TASK MATHEMATICS for ECONOMIC APPLICATIONS06/09/2017

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 06/09/2017

I M 1) A œ 3  D œ 3  #  3 œ #3  # œ 3  " "  3† œ #3 œ 3 o find the two 3  D 3  #  3 #3  # 3  " "  3 #

 

  .T

square roots of remember that A 3 œ cos  3sin , thus

# #

 1  1

3 œ   3   œ    5  3   5  5 œ !ß "

# # % %

cos 1 sin 1 cos 1 sin 1 .

1 1

The two roots are:

D" œ  3 œ  3 œ "  3

% % # # #

# # #

cos  sin     and

 

1 1

D_# œ &  3 & œ  #  #3 œ  # "  3 œ D_"

% % # # #

cos  sin     .

 

1 1

I M 2) From J B ß B ß B ß B œ B  B  B  B ß 7B  B  B  7B " # $ %  " # $ % " # $ % we get:

J — œ — with †  œ " " " "  J "ß "ß "ß " œ %ß #7  #    easily 7 " " 7 . From

we get 7 œ ". Since the dimension of the Image of a linear map is equal to the rank of the matrix  associated to the map, for the matrix œ" " " ", by elementary

" " " "

operations on the rows we get: " " " " " " " " so we easi-

" " " " ^{V  V}^#Ä ^" ! ! ! ! and ly get Rank  œ " œDim Imm . 

From Dim Imm œ " we get Dim Ker œ %  " œ $. Since the vector  %ß % belongs to the Image of , a basis for ImmJ  J is U_{M77 }_J œ "ß " . To find a basis for the

Kernel we solve the system: B  B  B  B œ ! .

B  B  B  B œ !^" ^# ^$ ^% Ê B œ  B  B  B

" # $ % % " # $

Every element —−Ker is —œ B B B ß  B  B  B "ß #ß $ " # $ and a basis for the Kernel is U_{O/< J}_{ } œ" ! !ß  " ß ! " !ß  " ß ! ! "ß  "ß ß   ß ß   ß ß .

I M 3) The characteristic polynomial of is  :  - œ -ˆœ

œ œ œ

   5

 

   

# # " " ! " 

! "  $ ! "  $

" # 5 " # 5

- - -

- -

V  V" $

œ "     5  œ

 - -    -  -

-

" !

"

 $ "

# 5 "  #

œ "  - - ^#- 5  5 -  " - 5- " œ œ "  - - ^# #  5 - #5  ".

So - œ " is an eigenvalue for the matrix for every value of the parameter . 5 To have multiple eigenvalues for there are two possibilities:

a) -œ " is a root also for -^# #  5 - #5  " œ !; b) -^# #  5 - #5  " œ ! has a multiple (double) root.

a) -œ " is a root for -^# #  5 - #5  " œ ! if "  #  5  #5  " œ !, satis- fied for 5 œ #. If 5 œ # we get -^# %  $ œ- - "- $ œ ! with roots -œ "

and - œ $. For -œ " and 5 œ # we have   " †ˆœ

 



" # "

! !  $

" # "

whose Rank

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is clearly equal to and so # 7 œ $  # œ "  7 œ #¹_" ⁺_" . For 5 œ # the matrix is not a diagonalizable one.

b) -^# - -

#

 #  5  #5  " œ ! Ê œ #  5 „ #  5  % #5  "

     #   

and we have a multiple (double) root if #  5^# % #5  " œ 5  %5  ) œ !  ^# . But this equation does not have real solutions and therefore case b) does not provide multiple solutions. So the matrix is not diagonalizable only for  5 œ #.

I M 4) Since ˜œ —† œ † œ

"  " " " "  5

" #  " 5 "  #5

"  " " ! "  5

     

      we get:

  ˜ œ"  5^# "  #5 ^# "  5 ^# œ'5  $^# . has modulus equal to if˜ $ and only if '5  $ œ *^# or 5 œ „ ".

II M 1) f œ #0 B  C  #C  #BC . Bß C # #C

#C  #  #B

 ^#ß  ‡  œ   .

M SG # œ ! Ê œ  #

œ #C B  " œ ! œ "

B œ !

:  B  C  œ ! C

#BC  #C C B

# #

  ∪ Impossible .

We have only the stationary point  !ß ! .

MM SG: ‡ !ß ! # !   

!  #

œ . Since 0_BB^ww  ! while 0_CC^ww  ! the point !ß ! is a sad- dle point.

II M 2) The problem    



 

 Max min

u c

Max min

u c .

Î 0 Bß C œ Þ Þ !

Î 0 Bß C œ Þ Þ  C Ÿ !

Ÿ ! BC

Ÿ C Ÿ #B  B Ê

BC C  #B  B

# #

The feasible region is red-drawed in the figure in the next page.

The constraints are qualified, the objective function is continous and the feasible region is a bounded and closed set, so by Weierstrass Theorem the problem admits the absolute maximum and minimum value.

In the feasible region we have 0 Bß C !  with 0 Bß C œ !  only for C œ ! Þ So all the points ! Ÿ B Ÿ #ß C œ ! are minimum points.

The function 0 Bß C œ  BC has a unique stationary point:  !ß ! , just studied.

On the points satisfying C  #B  B œ^# ! we apply the Kuhn-Tucker condition using the Lagrangian function ABß Cß-œ BC -C  #B  B^#.

We have to solve the system:

  

  

    

C  #  # B œ ! B  œ !

C œ #B  B

Ê Ê

œ B œ B

C œ #B  #B C œ #B  #B

#B  #B œ #B  B $B  %B œ B $B  % œ !

- -

-

- -

#

# #

# # #

and so we get the two solutions: just studied and which is, clearly,



 





 B œ !

C œ ! œ !

B œ C œ

- - œ

% )$

*%

$

the maximum point.

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We conclude: Max 0 œ 0% )ß  œ $#; Min 0 œ 0 Bß ! œ ! ! Ÿ B Ÿ #  , .

$ * #(

In the figure below are drawn zero level curve (yellow) and positive level curves (blue).

II M 3) 0 Bß C œ /  ^BC /^CB is a differentiable function, so H 0 T œ_@   f0 T †@ and H 0^#_@ßA  T œ †@ ‡ T † A^X .

f œ0 /^BC /^CBß /^BC /^CB, 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 true for α α 1 and α 1.

œ œ œ

% %

$ Given A œcos"ßsen" Bß C /  / /  /

/  / /  /

, since ‡  œ 

 ^BC^BC ^CB^CB ^BC^BC ^CB^CB, it re- sults ‡ "ß " œ ! and finally H 0 œ †‡ "ß " † A œ !, indepen-

!! 

! ^#_@ßA "ß " @ ^X

dently from .A

II M 4) The system   

   

0 Bß Cß D œ B C  /  / œ "

1 Bß Cß D œ BCD  /^# ^BD_BD ^DC_DC "ß "ß "ß

 / œ " at point is satisfi-

ed. From ‰   

   

Bß Cß D œ ` 0 ß 1 œ

` Bß Cß D

#BC  / B  / /  /

CD  / BD BC  /

BD # DC BD DC

BD  /DC BD /DC

we get ‰"ß "ß " œ " ! # . Since  ! œ # Á ! with this

# #  " #

"

# system we can

define, in a neighborhood of the point "ß "ß ", an implicit function D Ä B D ß C D    . For this function we get:

d

dB % and

D " œ B " # #

# !

#

" !

#

     

 

w œ   " œ  œ 

# d .

d

C & &

D " œ C " # #

" #

" !

#

     

 

w œ  #  " œ   œ

#

The equation of the tangent line at D œ " is D Ä B " ß C "     D B " ß C " ^w   ^w  or D Ä "ß "  D #ß& œ "  #Dß "  &D

# #

     , or, in cartesian form:

C œ  B &

% %

* .