QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS TASK 3/2/2020

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QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS

TASK 3/2/2020

I M 1) Transform D œ "  $  "  $ in trigonometric form and then calculate D

"  3 "  3

  _$

D œ "  $  "  $ œ œ

"  3 "  3 "  3 "  3

"  $ "  3  "  $ "  3

         

  

œ "  $  "  $   "  $  "  $ 3 œ œ "  $ 3 œ

"  " #

#  # $ 3

         

œ # "  $3 œ # &  3 &

# # $ $

  

cos 1 sin 1.

So D œ )^$ cos&  31 sin&1œ )cos1 3sin1œ  ).

I M 2) Given the matrix œ and the vector —œ , determine the va-

" # " "

 " 7 # "

" " 5 "

   

lues of and for which the vector 7 5 ˜œ —† is perpendicular to the vector !ß "ß  " and with modulus equal to #%.

˜œ  —† œ † œ

" # " " %

 " 7 # " 7  "

" " 5 " 5  #

     

      .

˜œ  —† perpendicular to !ß "ß  " implies %ß 7  "ß 5  # † !ß "ß  " œ ! Ê   Ê !  7  "  5  # œ ! Ê 7  5 œ " Ê 7 œ 5  " .

Modulus of equal to ˜ #% implies  ˜ œ%  7  "^#  ^# 5  # ^# œ#% Ê Ê%  5  #^#  ^# 5  # ^# œ%  # 5  #^#  ^# œ#% Ê

Ê #5  )5  #% œ #% Ê #5 5  % œ ! Ê 5 œ ! 7 œ 5  " œ "

5 œ  % 7 œ 5  " œ  $

#    and

and .

I M 3) Given the matrix  œ determine the two values of the parameter for5

" ! "

" 5 "

" ! "

 

which the matrix admits multiple eigenvalues. For these values of , check if the correspon-5 ding matrix is diagonalizable or not.

From   we get  

 

- ˆ œ ! œ 5 "

" ! "

" 5 "

" ! "



   " œ

-

-  -^#  œ5- - ^#  #- œ5- - -   # œ ! Ê-" œ 5ß-# œ !ß-$ œ #. To get multiple eigenvalues there are two possibilities.

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1) if 5 œ !Ê œ œ !ß œ # and so we get:  ! † œ !

" ! "

" "

" ! "

-_" -_# -_$  

 

 ˆ ;

from Rank ! †ˆœ " Ê 7 œ $  " œ # œ 7¹_! ⁺_! and the matrix, for 5 œ ! is a diagonalizable one.

2) if 5 œ #Ê œ !ß œ œ # and so we get:  # † œ !

 " ! "

" "

" !  "

-_" -_# -_$  

 

 ˆ ; since:

  " "  

" " œ  # Á !, from Rank  # †ˆ œ # Ê 7 œ $  # œ "  7 œ #¹_# ⁺_# and the matrix, for 5 œ # is not a diagonalizable one.

I M 4) Given the basis for ‘^$À –œ"ß "ß ! à "ß !ß " à !ß "ß "    , find the coordinates of the vector ˜ œ !ß "ß  "  in such basis.

To solve the problem we must simply solve the system:

˜œ– —† œ † œ Ê Ê

" " ! B ! B  B œ !

" " B " B  B œ "

! " " B  " B  B œ  "

     

   





!

" " #

# " $

$ # $

Ê Ê Ê

B œ  B B œ  B B œ "

B œ "  B B œ "  B B œ  "

B  B œ  "  B  "  B œ  " B œ !

  

  

  

# " # " "

$ " $ " #

# $ " " $

.

II M 1) Given the system    satisfied at P ,

   

0 Bß Cß D œ B /  C /  # / œ !

1 Bß Cß D œ B C  BD  #BCD œ ! œ "ß "ß "

C B D

# # !

verify that an implicit function B Ä Cß D  can be defined with it, and then calculate the first order derivatives of this function.

The functions 0 Bß Cß D  and 1 Bß Cß D  are differentiable functions a Bß Cß D −  ‘^$. To apply Dini's Theorem we calculate the Jacobian matrix:

` 0 ß 1

` Bß Cß D œ /  C / B/  /  # /

#BC  D  #CD B  #BD #BD  #BC

 

   ^C ^# ^B ^#^C ^B ^D 

and so ` 0 ß 1 . Since it is

` Bß Cß D "ß "ß " œ #/ #/  #/ #/  #/ œ  #/ Á !

"  " !  " !

 

      

possible to define an implicit function B Ä Cß D . For its derivatives we get:

d d

dC #/ and dD %/ .

B œ  œ   #/ œ " B œ  œ   #/ œ #

#/  #/ #/ #/

" !  " "

#/  #/ #/  #/

 " !  " !

   

II M 2) Solve the problem .

Max/min u.c.:









 





0 Bß C œ B  C  B C  $B B !

C ! C Ÿ %  B

2 2

The objective function of the problem is a continuous function, the feasible region is aX compact set and therefore there are certainly maximum and minimum values.

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Given the number of constraints, it is not convenient to use the Kuhn-Tucker conditions which would require the resolution of 8 systems.

We firstly study the possible points of free relative maximum and minimum.

Applying the first order conditions we get:

f Ê 0 œ #B  C  $ œ ! Ê Ê −

0 œ #C  B œ !

$C  $ œ ! B œ # B œ #C C œ "

0 Bß C œ    ^B_C^w^w   and  #ß " X.

For the second order conditions we get, using the Hessian matrix:

‡ ‡ ‡

     ‡

Bß C œ #  " œ  œ #  !

 " #  #ß " . Since ^" œ %  " œ $  ! , the point  #ß "

# is a

minimum point, with 0 #ß " œ  $  .

Now we study the objective function 0 Bß C œ B  C  B C  $B  ² ² at the points of the first constraint B œ !. It is 0 !ß C œ C  ² and for C  ! it is an ever increasing function.

Let's study the objective function 0 Bß C œ B  C  B C  $B  ² ² at the points of the second constraint C œ !. It is 0 Bß ! œ B  $B Ê 0 B œ #B  $ !  ² ^w  and therefore the function is increasing for B $.

#

Finally, let's study the objective function 0 Bß C œ B  C  B C  $B  ² ² at the points of the third constraint C œ %  B. It isÀ

0 Bß %  B œ B  "'  B  )B  %B  B  $B œ $B  "&B  "' Ê  ² ^# ^# ^#

Ê 0 B œ 'B  "& ! B &

#

w  and therefore the function is increasing for .

We therefore have the following situation:

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from which we see that:

-  !ß % is a maximum point, with 0 !ß % œ "'  , and it is the absolute maximumà -  %ß ! is a maximum point, with 0 %ß ! œ %  , and it is a relative maximumà -  #ß " is a minimum point, with 0 #ß " œ  $  , and it is the absolute minimum.

The point ^$_#ß ! is a minimum point, with 0$ß !œ  * , relatively to the points of the

# %

constraint C œ !. For a complete analysis we form the Lagrangian function only for this constraint. We get: ABß Cß-œB  C  B C  $B² ² - C. SoÀ



 





 A

A -

-

wB wC

œ #B  C  $ œ #C  B  C œ !

œ ! œ ! Ê

B œ C œ !

œ  !

$

#

$

#

and therefore the point ^$_#ß ! may be, since

- œ $  !

# , with respect to the interior points of the feasible region , a maximum point and X therefore it is neither a maximum point nor a minimum point.

The point ^{& $}_{# #}ß  is a minimum point, with 0& $ß œ  "" relatively to the points of

# # %

the constraint C œ %  B. For a complete analysis we form the Lagrangian function only for this constraint. We get: ABß Cß-œB  C  B C  $B² ² -B C  %. So À

 

  

 









A -

- -

wB wC

œ #B  C  $  œ #C  B  C œ %  B

œ !

œ ! Ê Ê

'B œ "&

C œ %  B œ )  $B

B œ C œ

œ  !

&

$#

#"

#

and therefore the point

^{& $}_{# #}ß  may be, since - œ "  !, with respect to the interior points of the feasible region ,

# X

a maximum point and therefore it is neither a maximum point nor a minimum point.

II M 3) Given 0 Bß C œ B C  and the vectors •œ "ß "  and –œ  "ß " , let and be@ A their unit vectors. If W_@0 B ß C _! _!œ# and W_A0 B ß C _! _!œ !, determine the coordinates of the point B ß C_! _!.

The function 0 Bß C œ B C  it is a polynomial and therefore it is differentiable of any order.

So H 0_@ B ß C_! _! œ f0B ß C_! _!† @ and H 0_A B ß C_! _!œ f0B ß C_! _!† A. Since f0 B ß C œ Cà B Ê f0 B ß C! !œ C B!ß !Þ

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Now we get @ œ " "  and A œ  " " . Then:

#ß #  #ß #









      

     

W_@ _! _! _! _! _! _!

! ! ! ! ! !

0 B ß C œ B ß C C B ß œ #

B ß C B ß C C B  ß œ !

Ê

f0 † @ œ ß †

H 0 œ f0 † A œ ß †

 

" "

# #

" "

# #

 

  A

Ê C B "ß " œ # Ê B  C œ # Ê B œ "

C B  "ß " œ ! B  C œ ! C œ "

    

 

 ^!! ^!! ^!! ^!! !^!

ß †

ß † .

II M 4) Given the function 0 Bß C œ B  BC  C  ^# ^$, determine the nature of its stationary points.

Applying the first order conditions we get:

f Ê 0 œ #B  C œ ! Ê Ê

0 œ $C  B œ !

C œ #B

"#B  B œ B "#B  " œ !

0 Bß C œ      

Bw

w #

C #

Ê B œ ! ∪ C œ !

B œ

 C œ

"

"#

"

'

.

For the second order conditions we use the Hessian matrix: ‡Bß C œ  #  "

 " 'C . Since ‡ !ß ! œ #  "Ê ‡ œ  "  ! the point  !ß !

 " ! ^# is a saddle point;

since ‡ ‡ ‡ the point

     ‡  

   

" "

"# 'ß œ #  " Ê œ #  !ß œ "  !

 " " œ #  " œ "  !

" "

#

" "

"# 'ß is a minimum point.