TASK MATHEMATICS for ECONOMIC APPLICATIONS 24/3/2018

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TASK MATHEMATICS for ECONOMIC APPLICATIONS 24/3/2018 I M 1) D œ /^{"% 3}1 † /^{"$ 3}1 œ /" " % 3( 3  1 1 œ /( 31 œcos(  31 sin(1œ

œcos1 3sin1œ  " . And so:

^$  " œ "cos1 1 sin1 1 Ÿ Ÿ

$  5 #$ $  5 #$

 3 à ! 5 # .

For 5 œ ! À cos 1 sin 1  à

$  3 $ œ "  3 $

# #

For 5 œ " À cos1 3sin1 œ  "à

For 5 œ # À cos & sin & " $Þ

 3 œ  3

# #

1 1

$ $

I M 2)

   

B  " B B    B  "  B B 

C C  " C C C  "  C

 C  C "  C  C  C "  C 

Ê -ˆ œ œ

-

G Ã G  G_" _" _# and G Ã G  G_$ _$ _#

œ œ "    " œ

"  B !

 " C  "   "

!  C " 

C  "   " B !

 C "   C " 

 

       

-

- - -

-

- - - -

- -

œ "  -C  " -" -  - " C  B "   -œ

œ "  -" -C  " - C  B œ "   -" -" - B Þ So the matrix ad- mits the eigenvalue - œ " with algebraic multiplicity equal to if $ B œ !ß a C.

I M 3) From 0 B ß B ß B " # $ œ +B ß B  ,B ß B  B  -B" " # " # $ we get, in matrix form:

$ß$ —$ß" ˜$ß"

" "

# #

$ $

† œ Ê † œ

+ ! ! B C

" , ! B C

" " - B C

     

     .

From  0 "ß "ß " œ !ß "ß #                            we get:  

 

 

 

+ ! ! " ! + œ ! + œ ! ! ! !

" , ! " " "  , œ " , œ ! " ! !

" " - " # "  "  - œ # - œ ! " " !

† œ Ê Ê . So œ .

Since Rank  œ #, we get Dim Imm   œ # and Dim Ker   œ $  # œ ".

To find a basis for the Image we apply the Rouchè-Capelli Theorem to the augmented matrix

 

! ! ! l C     

" ! ! l C

" " ! l C

œ l C œ !

"

#

$

"

. We get Rank  Rank  ˜ if . So the vectors belonging to the Image are the vectors of the form !ß C ß C# $. A basis for the Image may be:

– œ!ß "ß ! à !ß !ß "   . To find a basis for the Kernel we solve the homogeneous system: . So the vectors belonging to

     

 

 

 

! ! ! B ! ! œ ! B œ !

" ! ! B ! B œ ! B œ !

" " ! B ! B  B œ ! a B

† œ Ê Ê

" "

# " #

$ " # $

the Kernel are the vectors of the form !ß !ß B. A basis for the Kernel is –œ!ß !ß ".

I M 4) From œ  œ

" ! ! " # %

# " ! ! " #

% # " ! ! "

   

  we get ^X   and so:

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 œ  œ

" " " " # % # $ &

! " " ! " # ! # $

! ! " ! ! " ! ! #

X

     

     .

For the determinant we get  

 

 œ œ # † # † # œ ) Þ

# $ &

! # $

! ! #

X

Now we calculate the adjoint matrix and we get: Adj  .

 

 œ

% ! !

 ' % !

 "  ' %

X

Now we calculate the transpose of the adjoint matrix and we get: .

 

%  '  "

! %  '

! ! %

Finally, dividing by the determinant we get the inverse matrix:

 

 

 œ

 

! 

! !

X "

" $ "

# % )

" $

# %

"

#

.

II M 1) From the equation 0 Bß Cß D œ B/  ^CD  C/^DB D/^BC œ ! we get:

0 !ß !ß ! œ !  !  ! œ !  and so the point !ß !ß ! satisfies the equation. Then

f0 Bß Cß D œ /   ^CD C/^DB D/^BCß B/^CD  /^DB D/^BCß  B/^CD  C/^DB  /^BC and so we get f0 !ß !ß ! œ "ß  "ß "   . Since 0 œ " Á !D^w there exists an implicit function

Bß C Ä D Bß C   with partial derivativesÀ

`D " `D  "

`B !ß ! œ  " œ  " and `C !ß ! œ  " œ ". For the equation of the tangent plan at  !ß ! we get:

D  ! œ  " B  !  " C  ! Ê D œ  B  C Ê B  C  D œ !    .

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 an ellipse, a compactX set, and so maximum and minimum values surely exist. The constraints are qualified. The Lagrangian function of the problem is: ABß Cß-œB  C^# ^#-B  %C  %^# ^# .

1) case - œ ! À

 

 

       

A A

wB wC

œ #B œ  #C

œ !

œ ! Ê œ !ß ! œ

B œ ! C œ !

! %

!

! 

B  %C Ÿ % Ÿ

Bß C #

# # #

. Since ‡ ‡ the point

 !ß ! is a saddle point.

2) case - Á ! À

 

 

 

 

A - -

wB wC

œ #B  # B œ  #C  ) C

œ #B "  œ !

œ  #C "  % œ ! Ê

B œ ! C œ !

! %

B  %C œ %^# ^# Á

unacceptable solution;

∪ Ê à  !

B œ ! œ 

 

 

- ^"_% - ^"_% -

%C œ %^# C œ „"

since these points may be minimum points;

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∪ Ê à ! œ "

œ !

œ "

œ !

 

 

 

- -

C C - 

B œ %^# B œ „#

since these points may be maximum points;

∪

œ "

œ 



 -

- ^"_%

B  %C œ %^# ^#

there is no solution.

Since 0 !ß " œ 0 !ß  " œ  "    while 0 #ß ! œ 0  #ß ! œ %    we see that  !ß "

and !ß  " are minimum points while  #ß ! and  #ß ! are maximum points.

If we want to study our problem in the boundary points, only for exercise, since we have just completely solved the problem, we can use the parametric form for the ellipse:

B œ # >   C œ cos> Ê 0 BÞC

sin Ä 0 > œ %  cos^#> sin^#> œ &cos^#>  ". By deriving we get:

0 > œ "!^w  cos> †  sin> œ  "! cos sin> > œ  #sin#>.

Hence 0 > ! Ê Ê Ÿ > Ÿ ∪ Ÿ > Ÿ # and so we get the

# #

w  sin #> Ÿ ! 1  $1 

1 1

results that we have already found.

II M 3) Since 0 Bß C œ B /  C /  ^C ^B is a twice differentiable function, we get:

W_@0 !ß ! œ f0 !ß ! † @( )   and W_{@ @}², 0 !ß ! œ @ †( ) ‡ !ß ! † @^X . f0 Bß C œ /  C / ß B /  /   ^C ^B ^C ^BÊf0 !ß ! œ "ß "   .

W_@0 !ß ! œ f0 !ß ! † @ œ "ß " †( )     cosαßsinα œcosαsinαœ ! from which we get cosÀ œ sin œ $ and œ ( .

% %

α αÊα 1 α 1

Since ‡Bß C  C / /  /  ‡ œ! 

/  / B /

œ Ê !ß ! #

# !

B C B

C B C we get:

W α α α α α α

α α

2,

@ @0 !ß ! œ † ! † œ † # œ

( ) cos sin cos cos sin #sin

sin cos

   #      

# ! œ #sin cosα α #sin cosα αœ #sin#α.

II M 4) From 0 À‘Ä‘^$ß > Äsin > ß / ß^# ^#> log"  > we get:

— > œsin > ß / ß^# ^#> log"  > from which —  ! œ !ß "ß !. Then:

—^w > œ#>  > ß # / ß^# ^#> "  —^w  ! œ !ß #ß "

"  >

cos from which .

The equation of the tangent line to this curve at the point > œ ! is:

> Ä— !  > †—^w  ! œ !ß "ß !  > † !ß #ß "  Ê> Ä !ß "  #>ß > .