QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS
TASK 6/10/2020
I M 1) Calcolate " $ 3$Þ
Since " $ 3 œ # " $3 œ # cos 3sin we get:
# # $ $
1 1
" $ 3 $ 3 $ œ ) 3 œ )
$ $
$ œ )cos 1 sin 1 cos sin
1 1 .
And finally:
" $ 3 œ ) œ ) 5 3 5 ß ! Ÿ 5 Ÿ " Þ
# # # #
# #
$ cos 1 1 sin 1 1
For 5 œ ! we get D œ! ) 3 œ ## 3à
# #
cos1 sin1
for 5 œ " we get D œ" ) $ 3 $ œ ## 3Þ
# #
cos 1 sin 1
I M 2) Given the linear map 0 À Ä having matrix œ , determi-
" " # "
! # " #
# ! 5 7
‘% ‘$
ne the value of the parameters and if the Kernel and the Image have the same dimension.5 7 For such values find a basis for the Kernel of the linear map generated by Þ
To get Dim Imm œDim Ker we need Rank œ # Þ
By elementary operations on the rows V Ã V #V$ $ " we getÀ
" " # " " " # "
! # " # ! # " #
# ! 5 7 ! # 5 % 7 #
Ä à
By elementary operations on the rows V Ã V V$ $ # we getÀ
" " # " " " # "
! # " # ! # " #
! # 5 % 7 # ! ! 5 & 7 %
Ä Þ
So Dim Imm œRank œDim Ker œ # for 5 œ & and 7 œ % Þ
And so œ .
" " # "
! # " #
# ! & %
To find a basis for the Kernel of the linear map generated by we must firstly solve the sys-
tem: —† œÊ † œ .
" " # " !
! # " # !
# ! & % !
B B B B
"
#
$
%
So we get the system: which however, for the elementary opera-
B B #B B œ !
#B B #B œ !
#B &B %B œ !
" # $ %
# $ %
" $ %
tions on the rows that we have already done, is equivalent to the system:
B B #B B œ ! B œ B B #B
#B B #B œ ! Ê #B %B $B œ ! Ê B œ B B Þ
B œ B B
" # $ % % " # $
# $ % " # $
$ " #
% " #
# %
$ $
" &
$ $
Each vector of the Kernel can be represented as — œ B ß B ß #B B ß% "B &B .
$ $ $ $
" # " # " #
If we choose B œ $" and B œ !# we get the vector —" œ $ß !ß #ß " . If we choose B œ !" and B œ $# we get the vector —# œ !ß $ß %ß & . So a basis for the Kernel is: — œ$ß !ß #ß " ß !ß $ß %ß & .
I M 3) Given the matrix œ , find the value of the parameter knowing5
# # "
$ ! #
5 # "
that the matrix admits the eigenvalue - œ " and then determine, for this value of the para- meter, if the matrix is diagonalizable or not.
From - ˆœ ! we get:
# # " # ! #
$ # $ #
5 # " 5 # "
5
œ
- - -
- -
- -
œ
œ#- 5-# - % - #' 5-œ !Þ
If - œ " we get $ 5 ' $' 5 œ ") '5 ") $5 œ ! Ê 5 œ ! Þ
So œ . And also:
# # "
$ ! #
! # "
- ˆ œ #- - #- % - # ' œ#- - #- # œ ! Ê Ê#- - #- " œ ! Ê-" œ-# œ #ß-$ œ "Þ
To check if the matrix is diagonalizable or not we have to find the Rank of #ˆÀ
#ˆ œ œ ' Á ! #ˆ œ #
! # "
$ # #
! # "
! #
$ #
. Since it is Rank for
which it is 7 œ $ # œ " 7 œ #1# +# and so the matrix is not a diagonalizable one.
I M 4) Find the coordinates of the vector in the basis — • œ"ß !ß ! ß "ß "ß ! ß "ß !ß " if it has coordinates #ß !ß " in the basis –œ"ß "ß ! ß "ß "ß " ß "ß !ß " .
Since has coordinates — #ß !ß " in the basis –œ"ß "ß ! ß "ß "ß " ß "ß !ß " , it is:
— œ † œ
" " " # "
" " ! ! #
! " " " "
.
To find the coordinates of the vector in the basis — • œ"ß !ß ! ß "ß "ß ! ß "ß !ß " we have to solve the system:
" " " B " B C D œ " B œ !
! " ! C # C œ # C œ #
! ! " D " D œ " D œ "
† œ Ê Ê Þ
II M 1) Solve the problem : Max/min . u.c. :
0 Bß C œ B C B Ÿ C Ÿ "
# #
#
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 !
C " Ÿ !
# #
#
ABß Cß- -"ß #œB C # # -"B C # -#C ". By applying the first order conditions we have:
1) case -" œ !ß-# œ ! À
A A
wB wC
œ #B œ ! œ #C œ !
Ê
B œ ! C œ ! B C Ÿ !
Ÿ !
! Ÿ ! ! #
Ÿ !
# !
#
C " "
Bß C œ œ !ß !
Þ Since ‡ ‡ we have that, globally, the point !ß ! is a saddle point. We will study it more precisely later.
2) case -" Á !ß-# œ ! À
A - -
A -
-
wB " "
wC "
"
œ #B # B œ #B " œ ! œ #C œ !
Ê
B œ ! C œ ! œ !
C œ B
C Ÿ " ! Ÿ "
# À we will study it more precisely later;
A - -
A -
- -
wB " "
wC "
#
" "
œ #B # B œ #B " œ ! œ #C œ !
Ê Ê
B œ C œ
œ
B œ B œ
C œ C œ
œ
C œ B
C Ÿ " Ÿ " Ÿ "
#
"
"#
#
"
#
" "
# #
" "
# #
"
#
1 1
∪
-" œ 1
"
# Ÿ "
Þ
Since -" ! the points " ß " and " ß" may be maximum points.
# # # #
3) case -" œ !ß-# Á! À
A
A -
-
wB
wC #
#
œ #B œ !
œ #C œ !
C œ " Ê
B œ ! C œ "
B Ÿ C
œ # !
! Ÿ "
#
Þ
Since -# ! the point !ß " may be a minimum point.
4) case -" Á!ß-# Á! À
A -
A - -
- -
- - - -
wB "
wC " #
" "
" # " #
œ #B # B œ !
œ #C œ !
C œ "
Ê ∪ Ê
B œ " B œ "
C œ " C œ "
# # œ ! # # œ !
œ # œ #
C œ B#
Ê ∪
B œ " B œ "
C œ " C œ "
œ " œ "
œ " œ "
- -
- -
" "
# #
. The points "ß " and "ß " are neither maximum nor
minimum points.
Let's study the objective function on the constraint C œ B#.
It is 0 Bß B # œ B B Ê 0 B œ #B %B œ #B " #B# % w $ # ! . It is " #B ! B Ÿ "
#
# Ê # Ê " Ÿ B Ÿ "
# #
and so:
The points " " and " " are maximum points with " " .
#ß #ß #ß
# # #
0 „ œ "
% Let's study the objective function on the constraint C œ ".
It is 0 Bß " œ B " Ê 0 B œ #B ! # w for B !.
The point !ß " is a minimum point with 0 !ß " œ " .
Finally we study the point !ß ! . It is 0 !ß ! œ ! . Then 0 Bß C œ B C # # ! for C Ÿ B# # satisfied for B Ÿ C Ÿ B . In each neighborhood of the point !ß ! we have both positive and negative values and therefore !ß ! is a saddle point.
II M 2) Given the function 0 Bß Cß D œ B BC C D # # # determine the nature of its statio- nary point.
We determine the stationary points of the function. Applying the first order conditions we have:
f0 Bß Cß D œ Ê Ê Þ
0 œ #B C œ ! 0 œ B #C œ ! 0 œ #D œ !
B œ ! C œ ! D œ !
Bw Cw Dw
Then we apply the second order conditions: ‡Bß Cß D œ œ‡ .
# " !
" # !
! ! #
!ß !ß !
Since: the point
‡
‡
‡
"
#
$
œ # ! œ $ ! œ ' !
!ß !ß ! is a minimum point.
II M 3) Given the function 0 Bß C œ / B C# # and the unit vector @œcosαßsinα calculate its directional derivatives W@ @# 0 !ß ! .
The function 0 Bß C œ / B C# # it is clearly differentiable of every order a Bß C − ‘#. So W#@ß@0 !ß ! œ @ †‡0 !ß ! † @X .
From f0Bß C œ #B / B C# #à #C /B C# # we getÀ
‡Bß C œ ‡ œ
# " #B / %BC / # !
%BC / # " #C / Ê !ß !
! #
#
#
B C B C
B C B C
# # # #
# # # # and so:
W α α α α α
α
# # #
@ß@0 !ß ! # ! # # œ #
! #
œcos sin † † cos œ Þ
sin cos sin
II M 4) Given the equation 0 Bß C œ B sinC CcosB œ !, verify if at P! œ !ß ! it is pos- sible to define an implicit function having as dependent variable of and then calculate theC B partial derivative of the first order of C B at B œ !.
The function0 Bß C is a differentiable function a Bß C − ‘#, 0 !ß ! œ ! ! œ ! . It is then: f0 Bß C œ sinC CsinBà BcosC cosB Þ
So f0 !ß ! œ ! à ". Since 0 !ß ! œCw 1 Á ! we can define an implicit function B Ä C B whose first order derivative is:
d d
C 0 !ß !
B ! œ 0 !ß ! œ œ ! Þ
Bw Cw
!
"
