TASK MATHEMATICS for ECONOMIC APPLICATIONS 19/09/2019

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 19/09/2019

I M 1) Calcolate    . Since

  

# "  3 #  $3

&  3 "  3

# "  3 #  $3 # #  #3  $3  $ # &  3 &  3 &  3 #  $3

&  3 "  3 œ &  &3  3  " œ %  '3 œ #  $3 œ #  $3 #  $3† œ

      

  

œ "!  #3  "&3  $ œ "$  "$3 œ "  3 œ # † " "

%  * "$  #  3# œ

œ # † cos( sin ( , we get:

%1  3 %1

   

         

# "  3 #  $3

&  3 "  3 œ "  3 œ ^% # cos ( 5 # sin ( 5 # Ÿ Ÿ . )1  #1  3 )1  #1 ß ! 5 "

And so - œ" ^% # †cos( sin (  and - œ# ^% # †cos "& sin"& .

)1  3 )1 )1  3 )1

I M 2) Find eigenvalues and corresponding eigenvectors of the matrix  œ .

" # "

% ! !

" # "

 

From - ˆœ ! we get

" # " # "

% ! % !

" # " # "

   

 

œ œ

- -

- - -

œ ") -^#"- - ^# ) œ  -^$ #-^# ) œ !- Ê - - ^#  #  ) œ !-  .

So -" œ ! and -#ß$ œ "„"  ) œ "„$Ê-# œ %ß -$ œ  #.

For finding the corresponding eigenvectors we solve three systems.

-_" œ ! À 

     

 

  

 ! †ˆ †—œ Ê † œ Ê Ê

" # " B !

% ! ! B !

" # " B !

B  #B  B œ !

%B œ !

"

#

$

" # $

"

Ê B œ !

B œ  #B

 ^"_$ _# and so eigenvectors corresponing to -_" œ ! are —_" œ !ß Bß  #B .

-# œ % À 

     

 

 % †ˆ †—œ Ê † œ Ê

 $ # " B !

%  % ! B !

" #  $ B !

"

#

$

Ê Ê

 $B  #B  B œ ! B œ B

%B  %B œ ! B œ B

B  #B  $B œ ! B œ B

 

 

 

" # $ $ "

" # # "

" # $ $ "

and so eigenvectors corresponing to -# œ % are

—_# œ Bß Bß B Þ 

-$ œ  # À 

     

 

 # †ˆ †—œÊ † œ Ê

$ # " B !

% # ! B !

" # $ B !

"

#

$

Ê Ê

$B  #B  B œ ! B œ B

%B  #B œ ! B œ  #B

B  #B  $B œ ! B œ B

 

 

 

" # $ $ "

" # # "

" # $ $ "

and so eigenvectors corresponing to -$ œ  # are —$ œ Bß  #Bß B Þ 

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I M 3) Consider the linear map 0 À‘^% Ä‘^$, ˜œ  —† for which:

0 B ß B ß B ß B " # $ % œ B  B  B à B  B  B à B  B  B" # $ " $ % " # %.

Determine a basis for the Kernel and a basis for the Image of this linear map.

We have  œ . By elementary operations on the rows:

" " " !

" ! " "

" " ! "

 

   

 

V Ã V  V V Ã V  V À

" " " !

!  " ! "

! !  " "

# # " and $ $ " we get from which we see

that Rank  œ $ and so Dim Imm   œ $ and Dim Ker   œ %  $ œ ". To find a basis for the Kernel we must solve the system:

 —† œ Ê † œ Ê Ê

" " " ! ! B  B  B œ !

" ! " " ! B  B  B œ !

" " ! " ! B  B  B œ ! B

B B B

     

 





"

#

$

%

" # $

" $ %

" # %

Ê Ê

B  B  B œ ! B œ  #B

 B  B œ ! B œ B

 

 

 

" # $ " %

# % # %

$ % $ %

. So the vectors belonging to the Kernel are:

—œ  #Bß Bß Bß B  and a basis may be the vector —œ  #ß "ß "ß " .

To find a basis for the Image theoretically we should apply Rouchè-Capelli theorem to the system:

 —† œ˜ Ê † œ Ê

" " " ! C B  B  B œ C

" ! " " C B  B  B œ C

" " ! " C B  B  B œ C B

B B B

     

 





"

#

$

%

" " # $ "

# " $ % #

$ " # % $

and then we should study the Rank of the matrix and the Rank of the augmented matrix:

 

" " " ! l C      

" ! " " l C

" " ! " l C

œ $ œ

"

#

$

. But since Rank  Dim Imm  , the map is a surjective one and so a basis for the Image is simply the standard basis of ‘^$

I M 4) Since the vectors —_" œ "ß #ß  " , —2 œ #ß !ß "  e —_$ œ B ß B ß B _" _# _$ form a basis for

‘^$ and since in this basis the coordinates of the vector ˜œ "ß $ß !  are #ß  #ß ", determine

—_$.

To satisfy what is required, we must solve the system:

     

 

 

 

 

" # B # " #  %  B œ " B œ $

# ! B  # $ %  !  B œ $ B œ  "

 " " B " !  #  #  B œ ! B œ %

† œ Ê Ê

" " "

# # #

$ $ $

. And so —_$ œ $ß  "ß % .

II M 1) Given the equation 0 Bß C œ B C  BC  #C œ !  ^$ ^# and the point P! œ "ß "  satisfying it, determine first and second order derivatives of the implicit function B Ä C B  definable with it.

From f0Bß Cœ $B C  C à B  #BC  # ^# ^# ^$  we get f0   "ß " œ %à " and so:

C " œ % œ  %

"

w  . Then we have:

‡Bß Cœ 'BC $B  #C Ê‡ "à " œ ' &

$B  #C #B & #

 # ^#    and so, since:

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C œ  0  #0 C  0 C C " œ  œ # Þ

0 "

'  "! †  %  #  %

ww ww ww w ww w ww

BB BC CC #

Cw

  #

     

, we get

II M 2) Given the function 0 Bß C œ B C  $B  $B  #C  ^$ ^# and the point T œ "ß  #!  , compute W_@0 !ß ! , where represents the direction from the origin @  !ß ! to .T_!

0 Bß C œ B C  $B  ^$ ^# is a differentiable function a Bß C −  ‘^#. So H 0_@  !ß ! œ f0 !ß ! † @.

We get f0 B ß CœC  'B  $^$ à $BC  # Ê f0^#    !ß ! œ $ß  #.

Since     it is and therefore we have:

 

T @ œ " ß  #

! œ "  % œ &

& &

 

W_@0 !ß ! œ !ß ! " ß  # œ (

  f0 † @ œ $ß  # † 

& & &

       .

II M 3) S

Max/min u.c.:

olve the problem: .





   



0 Bß C œ C B  "

B  C  " Ÿ ! C Ÿ !

#

The objective function of the problem is a continuous function, the feasible region is a com-X pact set, and so surely exist maximum and minimum values.

Since in X it is C Ÿ ! and B Ÿ ", it is also 0 Bß C ! a Bß C −  ,   X.

Using Kuhn-Tucker's conditions, we form the Lagrangian function:

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

 

 

 

 

 

A A

wB wC

œ C œ ! œ B  " œ !

Ê

B œ "

C œ ! B  C  " Ÿ !

C Ÿ !

B  C  " Ÿ ! C Ÿ !

Bà C "à !

# # ; but ‡ œ‡ œ ! " which indicates a

" !

 

saddle point, to be re-checked anyway since  "à ! is on the boundary of X.

2) case -_" Á !ß-_# œ ! À

 

 

 

 

 

A -

- -

- - -

wB "

wC "

"

" " "

œ C  # B œ ! œ B  "  œ !

Ê

B œ "  C œ " 

"   # "  C œ B  "

C Ÿ !

 "

 " œ !

C Ÿ !

#

 

    which leads to the equation:

-^#_" #-_""  " #-_" #-^#_" œ $-_"^#  %-_" œ-_"$-_" % œ ! from which we get two solutions:

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 

 

 

 









 B œ "

C œ

B œ  C œ  0

-^" œ ! -_"

C Ÿ !à œ  !

C Ÿ !à

true true

already studied and , possible point of Max.

"

)$

% *

$

3) case -" œ !ß-# Á! À







 A

A -

wB

wC #

œ C œ !

œ B  "  œ !

C œ ! C œ ! 0 Bà ! œ !

B  C  " Ÿ !^#

. But if it is   i.e. a constant function.

4) case -" Á!ß-# Á! À

 

 

 

 

 

A -

A - -

- -

wB "

wC " #

"

#

œ C  # B œ !

œ B  "   œ ! C œ !

Ê

œ !

œ  #  ! C œ B  "^# B œ  "

C œ !

possible point of Min and

already studied. So by Weierstrass's Theorem is the Maximum point,









 

- -

"

#

œ !

œ !  à " )

$ *

B œ "

C œ !

with 0 œ $# while  "à ! , and all the points of the axis C œ !, are Minimum

 à " ) #(  

$ *

points, with 0Bà !œ !.

II M 4) Given the function 0 Bß C œ B  5BC  C  ^$ ^# nalyze the nature of its stationary pointsa on varying the parameter .5

To nalyze the nature of the stationary points of the function we apply first and second order a conditions. For the first order conditions we pose:

f Ê 0 œ $B  5C œ ! Ê

0 œ #C  5B œ !

$B  B œ B $B  œ !

C œ B 0 Bß C œ   







 

w #

Bw C

# 5 5

# #

5

#

# #

and so we get

two stationary points: T œ_"  !à ! and T œ_# 5 à5  ' "# Þ

# $

For the second order conditions we construct the Hessian matrix: ‡Bß C œ  'B  5.

 5 #

Since ‡ !à ! œ !  5, and since  ‡ œ  5  ! the point !à ! it is surely

 5 # ^#

# T œ_"  

a saddle point if 5 Á !; if 5 œ ! we get 0 Bß C œ B  C  ^$ ^# and since 0 Bß !  !  for B  ! while 0 Bß !  !  for B  !, also if 5 œ ! the point T œ_"  !à ! is a saddle point.

Then we study:

‡ ‡

5 5  ‡

' "# œ 5  5 T

 5 #

5

5 5

# $ #

" #

# # # #

à    

, and since  œ ! #  ! the point is surely a mi-

œ  !

; 2

nimum point if 5 Á !. If 5 œ ! we get the case, already studied, of the point  !ß ! .