TASK MATHEMATICS for ECONOMIC APPLICATIONS 11/6/2018 I M 1) D œ " 3 " 3 œ " 3 " 3 œ " " #3 " " #3 œ
" 3 " 3 " 3 " 3 " 3
# #
#
œ %3 œ #3 œ # Þ
# cos 1 sin 1
# 3 #
So D œ # cos1 sin1 Þ
1 1
% 5 3 % 5 ß !Ÿ5 Ÿ"
For 5 œ ! À #cos1 sin1 # œ " 3à
% 3 % œ # 3 #
# #
For 5 œ " À #cos& sin& # # #œ " 3Þ
3 œ 3
# #
1 1
% %
I M 2) From œ À
" " ! !
" " ! !
! ! " "
! ! " "
we get
-ˆ
- -
- - -
- -
- - -
œ
" " ! ! " ! !
" " ! ! " ! !
! ! " " ! ! "
! ! " " ! ! "
œ œ
G Ã G G" " # and G Ã G G$ $ % and then V Ã V V" " # and V Ã V V% % $
œ # œ
!
!
! !
! "
! !
# ! !
" ! !
! ! "
! ! #
œ
# -
- -
-
- -
-
-
-
œ- #-- #-œ ! for -" œ-# œ !à -$ œ #à -% œ # Þ
To find the eigenspace corresponding to the eigenvalue - œ ! we solve the system:
! †ˆ †—œ —† œ Ê † œ Þ
B !
C !
D !
A !
" " ! !
" " ! !
! ! " "
! ! " "
Since Rank ! †ˆœ Rank œ # the dimension of the eigenspace corresponding to the eigenvalue - œ ! is 7 œ % 1! Rank ! †ˆœ % Rank œ % # œ #. The system becomes B C œ ! C œ B and the eigenvectors corresponding to the ei-
D A œ ! Ê A œ D
genvalue - œ ! are •œ Bß Bß Dß D ß Bß D − ‘. A basis for the eigenspace may be:
•! œ"ß "ß !ß ! à !ß !ß "ß " .
To find the eigenspace corresponding to the eigenvalue - œ # we solve the system:
# †ˆ †—œ Ê † œ Þ
B !
C !
D !
A !
" " ! !
" " ! !
! ! $ "
! ! " $
Since - œ # is a simple eigenvalue, we get 7 œ 7 œ "1# +# and the system becomes:
B C œ ! C œ B
$D A œ ! D œ ! D $A œ ! A œ !
Ê . The eigenvectors corresponding to the eigenvalue - œ # are
•œ Bß Bß !ß ! ß B − ‘. A basis for the eigenspace may be the vector: •# œ"ß "ß !ß !. To find the eigenspace corresponding to the eigenvalue - œ # we solve the system:
# †ˆ †—œ Ê † œ Þ
B !
C !
D !
A !
$ " ! !
" $ ! !
! ! " "
! ! " "
Since - œ # is a simple eigenvalue, we get 71# œ 7+# œ " and the system becomes:
$B C œ ! B œ ! B $C œ ! C œ ! D A œ ! D œ A
Ê . The eigenvectors corresponding to the eigenvalue - œ # are
•œ !ß !ß Dß D ß D − ‘. A basis for the eigenspace may be the vector: •# œ!ß !ß "ß ". I M 3) Since the dimension of the Kernel of the map is equal to , we get that Rank" œ #, and so the determinant of the matrix must be equal to zero: œ !:
œ œ œ " † 5 " œ ! Ê 5 œ "
! !
! 5 ! 5
! "
1 1 1 1
1 1
1 1 2 1
. So:
-ˆ
-
-
-
œ Ê œ œ
! !
! " ! "
1 1 1 1
1 1
1 1 2 1 1 2
œ1-" -2- " " - " œ 1- - # $ # " " œ-
œ1- - # $- œ-1- - $ œ ! . So the matrix has three simple eigenvalues and it is a diagonalizable one.
I M 4) To solve the problem we can apply Rouchè-Capelli Theorem. So we study the aug-
mented matrix: . By elementary operations on the
˜l œ
" " " " l "
" ! " " l "
! " 7 # l 5 rows V Ã V V# # " and then V Ã V V$ $ # we get:
˜l
" " " " l " " " " " l "
! " ! # l # ! " ! # l #
! " 7 # l 5 ! ! 7 ! l 5 #
Ê Ê . So:
For 7 Á !ß a 5 we get: Rank œ Rank ˜l œ $: we have ∞" solutions;
For 7 œ ! and 5 œ # we get: Rank œRank ˜l œ #: we have ∞# solutions;
For 7 œ ! and 5 Á # we get: Rank œ # Rank ˜l œ $: we have no solutions; it is impossible to express the vector with a linear combination of the vectors ˜ — — —", #, $ and
—%.
II M 1) From the equation 0 ßB Cß Dœ B C D $BC $CD œ "$ $ $ we get:
0 "ß "ß " œ " " " $ $ œ " and so the point "ß "ß " satisfies the equation.
Then f0 Bß Cß D œ $B $Cß $C $B $Dß $D $C # # # and f0 "ß "ß " œ 'ß ß ' 3 . Since 0 œ ' Á !Dw there exists an implicit function Bß C Ä D Bß C with partial derivati- vesÀ `D "ß " œ ' œ " and `D "ß " œ $ œ ".
`B ' `C ' #
II M 2) To solve the problem Max/min we observe that u.c.:
0 Bß C œ B C B C Ÿ "
% %
# # objective fun-
ction of the problem is a continuous function, the feasible region is a circle, so a compactX set, and so maximum and minimum values surely exist. The constraints are qualified.
The Lagrangian function of the problem is: ABß Cß-œB C% %-B C # # 1. 1) case - œ ! À
A A
wB wC
œ %
œ % "#
"#
B C
B C Ÿ " Ÿ
Bß C B !
C
$
$
# #
#
#
œ !
œ ! Ê œ !ß ! œ
B œ ! C œ !
! "
! !
! ! !
. ‡ and ‡ ,
By the Hessian ‡ !ß ! we haven't any information about the point !ß ! , but trivialy we can see that !ß ! is a minimum point since 0 !ß ! œ ! and 0 Bß C !ß aBß C. 2) case - Á ! À
A - -
A - -
wB wC
œ % # B œ % # C
B C
B C œ " Á
$
$
# #
œ #B #B œ ! œ #C #C œ ! Ê
B œ ! C œ !
! "
#
# unacceptable solution;
∪ Ê à !
B œ ! œ #
B œ ! œ #
- - -
C œ "# C œ „"
since these points may be maximum points;
∪ Ê à !
œ # œ !
œ # œ !
- -
C C -
B œ "# B œ „"
since these points may be maximum points;
∪ Ê à !
œ
œ œ
B
C
œ "
B œ „ #
#
„ #
œ " #
#
#
- -
- -
-
-
#
#
# #
C
since these points may be maximum
points.
If we want to complete the study of our problem in the boundary points, we can use the parametric form for the circle:
B œ > C œ cos> Ê 0 BÞC
sin Ä 0 > œ cos%> sin%>. By deriving we get:
0 > œ %w cos$> † sin> % sin$> †cos> œ %sin> †cos>†cos#> sin#> . And so 0 > œ #w sin#> †cos#> œ sin%> .
Since sin> ! for ! Ÿ > Ÿ1 we get: sin%> ! for:
! Ÿ > Ÿ ∪ Ÿ > Ÿ ∪ Ÿ > Ÿ ∪ Ÿ > Ÿ
% # % % # %
$ & $ (
1 1 1 1 1 1
1 .
So 0 >w ! for:
1 1 1 1 1 1
1 1
% Ÿ > Ÿ # ∪ $% Ÿ > Ÿ ∪ &% Ÿ > Ÿ $# ∪ (% Ÿ > Ÿ # . So the four points „ # „ # are not maximum points.
# #
ß
II M 3) 0 Bß C œ log/B C# # /B C# # is a differentiable function aBß C − ‘#Þ
From f œ #B #B #C #C f œ
0 Bß C / / / / 0 !
/ / ß / / ß !
B C B C B C B C
B C B C B C B C
# # # # # # # #
# # # # # # # # : !ß !
and f œ # # # # œ # . And so:
0 "ß " / / ß / / ß #/ "
/ / / / / "
## ! ! ## !! ##
H 0 !ß ! œ@ f0 !ß ! † œ !ß ! †@ # #
# ß # œ ! ; WA
#
# #
0 "ß " "ß " A #ß #/ " #ß #
/ " # # / "
œf † œ † œ # #
0
.
II M 4) 0 Bß C œ B C C $B # # is a differentiable function aBß C − ‘ .#
0 Bß C œ B C # C $B # œ "! B #C )BC Ê f0 Bß C œ #!B )Cß %C )B# # .
M SG: #!B )C œ ! B œ !
)B %C œ ! Ê C œ ! . We have only one stationary point: !ß ! .
‡0 Bß C œ‡0 !ß ! #! )
) %
œ .
MM SG: .
‡ ‡
‡
" "
#
œ #! ! œ % !
Ê !ß !
or is a minimum point
œ "' !
