TASK MATHEMATICS for ECONOMIC APPLICATIONS09/01/2017

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 09/01/2017

I M 1) From B  3^$ œ 3 we get B  3œ ^$ 3 and so Bœ 3 ^$ 3. To find the three cubic roots of remember that 3 3 œ  3 , and so:

# #

cos 1 sin 1

^$ 3 œ   5  3   5  5 œ !ß "ß #

' $ ' $

# #

cos 1 sin 1

1 1 with . The three solutions are:

B œ 3   3 œ 3   3 œ  3

' ' # # # #

$ " $ $

! cos 1 sin 1  

; B œ 3  &  3 & œ 3  $  3 œ " $  3$

' ' # # # #

" cos 1 sin 1  

;

B œ 3  $  3 $ œ 3  !  3 œ !

# #

# cos 1 sin 1

. I M 2) The characteristic polinomial of is -

-

: œ œ

"   " "

! #  !

"  " " 

 

 

œ #  "  " œ #  "   " œ #   #

" " 

 - -   -  -   - - -

-

# # ;

from :_ - œ ! we have the three eigenvalues: -_" œ !, -_# œ-_$ œ #.

To find eigenvectors corresponding to the eigenvalue -" œ ! we must solve the linear homogeneous system  !ˆ†—œ to get:

     



 

"  " " B ! B  B  B œ !

! # ! B ! #B œ !

"  " " B ! B  B  B œ !

† ^" œ Ê ^" ^# ^$ Ê B œ

# #

$ " # $

$  B

B œ !

"

# and so the eigenvectors corresponding to -_" œ ! are —_! œ Bà !à  B œ B "à !à  "   .

To find eigenvectors corresponding to the eigenvalue -_#ß$ œ # we must solve the linear homogeneous system  #ˆ†—œ to get:

     



 

 "  " " B !  B  B  B œ !

! ! ! B ! ! œ !

"  "  " B ! B  B  B œ !

† ^" œ Ê ^" ^# ^$ Ê B

#

$ " # $

$ œ B

B œ !

"

# and

so the eigenvectors corresponding to -_#ß$ œ # are —_# œ Bà !à B œ B "à !à "   .

To check if  is or not a diagonalizable matrix we must check if 7 œ 7¹_- ⁺_- (the geome- tric multiplicity of -œ # , the dimention of Xf_-œ#, the eigenspace associated to eigenvalue , is equal or not to , its algebraic multiplicity).# #

Since 7 œ $ ¹_# Rank  #  and Rank  # œRank

 

 ˆ  ˆ

 

 "  " "

! ! !

"  "  "

œ # the matrix is not diagonalizable.

I M 3) The dimension of the Image of a linear map is equal to the Rank of the matrix  associated to the map; using elementary operations on the lines of the matrix we get:

     

    

   

" "

"  " !

! "

!  # 5 5

Ä Ä

"

#

V  V V  V

V  #V V  V

# " $ #

$ " % #

" " "

!  #  "

!  #





 

" " "

!  #  "

! ! !

! ! 5  "

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V$ÄV%

 

" " "

!  #  "

! ! 5  "

! ! !

, and from the last matrix it follows that the Rank of is equal to 2 iff 5 œ  ". To find vectors — for which  —† œ #ß "ß $ß  " we write the condition in matrix form:

 

" "

"  " !

! "

!  #  "  "

†

"

#

 

 

 

 

 







 B

B B

œ Ê Ê

# B  B  B œ #

" B  B œ "

$ B  B œ $

B  B œ  "

"

#

$

" # $

" #

" $

# $

#

  #







 

B  B  B œ #

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

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

Ê Ê

" # $

# $ " # $ " #

# $ # $ $ #

# $

 #

; vectors — are all

the vectors        

       

B "  B " "

B B ! "

B "  #B "  #

œ œ  B

" #

# #

$ #

# .

I M 4) The system is an omogeneus one and furthermore it has ∞^# solutions if and only if 8 Rank ‚ œ &Rank ‚ œ # , where 8 is the number of the unknown variables and ‚ is the coefficients matrix; to check the Rank we use elementary operations on the lines:

   

7 8 7 8

7 8 5 ! ! 5  "

7 8 5 2 ! ! ! 2  "

" " " " " "

" " ! !

" !

Ä Ä

V  V G G

# " & %

$ # % $



 

7 8

! ! ! 5  "

! ! ! 2  "

" " "

!

2 Á "

and it is easy to note that Rank ‚ is 3 iff and 5 Á " independently from and .7 8

II M 1) It's easy to verify that on point T the equation is satisfed, the gradient of the functions is f0 Bß Cß D œ CD  #BC  D ß BD  B  Dß BC  #BD  C   ^# ^# , from which we get f0 T œ  #ß "ß # ; since 0Dw T Á ! the equation defines an implicit function DÀ Bß C Ä D with fD "ß  " œ  T ß  œ "ß  " .

T T #

       T   

   

  0

0 0

Bw 0

Dw Dw

Cw

II M 2) 0 Bß C œ B  C  B C  ^# ^# ^# ^# and so f œ #0  B  #BC #C  #B C^#ß ^# .

M SÞGÞ # œ ! Ê

œ !

# œ !

: 0 œ  , with five solutions, the point

0 œ

Bw Cw

B  #BC

#C  #B C

B "  C C "  B

#

 

S œ !ß ! and the four points T œ    „ "ß „ ". MM SÞGÞ: ‡0  #C  %BC

 %BC #  #B

œ #

 ^# #,

l 0 l œ‡ #  #C^##  #B^#  %BC ^# œ% "  B  C  $B C ^# ^# ^{# #}.

‡0 œ Ê 0 ÐSÑ œ #  !‡ Ê

0 Ð Ñ œ %  !

 !ß ! !    !ß !

! # !ß !

#  _BB^ww   is a minimum point;

l 0 Ð Ñl œ  "'  !‡ T and so the four points are all saddle points.T

II M 3) Problem is equivalent to the problem:

Max min u c





 



Î 0 Bß C œ C  B Þ Þ B  C Ÿ "

B Ÿ C

# #

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



 

 Max min

u c ; the feasible region is red drawed in the figure in the Î 0 Bß C œ C  B

Þ Þ B  C  " Ÿ ! B  C Ÿ !

# #

second next page; the objective function is continuous, the feasible region is bounded and closed, so by Weierstrasse Theorem the problem admits absolute maximum and minimum.

The Lagrangian function is: AÐBß Cß ß Ñ œ- . C  B Ð- B  C  "^# ^# Ñ  Ð. B  CÑ

and its gradient is:f œ .

 "  # B 

"  # C 

 Ð Ñ

A

- .

 

 

 B  C  "  B  C

# #

KUNH-TUCKER CONDITIONS

First case -œ !ß.œ ! free optimization ( ):













- œ.œ !

 " œ !

" œ !

Ÿ ! Ÿ ! B  C  "

B  C

# #

, the system is impossible.

Second case -Á !ß.œ ! first constraint is active ( ):

 

  

  

  

 

  

   





- . - .

- -

- .

Á !ß œ ! Á !ß œ !

 "  # B œ ! B œ  "Î #

"  # C œ ! C œ "Î #

œ ! "Î %  "Î % œ "

Ÿ ! Ÿ !

Ê Ê

Á !ß œ

B  C  "

B  C B  C

# # # #

! B œ … #Î#

C œ „ #Î#

œ „ #Î#

… #Î# … #Î# Ÿ !



 

-

;

if - œ#Î#, condition #Î# #Î# Ÿ ! is satisfied; #Î#ß#Î# may be a maximum point -  !;

if - œ #Î#, condition #Î# #Î# Ÿ ! is not satisfied.

Third case -œ !ß.Á ! second constraint is active ( ):

  

  

  

  

    

- . - .

. .

- .

. .

œ !ß Á ! œ !ß Á !

 "  œ ! œ  "

"  œ ! œ  "

Ÿ ! # Ÿ "

œ ! œ

Ê Ê

œ !ß Á ! œ  "

œ  "

 #Î# Ÿ B Ÿ #Î#

œ B

, th

B  C  " B

B  C C B C

# # #

e system has

an infinite number of solutions, any point Bß B such that #Î# Ÿ B Ÿ#Î# may be a minimum point .  !.

Fourth case -Á !ß.Á ! both constraints are active ( ):

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 

  

  

  

 





- . - . 

- . - .

- .

Á !ß Á ! Á !ß Á !

 "  # B  œ ! # B  œ  "

"  # C  œ ! # C  œ "

œ ! # œ

œ ! œ B

Ê Ê

Á !ß Á !

„ #  œ  "

„

B  C  " B "

B  C C

# # #



#  œ "

œ „ #Î#

Ê

- .

B C















- .

-

. -

Á !ß Á ! œ !

œ  "

œ „ #Î#

, such solutions cannot be accepted since œ !. B

C

We conclude: Q E\ 0 œ 0    #Î#ß#Î# œ #; 738 0 œ 0 Bß B œ !    , for

#Î# Ÿ B Ÿ#Î#.

For constraints qualification we consider their Jacobian: Jœ #B #C .

"  "

 

At points #Î#ß#Î# and Bß B, #Î#  B #Î# only one constraint is active and trivia ly qualified, at points 6 Bß B, for B œ „#Î#, the Jacobian's deter- minant is … ## Á ! and constraints are qualified. On the figure below are drawn zero level curve (yellow), positive level curves (blue) and negative level curves (pink).

II M 4) 0 Bß C œ BC /  ^BC is a twice differentiable function.

f0 œC /^BC  BC /^BCßB /^BC BC /^BC œ C "  B /  ^BCßB "  C /  ^BC.

‡0 Bß C  œ œ

 C /  C "  B / "  B /  C "  B / 

"  C /  B "  C / B /  B "  C /

BC BC BC BC

BC   BC  BC  BC

     

œ œ

 C #  B / "  B "  C /  $ ! 

"  B "  C /    B #  C /  "ß " ! "

  ^BC BC   BC^BC , ‡0  .

H 0 "ß " œ @ † "ß " † A œ † $ ! † œ

! "

# X

@ßA   ‡0  cos sin  sin

α α cosα

ß     α

œ $sin cosα αsin cosα αœ #sin cosα αœsin #α .

H 0 "ß " œ ! # œ ! # œ ! # œ œ ! œ

#

@ßA#   if sin α Ê α  or  α 1 Êα  or α 1.