MATHEMATICS for ECONOMIC APPLICATIONS TASK 18/6/2020

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MATHEMATICS for ECONOMIC APPLICATIONS TASK 18/6/2020

I M 1) Calculate the cubic roots of the number D œ # 3 .

"  3



Since #3 œ # cos  3sin and "  3œ# cos  3sin we getÀ

# # % %

 1 1  1 1

D œ # 3 œ œ

"  3

#

 



 

cos sin     

cos sin cos sin

1 1

# #

% %

 3

 3 1#  1%  3 1#  %1 œ

D œ # #

cos1 sin1

%  3 % œ #  # 3

. For its cubic roots we get:

^$ D œ cos 1 1 sin 1 1

"#  5 #$  3 "#  5 #$ ß ! Ÿ 5 Ÿ # . For 5 œ ! we get D œ_! cos 1 sin 1 à

"#  3 "#

for 5 œ " we get D œ # # à

" cos * sin* cos$ sin$

"#1  3 "#1 œ %1  3 %1 œ  #  # 3 for 5 œ # we get D œ_# cos "( sin "( Þ

"#1  3 "#1

I M 2) Given the matrix  œ , determine, on varying the parameter , its eigen-5

" !  #

$ 5 "

# !  $

 

values and their multiplicity, then establishing if there are values of for which the given matrix5 is diagonalizable.

From - ˆœ ! we get:

 

" !  #     

$ 5 "

# !  $

 $



 "    % œ   #  " œ

-

- - - -

œ 5     5  ^# 

œ5- -  "^# œ ! if -" œ 5 and -# œ-$ œ  "Þ

If 5 œ  " we get the eigenvalue - œ  " whose algebraic multiplicity is 7⁺_" œ $; if 5 Á  " we get the eigenvalue - œ  " whose algebraic multiplicity is 7⁺_" œ #. For - œ  "ß a 5, the matrix  becomes   " œ .

# !  #

$ 5  " "

# !  #

     

 

 - ˆ  ˆ

But #  # and so Rank    .

$ " œ ) Á !   " ˆ œ # Ê 7¹_" œ $  # œ "  7⁺_"

So the matrix is not diagonalizable a 5.

I M 3) Determine, on varying the parameters and , the dimensions of the Image and the7 5 Kernel of the linear map ‘^& Ä‘ , 0 — œ —† , with œ .

" ! " ! "

" " " 7 5

! " ! " 7

3  

 

By Sylvester Theorem we need to calculate the Rank of the matrix .

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

   

" ! " ! " " ! " ! "

" " " 7 5 ! " ! 7 5  "

! " ! " 7 ! " ! " 7 Ä

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

   

" ! " ! " " ! " ! "

! " ! 7 5  " ! " ! 7 5  "

! " ! " 7 ! ! ! "  7 7  5  "

Ä

If 7 Á "ß a 5 ÀRank  œ $ ÊDim Imm  0 œ $ and Dim Ker  0 œ &  $ œ # à if 7 œ "ß 5 Á # ÀRank  œ $ ÊDim Imm  0 œ $ and Dim Ker  0 œ &  $ œ # à if 7 œ "ß 5 œ # ÀRank  œ # ÊDim Imm  0 œ # and Dim Ker  0 œ &  # œ $ Þ

I M 4) Given the vectors —_" œ "ß #ß " , —_# œ "ß  "ß #  and —_$ œ #ß #ß 5 , determine the value of the parameter for which the three vectors are linearly dependent.5

If the three vectors are linearly dependent they must form a singular matrix, and therefore the de- terminant of the matrix having the three vectors as its lines must be equal to .!

So we get    

   

 œ œ œ "  $5  '  # œ  $5  ) Þ

" " # " " #

#  " # !  $  #

" # 5 ! " 5  #

having used elementary operations on the rows V Ã V  #V_# _# _" and V Ã V  V Þ_$ _$ _" If  $5  ) œ ! Ê 5 œ ) the three vectors are linearly dependent;

$

if  $5  ) Á ! Ê 5 Á ) the three vectors are linearly independent.

$

II M 1) Solve the problem : Max/min .

u.c. :

  

 

0 Bß C œ B  C B  "  " Ÿ C Ÿ "

#

We write the problem in the form . The objective function of Max/min

s.v.:





 

  

0 Bß C œ B  C B  "  "  C Ÿ ! C  " Ÿ !

#

the problem is a continuous function, the constraints define a feasible region which is a compact set as it is limited and closed and therefore the maximum and minimum values certainly exist.

For a problem with inequality constraints we apply the Kuhn-Tucker conditions, we find the solutions and then we study the objective function on the boundary of .X

To apply the Kuhn-Tucker conditions we construct the Lagrangian function À ABß Cß- -_"ß _#œB  C ^# -_"B  "^# "  C  -_#C  ".

Applying the first order conditions we get:

1) case -_" œ !ß-_# œ ! À







 A A

wB wC

œ #B œ ! œ " Á !

B  "  "  C Ÿ ! Ÿ !

#

C  "

so there are no solutions.

2) case -_" Á !ß-_# œ ! À

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 

  

  

  

 









A -

A - - -

-

wB "

wC " " "

"

œ #B  # B  " œ ! œ "  œ !

Ê Ê  !

B  # œ ! œ 

B œ œ 

 

C œ B  "  "

C Ÿ "

"

C œ B  "  "

C Ÿ "

"  ! C œ 

 Ÿ "

 ^#  ^#

% ^"_#

$

$ %

%

: from it

follows that the point " 

#ß  $

% it is a possible Minimum point.

3) case -_" œ !ß-_# Á! À

 

 

 

 

 

  A

A -

- -

wB

wC #

# #

œ #B œ ! œ "  œ !

C œ " Ê  !

B œ ! C œ "

B  "  " Ÿ C

œ "  !

! Ÿ "

!ß "

#

; from it follows that the point is a possi-

ble Maximum point.

4) case -_" Á!ß-_# Á! À

  

  

  

  

  

A -

A - -

- -

- - - -

wB "

wC " #

" "

" # " #

œ #B  # B  " œ ! œ "   œ ! C œ "

Ê ∪

B œ "  # B œ "  #

C œ " C œ "

#B  # B  " œ ! #B  # B  " œ !

"   œ ! "   œ !

   

   

C œ B  " ^# " à

 

  

  

  

 





 B œ "  #

C œ "

#B  # B  " œ !

"   œ !

Ê Ê

B œ "  # C œ "

#  # #  # # œ ! œ " 

B œ "  # C œ "

œ  !



 



 



-

- -

-

- -

"

" #

"

# "

"

#

" #

#

"# #

#



À

from -"  !ß-#  ! it follows that the point "  ß "#  is a possible Maximum pointà

 

  

  

  

 





 B œ "  #

C œ "

#B  # B  " œ !

"   œ !

Ê Ê

B œ "  # C œ "

#  # #  # # œ ! œ " 

B œ "  # C œ "

œ  !



 



 



-

- -

-

- -

"

" #

"

# "

"

#



#"

#

# #"

#

à

from -"  !ß-#  ! it follows that the point "  ß "#  is a possible Maximum point.

Now we study the objective function on the constraint C œ ". It is 0 Bß " œ B  " Ê 0 B œ #B !  ^# ^w  for B !.

Then the function is decreasing for " # Ÿ B Ÿ ! and increasing for ! Ÿ B Ÿ " #. So it has at B œ ! a Minimum point.

But the point  !ß " had been reported as a possible maximum point and therefore it is neither a maximum point nor a minimum point

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We study the objective function on the constraint C œB  "^#  ". It is 0 Bß B  "  " œ #B  #B Ê 0 B œ %B  # ! B ".

  ^#  ^# ^w  for #

Then the function is decreasing for " # Ÿ B Ÿ " and increasing for " Ÿ B Ÿ " #.

# #

So it has at B œ " .

# a Minimum point

The point T œ "ß  $

# %

"   had been reported as a possible minimum point and therefore it is

the Minimum point with 0 T œ "

 " #à

The point T œ " _#  #ß ", with 0 T _# œ %  ## is a Maximum pointà the point T œ " _$  #ß ", with 0 T _$ œ %  ## is a Maximum pointà T# it is a relative maximum point, is the absolute maximum point.T$

II M 2) Given the function 0 Bß C œ /  ^BC, find the directions @œcosαßsinα for which it results W_@0 !ß ! œ ! Þ 

The function 0 Bß C œ /  ^BC it is clearly differentiable a Bß C −  ‘^#. So H 0@ 5ß 5œ f05ß 5† @. Since:

f0 B ß C œ /^BCà /^BCÊf05ß 5 œ "ß  ", to get W@0 5ß 5 œ !  it must be À

"ß  " ß œ  œ ! Ê œ Ê œ à œ

% %

&

cosα sinα cosα sinα cosα sinα α 1 α 1.

II M 3) Given the function 0 Bß Cß D œ B  C  D  BC  ^# 2 ^# ^# 2 , check the nature of its stationary points.

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By applying first order conditions we get:

f Bß Cß D Ê Ê

0 œ #B  #C œ ! 0 œ %C  #B œ ! 0 œ #D œ !

B œ ! C œ ! D œ !

0 œ Þ

 

 

 



Bw Cw Dw

Then:

‡  ‡ 

 

Bß Cß D œ !ß !ß ! œ

#  # !

 # % !

! ! #

.

Since the point it is a minimum point.





   

 ‡  ‡  

‡ ‡

‡

" "

# #

$

œ #  !à œ %  !

œ )  %  !à œ )  !  ! œ # )  %  !

!ß !ß !

II M 4) Given the equation 0 Bß C œ B /  ^BC C /^CB œ ! satisfied at the point  !ß ! , verify that an implicit function B Ä C B  can be defined with it, and then calculate the first order derivative of this function.

The function 0 Bß C  it is a differentiable function a Bß C −  ‘^#. Then it is:

f0 B ß C œ /^BC B /^BC C /^CBà B /^BC /^CB C /^CBœ

f0 B ß C œ "  B/^BCC /^CBà B /^BC "  C /  ^CBÊf0 !ß ! œ  "à  "Þ Since 0_C^w !ß ! œ  " Á ! it is possible to define an implicit function B Ä C B .

For its derivative we get: C ! œ  " œ " Þ

 "

w 