

  

  

" " "  "

" " !  "Î#

"  " " "

 #  œ

     

     

     

     

     

     .

II M 1) f œ #0  B  C  " #C  #BC^# ß .

I SG: # œ ! Ê # œ ! Ê ∪ œ " Ê

œ ! œ ! œ "

œ

 B  C  "  B  C  "  œ ! C

#C  #BC #C "  B B

B "Î#

C

# # #

 

B "Î# C

C B

œ œ „ "

œ ! ∪ œ " .

We get three stationary points: T" œ "Î# ß ! and T#ß$ œ " ß „ ".

‡ 0  # #C  ‡ 0  % C

#C #  #B

œ  à œ

  %B  % ^#.

II SG: 0 T œ  ! Ê T is a minimum point.

l T l œ  !

 ^{BßBww "}

" "

   #

0 #

‡

‡0T_#ß$œ  % ! Ê T_#ß$are saddle points.

II M 2) 0 Bß C œ BC  /  ^CBœ !. Since 0 "ß " œ !  , the equation satisfied at point is

 "ß " . f œ0 C  /^CBß B  /^CB and f0 "ß " œ #ß ! . Only 0 "ß " Á !B^w  ; thus the equation 0 Bß C œ !  defines in a neighborhood of Cœ" an implicit function

B œ B C B " œ  0 "ß " œ  œ ! 0 "ß " #

      !

with ^w   . For the second order derivative, sin-

Cw Bw

ce ‡Bß C œ   / "  /  and ‡ "ß " œ  " # , we get:

"  /  / #  "

CB CB

CB CB

B " œ  0 B  #0 B  0 œ 

0 0

 " † !  # † # † !   " "

ww ww w ww w ww

BB # BC CC

w w

C C

      œ 

#. Since B " œ œ !  the point is a minimum point.

#

w  and B " œ_ww "

 

II M 3) f œ0 /  "  /  "^B ß ^C , f0 !ß ! œ#ß # and so:

H 0 !ß ! œ_@   f0 !ß ! † œ@ #ß #†cosαßsenα œ#cosαsenα.

H 0 !ß ! œ !_@   if and only if cosα sen and this is true for α α 1 or α 1.

œ œ œ

% %

&

For H 0^#_@ß@  !ß ! we calculate ‡  0 /  ‡0 !ß ! " 

 / "

œ ! Ê œ !

! ! 

B

C . So

H 0_@ß@^#  !ß ! œ †@ 0 !ß ! †@^X ß † " †

"

‡ œ ! œ

! 

cos sen  cos 

α α senα

  α

œcos^#αsen^#αœcosαsenαcosαsenαœ ! since cosαœsenα. II M 4) Max min  

u c Î 0 Bß C œ B  C . The objective function of the problem is in- Þ Þ !

# #

Ÿ C Ÿ "  B^#

creasing on both variables from the origin point  !ß ! that belongs to the feasible region of the problem (red-drawed in the figure in the next page).

Easily we see that 738 0 œ 0 !ß ! œ !. For Q +B 0 again by increasing on both variables from the origin point  !ß ! we note that Q +B 0 must be found on the upper boundary of the admissible region, thus the problem: Max

u c 0 Bß C œ B  C  Þ Þ !

# #

Ÿ C Ÿ "  B^# is

equivalent to Max and by simmetry the last

u c 0 Bß œ B  

Þ Þ  "

"  B "  B Ÿ B Ÿ "

# # # #

problem is equivalent to Max .

u c 1 B œ B  B   Þ Þ !

% # "

Ÿ B Ÿ "

Since 1 B œ %B  #B œ #B #B  "^w  ^$  ^# , for !Ÿ B Ÿ " , 1 B   !^w  if and only if

#B  "   ! Í B   "Î# Í B  ^{# #} "Î#.

The function 1 B  is increasing in "Î#ß " while is decreasing in !ß"Î#.

We conclude that Q +B 0 œQ +B 1 œQ +BÖ1 ! ß 1 "   ל " at points  !ß " or

„ "ß ! .

On the figure below are drawed the positive level curves (blue).