TASK MATHEMATICS for ECONOMIC APPLICATIONS12/06/2017

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 12/06/2017

I M 1) D  $3 D  D  $3 D  # œ D  $3 D  #D  D  $3 D  # œ^% ^$ ^# ^% ^$ ^# ^# œ D D  $3 D  #  D  $3 D  # œ D  $3 D  # † D  " œ^# ^#   ^#   ^#   ^#  œD  #3 † D  3 † D  3 † D  3 œ D  #3 † D  3           ^# † D  3;

so D  $3 D  D  $3 D  # œ !^% ^$ ^# if and only if D  #3 † D  3   ^#† D  3 œ ! and the four roots are D œ #3" , D œ D œ 3# $ , D œ  3% with modulus  D œ #" ,

     D œ D œ D œ " D_# _$ _% . is the solution having the maximum modulus and its square_"

roots are D œ_" #3 œ # cos1Î#  3 sin1Î#œ œ # cos1Î%  51 3sin1Î%  51à 5 œ !ß "; so the two roots are:

A œ_"  # cos1Î%  3 sin1Î%œ "  3 and

A œ_#  # cos& Î%  31  sin& Î%1 œ  "  3 œ  A_". I M 2) The characteristic polynomial of is :  - œ -ˆ œ

œ œ   œ



 "

 

 

     

 

- -

-

- -

" ! !

" 5 !

! 5 "

! ! "

5 ! 5 !

5 " ! "

! " ! "

œ     œ

  

- - - -

- - -

# "  5 "   " 

" 5 ! "

œ- -^# ^# "  5^{# #}- -^# " œ -^# "^#5^{# #}- ;

since -œ " is an eigevalue of , : " œ  _  5^# œ ! and this implies 5 œ !; the characteristic polyinomial of becomes  :_ - œ-^# "^# œ- " ^#† - "^# with eigenvalues -_" œ-_# œ " and -_$ œ-_% œ  ". To find an orthogonal matrix which diagonalizes  we must find a matrix  œ i" i# i$ i%  where i i", #, i_$ and are four eigenvectors two by two orthogonal with and associated to thei_% i_" i_# eigenvalue - œ" and and associated to eigenvalue i_$ i_% - œ  ".

For and we consider a generic element i_" i_# i œ must satisfy the condi-

@

  

"

#

$

%

which

tion ˆ† i œ Ê

@  @ œ !

@  @ œ ! @ œ @

@  @ œ ! @ œ @

@  @ œ !

 or, in system form, ; so we get













" #

" # " #

$ % $ %

$ %

i_" œ i_# œ i_$ i_% i

" !

! "

   

, . Similarly, for and a generic must satisfy the conditi-

on ˆ† i œ Ê

@  @ œ !

@  @ œ ! @ œ  @

@  @ œ !

 or, in system form, ; so we get











" #

" # " #

$ % $ %

$ %

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i_$ œ i_% œ

" !

 " !

! "

!  "

   

, , and finally the orthogonal matrix, having the unit vec-

tors as columns, is  œ .

! !

!  !

! !

! ! 

 

" "

# #

" "

# #

" "

# #

" "

# #

 

I M 3) If vector ˜ has coordinates #ß  "ß " in the basis , it is:–

˜ œ   œ Í œ œ  "

" # 5 5  " 5

#  " 5 &  5 % &  5

" ! 5 #  5 " #  5

#

  

         

         , so 5 .

I M 4) By a reduction of the system with elementary operations on the rows we get:

   

" # !  " # " # !  " #

# "  " 7 5 !  $  " 7  # 5  %

 " % #5 7 % ! ' #5 7  " '

Ê Ê

l l

V  #V

V  V V  #V

# "

$ " $ #

Ê

" # !  " #

!  $  " 7  # 5  %

! ! # 5  " $ 7  " # 5  "

 

      

l l l

. For the last matrix we can ob- serve that when 5 Á " or 7Á  ", both matrix and augmented matrix have the same rank equal to and when $ 5 œ " and 7œ  " both matrix and augmented matrix have the same rank equal to ; we conclude that for any pair # 5 7, the system admits solutions and the number of solutions is ∞^# if 5 œ " and 7œ  ", ∞^" otherwise.

II M 1) 0 T œ $  and 1 T œ "  , conditions are satisfied at point . FT irstly we calcu- late the Jacobian matrix of 0 and 1 À

` 0 ß 1

` Bß Cß D  œ œ T œ

  ‰  #B #C #D  ‰    # #  #

C  D B  D  C  B with ! !  # .

Since ‰ T œ   # # œ ! ‰ T œ  #  #œ %

! ! !  #

 BßC ,  BßD and

‰       

 

T œ #  # œ  %

!  #

0 Bß Cß D œ B  C  D œ $ 1 Bß Cß D œ

 CßD

# # #

, the system at point

BC  CD  BD œ "

T œ  "ß "ß  "  can define an implicit function of the type Cß D È B Cß D   or

Bß D È C Bß D . If we consider the first type of function we get:  

`B `B

`C œ  œ " `D œ  œ !

 #  # #  #

!  # ! !

#  # #  #

!  # !  #

   

and , while for the second type

`C `C

`B œ  œ " `D œ  œ !

#  #  # #

!  # ! !

 #  #  #  #

!  # !  #

   

and .

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II M 2) For the Problem the admissible region is red Max min

u c









 





Î 0 Bß C œ B  C Þ Þ

B ! C ! C Ÿ "  B

# $

drawed in the figure in the next page; note that 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 value.

First case (free optimization): f # has the unique

# œ ! œ ! 0 œ Bß $C

B

$C B ! C ! C Ÿ "  B

 ^#

#

, the system













solution  !ß !       !ß !     !ß ! , so

'C !

. ‡ 0 œ # ! ‡ 0 œ # ! [0 œ !

! , ! and  

we cannot get any conclusion about point  !ß ! .

Second case (along the bo der of < >2/ admissible region):

 + if B œ ! we consider the function 1 C œ 0 ! ß CœC^$, for ! Ÿ C Ÿ " 1 C  has minimum for Cœ ! and maximum for Cœ ";

 , if C œ ! we consider the function 2 B  œ 0 B ! œ ß  B^#, for ! Ÿ B Ÿ " 2 B  has minimum for Bœ ! and maximum for Bœ ";

 - finally if C œ "  B, we consider the functionÀ

5 B œ 0 B ß "  Bœ B  "  B^#  ^$ œ  B  %B  $B  "^$ ^# ,

,5 B^w œ  $B  )B  $^# with 5 B^w  ! # %  ( Ÿ B Ÿ # %  (

$ $

for      .

For ! Ÿ B Ÿ " 5 B  has minimum for Bœ # %  ( and maximum for B œ ! or

$  

B œ " .

We conclude: Q E\ 0 œ Q E\ 0 !ß " ß 0 "ß !      œ " at points  !ß " and  "ß ! and 738 0 œ 738 0 !ß ! 0 # %  ( ß" (  " œ ! !ß !

$ $

    ,       at point  .

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On the figure below we draw zero level curve (yellow), positive level curves (blue) and negative level curves (pink).

II M 3) Since 0 Bß C  is a differentiable function:

H 0_@  T_! œf0 T _! †@ and H 0^#_@ß@  T_! œ@ †‡ T! † @T. Now f0 œ#BC  C ß B  #BC^# ^# , f0 T ! œ"ß  ", and so H 0@  T! œ"ß  " † cosαßsinαœ cosαsin .α Since H 0_@  T_! œ ! it follows cosαsinαœ !. Then

‡Bß C œ  # #  ‡ œ 

#

C B  #C

B  #C  #B , T! #  #! .

!

H 0_@ß@^#  T_! œ #cos^#α #sin^#αœ #cosαsinαcosαsinαœ ! since cosαsinαœ !.

II M 4) For 0 Bß Cß D œ  logB  C  D/^#  ^DBcosC  D we get:

f0 œ ß 

 #B " 

B  C_#  D /^{# DB} B  C_# sinC  D ß / ^DB BD/^DBsinC  D , and so f0 "ß !ß " œ  #  /ßsin"  "ß #/ sin".