QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS TASK 13/1/2021

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QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS

TASK 13/1/2021

I M 1) If / œ / $  3 , find .D

#

D 

 

$

Applying the definition of complex exponential: /^B3C œ /^BcosC  3senC and sinceÀ

/ œ / $  3 œ /  3 œ /  3 À

# # # ' '

$ " "" ""

D 

   

$ " "

$ $

cos 1 sen 1 we get / œ / D œ "  3""

$ '

D ^"^$^3^""^'¹ and so 1 .

I M 2) Given the vectors —_" œ "ß !ß  #ß # , —_# œ #ß "ß  "ß #  and —_$ œ "ß #ß %ß 5 , check, depending on the parameter , when they are linearly independent.5

We solve the problem by calculating the rank of the matrix Œ œ .

" !  # #

# "  " #

" # % 5

 

If the vectors are linearly independent, the rank must be equal to ; if, on the other hand, the$ vectors are linearly dependent, the rank must be less than .$

Since " ! , the rank is at least equal to .

# " œ " Á ! #

By elementary operations on the rows V Ã V  #V_# _# _" , V Ã V  V_$ _$ _", we getÀ

   

" !  # # " !  # #

# "  " # ! " $  #

" # % 5 ! # ' 5  #

Ä Þ

By elementary operations on the rows V Ã V  #V_$ _$ _#, we getÀ

   

" !  # # " !  # #

! " $  # ! " $  #

! # ' 5  # ! ! ! 5  #

Ä Þ

Therefore, if 5 Á  # the rank is equal to and the vectors are linearly independent, if inste-$ ad 5 œ  # the rank is equal to and the vectors are dependent.#

I M 3) Determine the only value of the parameter for which the matrix 5 is dia-

" # $

! 5 #

! ! "

 

gonalizable.

From - ˆœ ! we get:

 

" # $     

! 5 #

! ! "

" 5 "



   œ ! œ œ "ß œ 5

-

- - - -

œ   Þ So _" _# _$ Þ

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If 5 œ " we get 7 œ $ and  " † œ and since # $ œ % Á !, it is

! #

+"  

 

   

 ˆ

! # $

! ! #

! ! !

Rank " †ˆœ # ß so 7 œ $  # œ "  7 œ $¹_" ⁺_" and the matrix is not diagonalizable.

If 5 Á " it is 7 œ # and  " † œ à if # $ œ (  $5 œ !

#

+"  

 

   

 ˆ

! # $

! 5 #

!  "! ! 5

 "

then Rank " †ˆœ " ß so 7 œ $  " œ # œ 7¹_" ⁺_" and the matrix is diagonalizable.

If 5 Á ( then Á !, Rank  " † œ # ß so 7 œ $  # œ "  7 and the

$

# $

5  " #  ˆ ¹_" ⁺_"

matrix is not diagonalizable. Therefore the only value of the parameter for which the matrix5 is diagonalizable is 5 œ (Þ

$

I M 4) Consider the linear map 0 À Ä , œ † whith œ .

"  " 7 "

! # " 5

" " #5 7

‘^% ‘^$ ˜  — 

 

Since 0 "ß "ß "ß " œ %ß &ß *   , determine the values of the parameters and and then find7 5 the dimensions of the Kernel and of the Image of this linear map.

From 0 "ß "ß "ß " œ %ß &ß *    we get:

     

 

 

 

 

 

"  " 7 " % "  "  7  " œ % 7 œ $

! # " 5 & !  #  "  5 œ & 5 œ #

" " #5 7 * "  "  #5  7 œ * #  %  $ œ *

† œ Ê Ê Þ

"

So  œ . By elementary operations on the rows V Ã V  V and

"  " $ "

! # " #

" " % $

 

   $ $ "

then V Ã V  V_$ _$ _#, we getÀ

     

       

"  " $ " "  " $ " "  " $ "

! # " # ! # " # ! # " #

" " % $ ! # " # ! ! ! !

Ä Ä Þ "  "

! #

Since Á !

we get Rank  œ # and so Dim Imm œ # and Dim Ker œ %  # œ # Þ

II M 1) Given 0 Bß C œ B  $BC  #C  ^# ^#, and the unit vectors of @ A  "ß " and "ß  ", since W_@0 P_! œ# and W_A0 P_! œ ##, determine P and then calcolate _! H_@^#_ßA0 P_! . The function 0 Bß C œ B  $BC  #C  ^# ^# ist twice differentiable a Bß C −  ‘^#.

So W_@0 P_! œ f0 P_! † @ and W^#_@ßA0 P_! œ @ †‡0 P_! † A^X . From f0Bß C œ #B  $Cà %C  $B   we getÀ









        

W W

@ ! !

A ! !

0 #B  $Cà %C  $B œ #

0 #B  $Cà %C  $B œ # #

P P

œ f0 † @ œ †

œ f0 † A œ †

" "

# #

" "

# #

 

ß ß 

Ê

#B  $C  %C  $B # C  B # C # B

#B  $C  %C  $B % &B  (C % &B  (B  "% % C (

œ œ œ B  œ  *

œ Ê œ Ê œ Ê œ  .

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Therefore P_! œ  *ß  ( . It is then #  $  *ß  (

 $ %

  ‡Bß C œ  œ ‡  and so:

W_@ßA^# 0 _! #  $ œ

 $ %  

   

   

P œ ^"_# ^"_# † † œ ^"_# ^"_# †

" &

# #

"

#

(

#

     

 

œ &  ( œ  "

# # Þ

II M 2) Solve the problem .

Max/min u.c.:





   



0 Bß C œ B C  "

B Ÿ "  C

" Ÿ B  C

#

The objective function of the problem is a continuous function, the constraints define a feasi- ble region which is a compact and bounded set, and so we can apply Weierstrass Theorem.

Surely the function admits maximum value and minimum value.

To solve the problem we use the Kuhn-Tucker conditions.

We write the problem as .

Max/min u.c.:





   



0 Bß C œ B C  "

B  C  " Ÿ !

"  B  C Ÿ !

#

We form the Lagrangian function:

ABß Cß- -_"ß _#œB C  "   -_"B  C  " ^#  -_#"  B  C. By applying the first order conditions we have:

1) case -_" œ !ß-_# œ ! À

 

 

 

 

 

 

A A

wB wC

œ C  " œ ! œ B œ !

Ê Ê

B œ ! C œ  "

B  C  " Ÿ !

"  B  C Ÿ !

!  "  " Ÿ ! Ÿ ! À

#

"

!ß  " Â Þ

 !  " NO

X

2) case -_" Á !ß-_# œ ! À

 

 

 

 

 

A -

-

- - - -

- - -

wB "

wC "

"

" " # "

"

#" " "

œ C  "  œ ! œ B  # C œ !

Ê Ê

C œ  "

B œ #  " œ #  #

#  # œ  "

B œ "  C

"  B  C Ÿ !

" 

"  B  C Ÿ !

#   #

 

Ê Ê

C œ  "

B œ #  " œ #  #

$  % œ  %

B œ ! C œ  "

B œ C œ

 

  

  

  

 







 -

- - - -

"

" " # "

"

#" " " " " "

 

$ œ !

"  B  C Ÿ !

œ !

Ÿ !à œ  !

"   Ÿ !

"

∪ à

 !  " NO

)

"*

$% )$ "

* $

Since -" œ %  ! the point ) "ß could be a Maximum point.

$ * $

3) case -" œ !ß-# Á! À

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  

  

  

  

  

A -

- -

- - -

wB #

wC #

#

# # #

œ C  "  œ !

œ B  œ !

C œ "  B Ê Ê

B œ  C œ  " 



B œ "

C œ ! B  C  " Ÿ !

œ "

B  C  " Ÿ !

œ  "  !

"  !  " Ÿ !

# #

 "  Þ

Since -_# œ  "  ! the point  "ß ! could be a minimum point.

4) caso -" Á!ß-# Á! À

  

  

  

  

  

A - -

- - - -

- - -

wB " #

wC " #

" # " #

# " #

œ C  "   œ ! œ B  # C  œ !

"  B B œ " 

Ê ∪ Ê

B œ " B œ !

C œ ! C œ "

"   œ ! #   œ !

"  œ !  œ !

C œ

C^#  #

Ê ∪

B œ " B œ !

C œ ! C œ "

œ ! œ  #  !

œ  "  ! œ  %  !

 

 

 

 

- -

- -

" "

# #

.

Since the values are not positive, points -  "ß ! and  !ß " could be minimum points.

Let's study the objective function on the constraint B œ "  C^#.

It is 0 "  C ß ^# C œ "  C^#C  " œ C  "  C  C Ê 0 C œ "  $C  #C ! ^$ ^# ^w  ^# . So $C  #C  " Ÿ ! Ê C œ  "„ "  $ œ  "„# . Therefore:

$ $

# 

0 C !  " Ÿ C Ÿ " Ê ! Ÿ C Ÿ " C œ " Ê B œ )

$ $ $ *

w  for . If .

At the point ) " there is the Maximum with ) " $#.

* $ß 0 * $ß œ #(

Let's study the objective function on the constraint C œ "  B.

It is 0 Bß "  Bœ B"  B " œ #B  B Ê 0 B œ #  #B ! ^# ^w  for ! Ÿ B Ÿ ". If B œ ! Ê C œ ". The point  !ß " is the minimum point with 0 !ß "  œ0.

If B œ " Ê C œ !. The point  "ß ! it is neither a minimum nor a maximum point.

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II M 3) Given the equation 0 Bß Cß D œ #BC  /  ^DB  /^DC œ !, satisfied at "ß "ß ", veri- fy that an implicit function Bß C Ä D Bß C   can be defined with it and then calcolate the first order derivatives of such function.

The function 0 Bß Cß D  is differentiable a Bß Cß D −  ‘^$, and 0 "ß "ß " œ #  "  " œ !  . It is: f0Bß Cß Dœ #C  /^DBà #B  /^DCà  /^DB /^DCÞ

So f0"ß "ß "œ $à $ à  #.

Since 0 "ß "ß " œ_D^w   #Á ! it is possible to define an implicit function Bß C Ä D Bß C   whose first order derivatives will be equal to:

`D $ $ `D $ $

`B     # # `C   # #

   

"à " œ  0 "ß "ß " œ  œ à "à " œ  œ  œ Þ 0 "ß "ß " 0 "ß "ß "

0 "ß "ß "

Bw

w w

D D

Cw

II M 4) Given the function 0 Bß Cß D œ B  C  D  B C  CD  ^# ^# ^# ^# , analyze the nature of its stationary points.

We determine the stationary points of the function. Applying the first order conditions we have:

f0 Bß Cß D œ   Ê Ê

0 œ #B  #BC œ #B "  C œ ! 0 œ #C  B  D œ !

0 œ #D  C œ !

B œ ! C œ ! D œ !

 

 

 

 

Bw

w #

C Dw

and also:

  

  

  



 

 

 

 

B œ #  œ

C œ "

D œ

Ê

B œ B œ 

C œ " C œ "

D œ D œ

# " $

# #

"

#

$ $

# #

" "

# #

and Þ

We have then found three stationary points: !ß !ß ! , _#^$ß "ß ^"_# and ^$_#ß "ß^"_#Þ

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We apply the second order conditions. From ‡Bß Cß D œ we get:

 

#  #C  #B !

 #B #  "

!  " #

‡!ß !ß ! œ à

 













 

       

   

# ! !

! #  "

!  " #

Ê

œ #  !

œ # !  !à œ #  "  !

! #  " #

œ # † #  "  !

 " #

‡

‡ ‡

‡

"

# #

$

since the point





  

 ‡  

‡

"

#

$

 !

!ß !ß ! is a minimum pointà

‡   

 

   

 



 



$ "

#ß "ß# Ê œ  !

!  # !

 # #  "

!  " #

!  #

 # #

œ à

$

#

$

# #

$

#

$

#

‡

since  ‡_#  ! the point $ß "ß" à

# #

  is a saddle point

‡ œ   à

 

   

 



 



$ "

#ß "ß # Ê œ  !

! # !

# #  "

!  " #

! #

# #

$

#

$

# #

$

#

$

#

‡

since  ‡_#  ! the point $ß "ß"

# #

  is a saddle point.