QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS
TASK 13/1/2021
I M 1) If / Ε / $ ο 3 , find .D
#
D ο
οο ο
$
Applying the definition of complex exponential: /Bο3C Ε /Bο‘cosC ο 3senCο’ and sinceΓ
/ Ε / $ ο 3 Ε / ο 3 Ε / ο 3 Γ
# # # ' '
$ " "" ""
D ο
οο ο οο ο
$ " "
$ $
οcos 1 sen 1ο we get / Ε / D Ε " ο 3""
$ '
D "$ο3""'1 and so 1 .
I M 2) Given the vectors β" Ε "Γ !Γ ο #Γ #ο‘ ο’, β# Ε #Γ "Γ ο "Γ #ο‘ ο’ and β$ Ε "Γ #Γ %Γ 5ο‘ ο’, check, depending on the parameter , when they are linearly independent.5
We solve the problem by calculating the rank of the matrix Ε Ε .
" ! ο # #
# " ο " #
" # % 5
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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 οΊ" !οΊ , the rank is at least equal to .
# " Ε " Γ ! #
By elementary operations on the rows ο‘V Γ V ο #V# # "ο’ ο‘, V Γ V ο V$ $ "ο’, we getΓ
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" ! ο # # " ! ο # #
# " ο " # ! " $ ο #
" # % 5 ! # ' 5 ο #
Γ Γ
By elementary operations on the rows ο‘V Γ V ο #V$ $ #ο’, we getΓ
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" ! ο # # " ! ο # #
! " $ ο # ! " $ ο #
! # ' 5 ο # ! ! ! 5 ο #
Γ Γ
Therefore, if 5 Γ ο # the rank is equal to and the vectors are linearly independent, if inste-$ ad 5 Ε ο # 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 #
! ! "
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gonalizable.
From ο«οο- Λο«Ε ! we get:
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ο’" # $ ο’ ο‘ ο’ ο‘ ο’
! 5 #
! ! "
" 5 "
ο
ο
ο
ο ο ο Ε ! Ε Ε "Γ Ε 5
-
-
-
- - - -
Ε ο‘ ο’ Γ So " # $ Γ
If 5 Ε " we get 7 Ε $ and ο " β Ε and since # $ Ε % Γ !, it is
! #
+" ο¬ ο¬
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ο£ ο£ οΊ οΊ
ο Λ
! # $
! ! #
! ! !
Rankο‘οο " β Λο’Ε # Γ so 7 Ε $ ο # Ε " ο 7 Ε $1" +" and the matrix is not diagonalizable.
If 5 Γ " it is 7 Ε # and ο " β Ε Γ if # $ Ε ( ο $5 Ε !
#
+" ο¬ ο¬
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ο£ ο£
ο£ ο£
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ο£ ο£
ο£ ο£ οΊ οΊ
ο Λ
! # $
! 5 #
! ο "! ! 5
ο "
then Rankο‘οο " β Λο’Ε " Γ so 7 Ε $ ο " Ε # Ε 71" +" and the matrix is diagonalizable.
If 5 Γ ( then Γ !, Rank ο " β Ε # Γ so 7 Ε $ ο # Ε " ο 7 and the
$
# $
οΊ5 ο " #οΊ ο‘ο Λο’ 1" +"
matrix is not diagonalizable. Therefore the only value of the parameter for which the matrix5 is diagonalizable is 5 Ε (Γ
$
I M 4) Consider the linear map 0 Γ Γ , Ε β whith Ε .
" ο " 7 "
! # " 5
" " #5 7
β% β$ Λ ο β ο
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Since 0 "Γ "Γ "Γ " Ε %Γ &Γ *ο‘ ο’ ο‘ ο’, 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 "Γ "Γ "Γ " Ε %Γ &Γ *ο‘ ο’ ο‘ ο’ we get:
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" ο " 7 " % " ο " ο 7 ο " Ε % 7 Ε $
! # " 5 & ! ο # ο " ο 5 Ε & 5 Ε #
" " #5 7 * " ο " ο #5 ο 7 Ε * # ο % ο $ Ε *
β Ε Γ Γ Γ
"
"
"
"
So ο Ε . By elementary operations on the rows V Γ V ο V and
" ο " $ "
! # " #
" " % $
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ο’ ο’ ο‘ $ $ "ο’
then ο‘V Γ V ο V$ $ #ο’, we getΓ
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" ο " $ " " ο " $ " " ο " $ "
! # " # ! # " # ! # " #
" " % $ ! # " # ! ! ! !
Γ Γ Γ " ο "
! #
Since Γ !
we get Rankο‘ ο’ο Ε # and so Dim Immο‘ ο’Ε # and Dim Kerο‘ ο’Ε % ο # Ε # Γ
II M 1) Given 0 BΓ C Ε B ο $BC ο #Cο‘ ο’ # #, and the unit vectors of @ A ο‘ ο’"Γ " and ο‘"Γ ο "ο’, since W@0ο‘ ο’P! Εο# and WA0ο‘ ο’P! Ε #ο#, determine P and then calcolate ! H@#ΓA0ο‘ ο’P! . The function 0 BΓ C Ε B ο $BC ο #Cο‘ ο’ # # ist twice differentiable a BΓ C βο‘ ο’ β#.
So W@0ο‘ ο’P! Ε f0ο‘ ο’P! β @ and W#@ΓA0ο‘ ο’P! Ε @ β β‘0ο‘ ο’P! β AX . From f0ο‘BΓ C Ε #B ο $CΓ %C ο $Bο’ ο‘ ο’ we getΓ
οο
ο
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οο
ο‘ ο’ ο‘ ο’ ο‘ ο’ ο ο ο
ο‘ ο’ ο‘ ο’ ο‘ ο’ ο ο ο
W W
@ ! !
A ! !
0 #B ο $CΓ %C ο $B Ε #
0 #B ο $CΓ %C ο $B Ε # #
P P
P P
Ε f0 β @ Ε β
Ε f0 β A Ε β
" "
# #
" "
# #
ο ο
ο ο
Γ Γ ο
Γ
ο#B ο $C ο %C ο $B # οC ο B # οC # οB
#B ο $C ο %C ο $B % &B ο (C % &B ο (B ο "% % C (
Ε Ε Ε B ο Ε ο *
Ε Γ Ε Γ Ε Γ Ε ο .
Therefore P! Ε ο *Γ ο ( . It is then # ο $ ο *Γ ο (
ο $ %
ο‘ ο’ β‘ο‘BΓ C Εο’ οΎ οΎΕ β‘ο‘ ο’ and so:
W@ΓA# 0 ! # ο $ Ε
ο $ % ο ο
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P Εο½ "# "# ο½β β Εο½ "# "# ο½β
" &
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Ε & ο ( Ε ο "
# # Γ
II M 2) Solve the problem .
Max/min u.c.:
ο
οο
ο‘ ο’ ο‘ ο’
ο
0 BΓ C Ε B C ο "
B ΕΈ " ο C
" ΕΈ B ο 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.:
ο
οο
ο‘ ο’ ο‘ ο’
ο
0 BΓ C Ε B C ο "
B ο C ο " ΕΈ !
" ο B ο C ΕΈ !
#
We form the Lagrangian function:
Aο‘BΓ CΓ- -"Γ #ο’ΕB C ο " οο‘ ο’ -"οB ο C ο " ο# ο -#ο‘" ο B ο Cο’. By applying the first order conditions we have:
1) case -" Ε !Γ-# Ε ! Γ
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A A
wB wC
Ε C ο " Ε ! Ε B Ε !
Γ Γ
B Ε ! C Ε ο "
B ο C ο " ΕΈ !
" ο B ο C ΕΈ !
! ο " ο " ΕΈ ! ΕΈ ! Γ
#
"
!Γ ο " Γ Γ
ο ! ο " NO
X
2) case -" Γ !Γ-# Ε ! Γ
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A -
A -
-
- - - -
- - -
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wC "
"
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"
#" " "
Ε C ο " ο Ε ! Ε B ο # C Ε !
Γ Γ
C Ε ο "
B Ε # ο " Ε # ο #
# ο # Ε ο "
B Ε " ο C
" ο B ο C ΕΈ !
" ο
" ο B ο C ΕΈ !
# ο‘ ο’ #
ο‘ ο’
Γ Γ
C Ε ο "
B Ε # ο " Ε # ο #
$ ο % Ε ο %
B Ε ! C Ε ο "
B Ε C Ε
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* $
Since -" Ε % ο ! the point ) "Γ could be a Maximum point.
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3) case -" Ε !Γ-# Γ! Γ
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ο ο ο
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ο ο ο
A -
A -
- -
- - -
wB #
wC #
#
#
# # #
Ε C ο " ο Ε !
Ε B ο Ε !
C Ε " ο B Γ Γ
B Ε ο C Ε ο " ο
ο
B Ε "
C Ε ! B ο C ο " ΕΈ !
Ε "
B ο C ο " ΕΈ !
Ε ο " ο !
" ο ! ο " ΕΈ !
# #
ο " ο Γ
Since -# Ε ο " ο ! the point ο‘ ο’"Γ ! could be a minimum point.
4) caso -" Γ!Γ-# Γ! Γ
ο ο ο
ο ο ο
ο ο ο
ο ο ο
ο ο ο
ο ο ο
ο ο ο
A - -
A - -
- - - -
- - -
wB " #
wC " #
" # " #
# " #
Ε C ο " ο ο Ε ! Ε B ο # C ο Ε !
" ο B B Ε " ο
Γ βͺ Γ
B Ε " B Ε !
C Ε ! C Ε "
" ο ο Ε ! # ο ο Ε !
" ο Ε ! ο Ε !
C Ε
C# ο #
Γ βͺ
B Ε " B Ε !
C Ε ! C Ε "
Ε ! Ε ο # ο !
Ε ο " ο ! Ε ο % ο !
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
ο- ο-
- -
" "
# #
.
Since the values are not positive, points - ο‘ ο’"Γ ! and ο‘ ο’!Γ " could be minimum points.
Let's study the objective function on the constraint B Ε " ο C#.
It is 0 " ο C Γο # Cο οΕ " ο C#οο‘C ο " Ε C ο " ο C ο C Γ 0 C Ε " ο $C ο #C Β !ο’ $ # wο‘ ο’ # . So $C ο #C ο " ΕΈ ! Γ C Ε ο "β " ο $ Ε ο "β# . Therefore:
$ $
# ο
0 C Β ! ο " ΕΈ C ΕΈ " Γ ! ΕΈ C ΕΈ " C Ε " Γ B Ε )
$ $ $ *
wο‘ ο’ for . If .
At the point ο) "ο there is the Maximum with ο) "ο $#.
* $Γ 0 * $Γ Ε #(
Let's study the objective function on the constraint C Ε " ο B.
It is 0 BΓο‘ " ο Bο’Ε Bο‘" ο Bο " Ε #B ο B Γ 0 B Ε # ο #B Β !ο’ # wο‘ ο’ for ! ΕΈ B ΕΈ ". If B Ε ! Γ C Ε ". The point ο‘ ο’!Γ " is the minimum point with 0 !Γ "ο‘ ο’ Ε0.
If B Ε " Γ C Ε !. The point ο‘ ο’"Γ ! it is neither a minimum nor a maximum point.
II M 3) Given the equation 0 BΓ CΓ D Ε #BC ο /ο‘ ο’ DοB ο /DοC Ε !, satisfied at ο‘"Γ "Γ "ο’, veri- fy that an implicit function ο‘BΓ C Γ D BΓ Cο’ ο‘ ο’ can be defined with it and then calcolate the first order derivatives of such function.
The function 0 BΓ CΓ Dο‘ ο’ is differentiable a BΓ CΓ D βο‘ ο’ β$, and 0 "Γ "Γ " Ε # ο " ο " Ε !ο‘ ο’ . It is: f0ο‘BΓ CΓ Dο’Ε #C οο‘ /DοBΓ #B ο /DοCΓ ο /DοBο /DοCο’Γ
So f0ο‘"Γ "Γ "ο’Ε $Γ $ο‘ Γ ο #ο’.
Since 0 "Γ "Γ " ΕDwο‘ ο’ ο #Γ ! it is possible to define an implicit function ο‘BΓ C Γ D BΓ Cο’ ο‘ ο’ whose first order derivatives will be equal to:
`D $ $ `D $ $
`Bο‘ ο’ ο‘ ο’ ο # # `Cο‘ ο’ ο # #
ο‘ ο’ ο‘ο‘ ο’ο’
"Γ " Ε ο 0 "Γ "Γ " Ε ο Ε Γ "Γ " Ε ο Ε ο Ε Γ 0 "Γ "Γ " 0 "Γ "Γ "
0 "Γ "Γ "
Bw
w w
D D
Cw
II M 4) Given the function 0 BΓ CΓ D Ε B ο C ο D ο B C ο CDο‘ ο’ # # # # , analyze the nature of its stationary points.
We determine the stationary points of the function. Applying the first order conditions we have:
f0 BΓ CΓ D Εο‘ ο’ ο Γ Γ
0 Ε #B ο #BC Ε #B " ο C Ε ! 0 Ε #C ο B ο D Ε !
0 Ε #D ο C Ε !
B Ε ! C Ε ! D Ε !
ο ο
ο ο
ο ο
ο‘ ο’
Bw
w #
C Dw
and also:
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ο ο ο
ο ο ο
ο
ο ο
ο ο
ο ο
ο ο
ο ο
B Ε # ο Ε
C Ε "
D Ε
Γ
B Ε B Ε ο
C Ε " C Ε "
D Ε D Ε
# " $
# #
"
#
$ $
# #
" "
# #
and Γ
We have then found three stationary points: ο‘!Γ !Γ !ο’ ο, ο#$Γ "Γ "#ο and οοο$#Γ "Γ"#οΓ
We apply the second order conditions. From β‘ο‘BΓ CΓ D Εο’ we get:
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# ο #C ο #B !
ο #B # ο "
! ο " #
β‘ο‘!Γ !Γ ! Εο’ Γ
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ο« ο«
ο« ο« οΊ οΊ ο« ο« οΊ οΊ
ο« ο« οΊ οΊ
# ! !
! # ο "
! ο " #
Γ
Ε # ο !
Ε # ! ο !Γ Ε # ο " ο !
! # ο " #
Ε # β # ο " ο !
ο " #
β‘
β‘ β‘
β‘
"
# #
$
since the point
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ο« ο«ο« ο«
ο« ο«β‘ ο‘ ο’
β‘
β‘
"
#
$
ο !
ο !
ο !
!Γ !Γ ! is a minimum pointΓ
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$ "
#Γ "Γ# Γ Ε ο !
! ο # !
ο # # ο "
! ο " #
! ο #
ο # #
Ε Γ
$
#
$
# #
$
#
$
#
β‘
since ο« ο«β‘# ο ! the point ο$Γ "Γ" Γ
# #
ο ο is a saddle point
β‘οοο οΕ ο« ο« Γ
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$ "
#Γ "Γ # Γ Ε ο !
! # !
# # ο "
! ο " #
! #
# #
$
#
$
# #
$
#
$
#
β‘
since ο« ο«β‘# ο ! the point ο$Γ "Γ"
# #
οο ο is a saddle point.
