QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS TASK 15/3/2021

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QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS

TASK 15/3/2021

I M 1) If D œ /cosα 3sinα, since /^"$3† D œ /^#3, determine .α

/ † D œ / Ê D œ / œ / œ / œ / %  3 %

/

"$3 #3 #3 #3 "$3 "%3

"$3   cos sin .

Therefore α œ % Þ

I M 2) In the basis for ‘^$ formed by the vectors •" œ "ß #ß  " , •² œ #ß "ß "  and

•_$ œ B ß B ß B _" _# _$, the coordinates of the vector ˜œ "ß $ß !  are #ß  #ß ". Determine —_$. The problem is equivalent to solving the following linear system:

     

 

 

 

 

" # B # " #  %  B œ " B œ $

# " B  # $ %  #  B œ $ B œ "

 " " B " !  #  #  B œ ! B œ %

† œ Ê Ê Þ

" " "

# # #

$ $ $

And so •$ œ &ß $ß ! Þ 

I M 3) Given the matrix  œ , determine, on varying the parameter , the7

# ! "

# " #

" 7 %

 

existence of multiple eigenvalues and determine if, for these values, the matrix is diagonalizableÞ

Let's determine the eigenvalues of the matrix:

From   we get:

   

- ˆ œ œ œ !

# ! " " ! "

# " # ! " #

" 7 % 7 %

 

  $ 

- -

- - -

"-  "-%- #7  - $ !  "-œ

œ"- - ^# &  %  #7 - - $ œ "- - ^# '  #7  ( œ !-  Þ There are two ways to obtain multiple eigenvalues À

 The polynomial -^# '  #7  (-  vanishes for - œ " à

 The polynomial -^# '  #7  (-  admits two identical roots.

First case For Þ - œ " we get: 1 '  #7  ( œ !Ê #7 œ # Ê 7 œ " and so:

-^# '  & œ !- Ê- œ $„*  & œ $„# and so we get - œ " and - œ &Þ For 7 œ " and - œ " we get:

   

 

     

 " †ˆ œ Ê œ # Á !  " †ˆ œ #

" ! "

# ! #

" " $

# !

" " and so Rank and therefore 7 œ $  # œ "  7 œ #¹_" ⁺_" and si the matrix is not diagonalizable.

Second case Þ-^# '  #7  ( œ !- Ê- œ $„*  #7  ( œ $„#  #7Þ

#  #7 œ ! for 7 œ  " and the eigenvalues are -_" œ "ß-_# œ-_$ œ $ Þ For 7 œ  " and - œ $ we get:

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   

 

     

 $ †ˆ œ Ê œ # Á !  $ †ˆ œ #

 " ! "

#  # #

"  " "

 " !

#  # and so Rank

and therefore 7 œ $  # œ "  7 œ #¹_$ ⁺_$ and so the matrix is not diagonalizable.

For any other value of we will have only simple eigenvalues and therefore the matrix will7 be diagonalizable.

I M 4) Given the linear system , find the values of the parameter





B  B  #B  B œ #

#B  B  B  #B œ "

B  %B  (B  B œ 5

" # $ %

5 for which the system has solutions.

To check if the linear system admits solutions we use the Rouchè-Capelli Theorem and we calculate the rank of the matrix and the rank of the augmented matrix À

 

 

 ˜l œ Þ

"  " # " l #

# "  " # l "

"  % ( " l 5

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

   

"  " # " l # "  " # " l #

# "  " # l " ! $  & ! l  $

"  % ( " l 5 !  $ & ! l 5  #

Ä Þ

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

   

"  " # " l # "  " # " l #

! $  & ! l  $ ! $  & ! l  $

!  $ & ! l 5  # ! ! ! ! l 5  &

Ä Þ

From this it follows that Rank  œ # œ Rank ˜l  only for 5 œ &, and for this value the system has ∞^%# œ ∞^# solutions For Þ 5 Á & it is Rank  œ # Rank ˜l œ $ and the system has no solutions.

II M 1) Given the equation 0 Bß C œ B  $BC  C  $ œ !  ^$ ^$ and the point T œ  "ß "  that satisfies such equation, find the equation of the tangent line to the graphic of the implicit function C œ C B  defined at the point B œ  ".

The function 0 Bß C œ B  $BC  C  $  ^$ ^$ is differentiable a Bß C −  ‘^#. From f0Bß Cœ $B ^# $Cà $C^#  $B we get f0 "ß "œ !à ' Þ 

Since 0_C^w "ß " œ 'Á ! it is possible to define an implicit function CœC B whose first

order derivative is equal to: So the equation of the

d

dC !

B    '

 

 " œ  0  "ß " œ  œ ! Þ 0  "ß "

Bw Cw

tangent line to the function C œC B at B œ  "! is C  " œ ! † B  "  that is C œ " Þ II M 2) Solve the problem Max/min .

u.c.:

 0 Bß C œ B  C  C  " Ÿ B Ÿ "^#

The objective function of the problem is a continuous function, the constraint defines a feasi- ble region which is a closed and bounded (compact) set, and so we can apply Weierstrass Theorem. Surely the function admits maximum value and minimum value.

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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 C  B  " Ÿ ! B  " Ÿ !

# Þ We form the Lagrangian function:

ABß Cß-œ B  C -"C  B  " #  -#B  ". By applying the first order conditions we have:

1) case -" œ !ß-# œ ! À







 A A

wB wC

œ " Á ! œ  " Á ! C  B  " Ÿ ! Ê B  " Ÿ !

# no solutions Þ

2) case -_" Á !ß-_# œ ! À

 

 

 

 











A -

w -

B "

wC "

"

œ "  œ ! œ  "  # C œ !

Ê

œ  "  ! C œ

B œ C  " B œ B Ÿ "

 " œ 

 Ÿ "

#

"

#" $

% %

$

%

à

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

 % #

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3) case -_" œ !ß-_# Á! À









A -

A

wB #

wC

œ "  œ ! œ  " Á ! B œ " Ê

C  B  " Ÿ !^#

no solutions Þ

4) case -" Á!ß-# Á! À

 

 

 

 

 

A - - A - -

A - A -

w w

B " # B " #

w w

C " C "

œ "   œ ! œ "   œ !

œ  "  # C œ ! œ  "  # C œ !

" "

B œ

Ê Ê

B œ B œ

C  "^# C œ #^#

Ê ∪

B œ " B œ "

C œ # C œ  #

"   œ ! "   œ !

 "  # C œ !  "  # C œ !

 

 

 

 

 

 

- - - -

- -

" # " #

" "

Þ

 

  

  

  

 





 B œ "

C œ #

"   œ !

 "  # C œ !

Ê Ê

B œ "

C œ #

"   œ !

 "  # # œ !

B œ "

C œ #

œ   !

œ "  œ  !

 



- -

-

- -

-

- -

" #

"

" #

"

# "

"

# #

# #"

# #





Þ

Since -_" œ  "  ! and -_# œ # #  "  ! the point "ß # it is neither a maxi-

## ##

   

mum nor a minimum point .

 

  

  

  

 





 B œ "

C œ  #

"   œ !

 "  # C œ !

Ê Ê

B œ "

C œ  #

"   œ !

 "  # # œ !

B œ "

C œ  #

œ  !

œ "  œ  !

 



- -

-

- -

-

- -

" #

"

" #

"

# "

"

# #

# #"

# #





Þ

Since -_" œ " and -_# œ # #  " the point " could be a maximum

# #  ! # #  ! ß  #

 

   

point.

However, having found only two solutions, for the Weierstrass Theorem, the point

 $ "  $ "  & " 

% #ß is the minimum point, with 0 % #ß œ % while the point ß # is the maximum point with 0"ß #œ" #.

II M 3) Given 0 Bß C œ BC  #B  C  , let and be the unit vectors of @ A  "ß " and "ß  "; determine if there are points where T W@0 T œ  # and WA0 T œ   #.

The function 0 Bß C œ BC  #B  C  is differentiable a Bß C −  ‘^#. So W_@0 T œ f0 T † @ and W_A0 T œ f0 T † A Þ

From f0Bß C œ C  #à B  "  , @ œ " ß " , A œ" ß  "  we get À

# # # #

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







 

    

C  #à B  " † œ #

C  #à B  " † œ  #

B  C  " œ # B  C œ $ C  B  $ œ  # C œ  "

" "

# #

" "

# #

 

ß ß 

Ê Ê

B 

ÊC  "  C œ $ Ê #C œ % Ê œ "

œ C  " œ C  " œ # T œ "ß #

B B C

B Þ So  Þ

II M 4) Find the pair Bß C for which the determinant of the matrix œBC C  has B B  C

 a minimum value.

From   œBC C œ BC B  C  BC  we get 0 Bß C œ B C  BC  BC Þ  B B  C

# #

We determine the stationary points of the function. Applying the first order conditions f0 Bß C œ   we have:

0 œ #BC  C  C œ !   

0 œ B  #BC  B œ !

C #B  C  " œ ! B  #C  " œ !

w #

B

w #

C

Ê  

B Þ So four possible cases:

C œ ! C œ ! #B  C  " œ ! #B  C  " œ ! œ ! B  #C  " œ ! œ ! B  #C  " œ !

B ∪ ∪ B ∪ and so:

B œ ! B œ " B œ ! C œ "  #B  œ ! C œ ! œ " B  #  %B  " œ !

B œ

C ∪ ∪ C ∪ œ Þ

Ê C

"

"$

$

From ‡Bß C œ  #C #B  #C  " we get:

#B  #C  " #B

‡ !ß ! œ !  " À  ‡ œ  "  !  !ß !

 " ! since _# the point is a saddle point;

‡ "ß ! œ! " À  ‡ œ  "  !  "ß !

" # since _# the point is a saddle point;

‡ !ß " œ# " À  ‡ œ  "  !  "ß !

" ! since _# the point is a saddle point;

‡ ‡

     ‡  

 

" " " "

$ $ $ $

 !

 œ  !

ß œ À œ ß

œ

# "

$ $

" #

$ $

#

$% " "

* * $

since ^" the point

#

is a minimum point, the solution of the problem.