TASK MATHEMATICS for ECONOMIC APPLICATIONS 12/2/2018
I M 1) / œ œ # " 3 " œ # 3 Ê
# #
D " 3 cos1 sin 1
% %
Ê / œ / œ / / œ # 3 Ê / œ # Ê B œ #
C œ
D B3C B 3C B
cos 1 sin1 log .
% % 1%
So D œlog# 31 .
%
I M 2) To find the eigenvalues of the matrix , we need the determinant -ˆœ !:
$ # " ! %
" % * " % *
" # $ " # $
œ
- - -
- -
- -
V V"œ $
2
œ# -% - $ - ") % -# % -œ
œ# - - #- "# ") % -œ# - - # # # œ !- . There is a real eigenvalue: -" œ #. From -# # # œ !- we get:
- œ "„" # œ "„ " œ "„3. So there are two complex eigenvalues:
-# œ " 3 and -$ œ " 3. Three simple (distinct) eigenvalues, one real and two complex.
To find eigenvectors corresponding to -" œ # we solve the system:
#ˆ†—œ † œ Ê † œ Ê
" # " B ! " # " B !
" # * C ! ! ! "! C !
" # D ! ! ! D !
& '
Ê B #C D œ ! Ê B œ #C
"!D œ ! D œ !
so the eigenvectors are: • œ #5ß 5ß ! .
I M 3) From B 7B 7B #B œ ! B B B #7B œ !
" # $ %
" # $ % we get:
" 7 7 # " 7 7 #
" " " #7 V V#Ê " ! " 7 " 7 #7 # and so:
for 7 œ "ÊRank œ "Êthe system has ∞%" œ ∞$ solutions ; for 7 Á "ÊRank œ #Êthe system has ∞%# œ ∞2 solutions . For 7 œ " " " " #
! ! ! !
we get † œ Ê B C D #A œ ! Ê
B
C !
D !
A
Ê D œ B C #A . So a basis for the vector space of the solutions may be:
•œ •" œ "ß !ß "ß ! à •# œ !ß "ß "ß ! à •$ œ !ß !ß #ß " .
I M 4) We have — —"† # œ "ß #ß $ † "ß "ß " œ " # $ œ ! . To find a vector orthogonal to —" we put:
—" †—$ œ "ß #ß $ † Bß Cß D œ B #C $D œ ! Ê B œ $D #C .
To get the vector —$ to be orthogonal to —# we put:
—# †—$ œ "ß "ß " † Bß Cß D œ "ß "ß " † $D #Cß Cß D œ $D #C C D œ ! Ê Ê %D C œ ! Ê C œ %D and so B œ $D #C œ $D )D œ &D.
So —$ œ &Dß %Dß D . If we choose D œ " we get: —$ œ &ß %ß " .
And so we get the basis —œ—" œ "ß #ß $ à —# œ "ß "ß " à —$ œ &ß %ß " . To find the coordinates of the vector ˜ œ "ß !ß " in the basis we must solve the system:—
" " & B "
# " % C !
$ " " D "
† œ .
By elementary operation on the rows of the augmented matrix we get:
" " & l " " " & l " " " & l "
# " % l ! ! " "% l # ! " "% l #
$ " " l " ! % "% l % ! $ ! l #
Ê Ê .
So we solve the system:
B C &D œ "
! B C "%D œ #
! B $ C !D œ # Ê
B œ " C &D œ D œ # C œ C œ
"
" # (
"% #"
#
$
.
The coordinates of the vector ˜œ "ß !ß " — ß ß" # # Þ ( $ #"
in the basis are
II M 1) For the system we get:
0 Bß Cß D œ B C D $BC œ ! 1 Bß Cß D œ B C $D BCD œ !
# # $
$ $ #
0 "ß "ß " œ $ œ !
1 "ß "ß " œ 1 $ 1 1 œ ! "ß "ß "
1 1 1 , so the system is satisfied at point .
We construct the Jacobian matrix ` 0 ß 1 from
` Bß Cß D œ #B $C #C $B $D
$B CD $C BD BC 'D
# # #
which we get: ` 0 ß 1 .
` Bß Cß D "ß "ß " œ " " $
% % &
As " $ it is possible to define an implicit function like
% & œ & "# œ ( Á !
B Ä C B ß D B and also an implicit function like C Ä B C ß D C .
As " " it is not possible to define an implicit function like .
% % œ! D Ä B D ß C D
We choose B Ä C B ß D B and we get:
.C .D
.B œ œ œ " .B œ œ œ !
" $ " "
% & % %
" $ " $
% & % &
( !
( ; ( .
II M 2) Firstly we write the problem in the form:
Max/min u.c.:
0 Bß C œ B C "
B C Ÿ "
! Ÿ B
#
Max/min
u.c.: .
0 Bß C œ B C "
B C " Ÿ !
B Ÿ !
#
The objective function of the problem is a continuous function, the feasible region is aX compact set, and so maximum and minimum values surely exist. The constraints are qualified.
Since 0 Bß C œ B C " !ß a Bß C − X and for all the points !ß C it is 0 !ß C œ ! , all the points Bß C À B œ !ß " Ÿ C Ÿ " are minimum points for the problem.
The Lagrangian function of the problem is:
ABß Cß- -"ß #œB C " -"B C "# -#B. 1) case -" œ !ß-# œ ! À
A A
wB wC
œ C " œ ! œ B œ !
Ê
œ ! œ "
B C Ÿ "
! Ÿ B
! Ÿ "
! Ÿ !
#
B
C : the point has already been studied.
2) case -" Á !ß-# œ ! À
A -
A -
- -
- - - -
- - - -
wB "
wC " " " " "
" "
" "
# #
" " "
# # #
" "
œ C " œ ! œ B # C œ !
Ê Ê
œ " œ "
œ # " œ # # œ # #
# # " # $ %
B C œ "
! Ÿ B
œ "
! Ÿ B
#
C C
B B
" œ !
! Ÿ B
Ê Ê
œ # # œ # # œ !
œ " œ " œ "
$ %
B B
C C
- - - -
- -
- - -
"# "#
" "
" "
" " œ ! " œ !
! Ÿ B ! Ÿ !
the point has already been studied
or Ê ; since
œ # # œ
œ
B C
"' % )
$ $ *
"
$
"
)
*
"
- -
œ
! Ÿ
œ % !
% $
$
the point may be a maximum point.
3) case -" œ !ß-# Á! À
A -
A
wB #
wC
œ C " œ ! œ B œ !
B œ ! B C Ÿ "#
: points !ß C have already been studied.
4) case -" Á!ß-# Á! À
A - -
A -
wB " #
wC "
œ C " œ ! œ B # C œ !
B C œ "
œ !
B C œ "
! œ B
#
#
B
!ß "
. Since the points solutions of are and !ß ", such points have already been studied.
So the point ) " is the maximum point, with ) " while all the points
* $ß 0 * $ß œ $#
Bß C À B œ !ß " Ÿ C Ÿ " with 0 !ß C œ ! are minimum points for the problem.#(
II M 3) For the existence field of the function 0 Bß C œ B we must
log C " C B "#
satisfy the condition: B C " whose solution is given by the union of the solu- C B "# !
tions of the two inequality systems:
B C " ! C B " B C " ! C B "
C B " ! C B " C B " ! C B "
# Ê # or # Ê # .
We graphically represent the situation as follows:
The vertical green lines indicate the solution of the first system, the horizontal red lines the solution of the second system.
For the gradient of the function we get:
`0 C B " " C B " #B C " B #BC #B C "
`B œ C " † œ C " C B "
C B "
# # #
# # #
B B
B
`0 C B " " C B " " C " B B #
`C œ C " † œ C " C B "
C B "
# # #
# # #
B B
B
.
So f0 !ß ! œ "ß # .
II M 4) The function 0 Bß C œ 5 B # C#B C is a polynomial, so it is twice differen- tiable. So, to calculate the first order directional derivatives we get:
H 0/" "ß ! œf0 "ß ! † "ß ! œ 0 "ß ! Bw and H 0/# !ß " œf0 !ß " † !ß " œ 0 !ß " Cw Þ
Then from H 0/" "ß ! œH 0/# !ß " we get 0 "ß ! œ 0 !ß "Bw Cw . Since: 0 Bß C œ #5B "Bw Ê0 "ß ! œ #5 "Bw and
0 Bß C œ #5C "Cw Ê0 !ß " œ #5 "Cw from 0 "ß ! œ 0 !ß "Bw Cw we get:
#5 " œ #5 " Ê %5 œ # Ê 5 œ "
#. The second order directional derivative it is obtained as:
H/ ß/#" #0 "ß " œ /"† "ß " † /#X Bß C #5 "
‡ . Since ‡ œ !#5 œ !
! ! " we get:
H/ ß/#" #0 "ß " œ " !† "ß " † ! œ" !† " † ! œ !
" " "
‡ !
! .
