QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS
TASK 17/9/2020
I M 1) Calculate the cubic roots of the number D œ " "" 3 " 3.
Since D œ " " œ " 3 " 3 œ #3 œ 3 œcos 3sin we get:
" 3 " 3 " 3 " 3 # # #
1 1
$ 3 œ cos1 1 sin1 1
' 5 #$ 3 ' 5 #$ ß ! Ÿ 5 Ÿ # . For 5 œ ! we get D œ! cos 1 sin1 à
' 3 ' œ #$ 3 #"
for 5 œ " we get D œ" cos & sin& $ "à '1 3 '1 œ # 3 #
for 5 œ # we get D œ" cos * sin* cos$ sin$ Þ '1 3 '1 œ #1 3 #1 œ 3
I M 2) Given the matrix œ , determine the values of the parameter for5
5 ! 5
! # !
5 ! 5
which the matrix admits a multiple eigenvalue.
The matrix is a symmetric one, therefore certainly it has only real eigenvalues.
From - ˆœ ! we get:
5 ! 5
! # !
5 ! 5
5
5 œ #5 œ
-
-
-
- - - - -
œ # # # # #
œ-#- - #5 œ ! and so we get the eigenvalues -" œ !ß-# œ # and -$ œ #5Þ So the matrix admits a multiple eigenvalue when #5 œ ! Ê 5 œ ! Ê -" œ -# œ ! or when
#5 œ # Ê 5 œ " Ê -# œ-$ œ " .
I M 3) Given the linear system , check existence and number of its
B #B œ "
#B B 5 B œ # B #B 'B œ 5
" $
" # $
" # $
solutions on varying the parameter .5
By Rouchè-Capelli Theorem, from —† œ˜ we need to calculate the Rank of the matrix and the Rank of the augmented matrix ˜l .
By elementary operations on the rowsV Ã V #V# # " and V Ã V V$ $ " we get À
" ! # l " " ! # l "
# " 5 l # ! " 5 % l !
" # ' l 5 ! # ) l 5 "
Ä Þ
By elementary operations on the rowsV Ã V #V$ $ # we getÀ
" ! # l " " ! # l "
! " 5 % l ! ! " 5 % l !
! # ) l 5 " ! ! #5 l 5 "
Ä Þ
And so:
if #5 œ ! Ê 5 œ ! ÊRank œ # Rank ˜l œ $ and so the system has no solutions;
if #5 Á ! Ê 5 Á ! ÊRank œ $ œRank ˜l and so the system has one and only one solution.
I M 4) Determine the matrix œ + + knowing that "ß " is an eigenvector rela-
+ +
"" "#
#" ##
tive to the eigenvalue - œ ! and that "ß " is an eigenvector relative to -œ #. Since "ß " is an eigenvector relative to the eigenvalue -œ ! weget:
+ + " " ! + + œ ! + œ +
+"" +"# † " œ ! † " œ ! Ê + +"" "# œ ! Ê +"# œ +"" à
#" ## #" ## #" ##
Since "ß " is an eigenvector relative to the eigenvalue -œ # weget:
+ + " " # + + œ # + œ # +
+"" +"# † " œ # † " œ # Ê + +"" "#œ # Ê +"# œ # +"" à
#" ## #" ## #" ## from these
we get: + œ # + + œ " + œ " " "
+""œ # +"" Ê +"" œ " Ê +"#œ " Ê œ " " Þ
## ## ## #"
II M 1) Solve the problem :
Max/min u.c.:
0 Bß C œ B C C C Ÿ " #B
B ! C !
Þ
# #
The objective function of the problem is a continuous function, the constraints are linear func- tions and they define a feasible region which is a compact and bounded set as it is a triangle, and so we can apply Weierstrass Theorem. To solve the problem we don't use the Kuhn- Tucker Theorem.
We determine the stationary points of the function.
Applying the first order conditions we have:
f0 Bß C œ Ê 0 œ #B œ ! Ê Þ ‡Bß C œ œ‡!ß
0 œ #C " œ !
B œ ! C œ
# !
! #
"
BCww " #
#
Since ,
immediatly we see that !ß "# is a minimum point.
For B œ ! we get 0 !ß C œ C C Ê 0 !ß C œ #C " C "
# w ! for # and so the result that we have already found: !ß "# is a minimum point.
For C œ ! we get 0 Bß ! œ B Ê 0 Bß ! œ #B # w ! for B ! and so the function is always increasing from B œ ! to B œ "
#.
For C œ " #B we get 0 Bß " #B œ &B #B Ê 0 Bß " #B œ "!B # # w ! for
B " B œ ! B œ "
& &
and so the function is decreasing from to and then is increasing from B œ " B œ " B œ "
& to #, so & is a minimum point and so " $& &ß is a minimum point.
By representing with a figure the results found we have:
So !ß " is a maximum point, with 0 !ß " œ ! à relative maximum point;
" " "
#ß ! is a maximum point, with 0 #ß ! œ %; absolute maximum point;
!ß" 0 !ß " œ "
# is a minimum point, with # %; absolute minimum point;
" $ " $
& &ß 0 & &ß œ " Þ
&
is a minimum point, with ; relative minimum point
II M 2) Given the function 0 Bß C œ B BC # #, and the unit vector @œcosαßsinα, cal- culate W and W . And then calculate W when α 1 .
@ # #
@ß@ @ß@
0 "ß " 0 "ß " 0 "ß " œ
%
The function 0 Bß C œ B BC # # it is clearly differentiable of every order a Bß C − ‘#. So H 0@ "ß " œ f0 "ß " † @ and W#@ß@0 "ß " œ @ †‡0 "ß " † @X .
From f0Bß C œ #B C ß #BC # we get f0 "ß " œ "ß # and so:
W@0 "ß " œ "ß # † cosαßsinαœcosα #sinαÞ Then we getÀ
‡Bß C œ # #C ‡ œ # #
#C #B Ê "ß " # # and so:
W α α α
α
@ß@# 0 "ß " # # Ê
# #
œcos sin † † cos
sin
W@ß@# 0 "ß " œ cos# #α #sin#α %cos sinα αœ #cos# #α sin#α.
Finally, for α 1 W 1 1 .
œ 0 "ß " # # œ #
% we get: #@ß@ œ cos # sin #
II M 3) Given the function 0 Bß C œ B B C C # # # determine existence and nature of its stationary points.
Applying the first order conditions we have:
f0 Bß C œ Ê 0 œ #B #BC œ #B " C œ ! Ê ∪ 0 œ B #C œ !
B œ ! C œ !
B œ # C œ "
Bw
w #
C
# and so the
solutions B œ ! and
C œ !
B œ # B œ #
C œ " C œ "
ß Þ
Since ‡Bß C œ # #C #B we get:
#B #
- ‡ !ß ! œ and so !ß ! is a minimum point;
# ! œ # !
! # Ê ‡ œ % !
‡
"
#
- ‡ so is a saddle point;
# " ! # Ê œ # ! # "
# #
ß œ ‡# ß
- ‡ ß œ so ß is a saddle
# " ! # Ê œ # ! # "
# # ‡#
point.
II M 4) Given the equation 0 Bß C œ B / BC C /BC 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 functions0 Bß C is a differentiable function a Bß C − ‘#, 0 !ß ! œ ! ! œ ! . It is then: f0 Bß C œ /BC B /BC C /BCà B /BC /BC C /BC from which:
f0 Bß C œ " B / BC C /BCà B /BC " C / BC.
So f0 !ß ! œ1 !à ! " œ "ß ". Since 0 !ß ! œCw 1Á ! we can define an im- plicit function B Ä C B whose first order derivative is:
d d
C 0 !ß !
B ! œ 0 !ß ! œ œ " Þ
Bw Cw
"
"
