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Rewritten the limit in polar coordinates with X ÐBß CÑ œ +B  ,C, we have 3637 3 α 3 α 3 Ä ! 3 α3 α 3 α3 α 3 cos sin l cos sin l

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(1)

") The vector has norm one while gradient and hessian of are@ 0

f0 œ Ð/ ß  / Ñ 0 œ /  /

 / /

BC BC BC BC

BC BC

; . [   Thus

W@@0 ÐSÑ œ @ †X [0 ÐSÑ † @ œ

Ð ß Ñ † "  " † œ  #  œ

 " "

cos sin cos cos cos sin sin

α α sinα α α α α

   α # #

" sin#α. The condition is satisfied iff sin# œ !α and this is true for αœ ! ”αœ Î# ”1 αœ1”αœ $ Î#.1

# 0) is differentiable at point Ð!ß !Ñ iff exist an affine function X ÐBß CÑ such that:

ÐBß CÑ637Ð ß Ñ

0 ÐBß CÑ  0 Ð!ß !Ñ  X ÐBß CÑ

B  C œ !

Ä ! !# # . Rewritten the limit in polar coordinates with X ÐBß CÑ œ +B  ,C, we have

3637

3 α 3 α

3

Ä !

3 α3 α 3 α3 α

3

cos sin l cos sin l

#  Ð+  , Ñ

cos sin œ

3637 3 α α α α α α

Ä ! cos sin lcos sin l  Ð+cos  ,sin Ñ and the limit is only if! + œ , œ !. To conclude that is differentiable at point 0 Ð!ß !Ñ we must verify that the result is uniformly on , for this goal considerα

l3cos sinα αlcos sinα αll œ Ð3 cos sinα αÑ Ÿ ß a# 3 α and if + œ , œ ! we can conclude that a  !ß% 3 Ê l% 3cos sinα αlcos sinα αll %, convergence is uniformly and is differentiable at point 0 Ð!ß !Ñ.

$) Lagrange function of the problem is

_ÐBß Cß Dß Ñ œ- B"C""D  ÐB  C  D  $Ñ- # # # .

f œ _  B"#  # Bß - C"#  # Cß - D"#  # Dß  ÐB  C  D  $Ñ- # # # . Put the first order condition we have the sistem:

 

 

  

  

  

  

  

 

 



  # B œ !

  # C œ !

  # D œ ! B  C  D œ $

Ê Ê

"  # B œ !

"  # C œ !

"  # D œ ! B  C  D œ $

B œ  "Î#

C œ  "Î#

D œ  "Î#

"

B"

C

"

# D # #

$

$

$

# # #

#

#

#

$

$

$

- - -

- - -

- - - B  C  D œ $

Ê

# # #



 

 

 

 

 





 

 B œ  "Î#

C œ  "Î#

D œ  "Î#

$ "Î% œ $ Ê

B œ …"

C œ …"

D œ …"

œ „"Î#

$

$

$

$

- - -

-# -

. Two points satisfied first order conditions

T œ Ð"ß "ß "ß  "Î#Ñ" and T œ Ð  "ß  "ß  "ß "Î#Ñ# .

(2)

[ -

-

- µ œ

! #B #C #D

#B  # ! !

#C !  # !

#D ! !  #

 

 

 

 

 

#

B #

C

# D

$

$

$

, becouse there's only one constraint

we must analyse third and fourth principal minor of border hessian:

[ -

- -

-

µ œ œ  #B  #C œ

! #B #C

#B  # !

#C !  #

#B !

#C  #

#B  #

#C !

$ #

B

# C

# C

# B

 

 

 

 

 

 

$     

$

$

$

 #B#C#$  #- #C#B#$  #- ;

[ -

-

-

- µ -

œ œ  #D 

! #B #C #D

#B  # ! !

#C !  # !

#D ! !  #

#B  # !

#C !  #

#D ! !

%

#

B #

C

# D

# B

# C

 

   

   

   

   

   

   

   

 

$

$

$

$

$

  # µ œ  #D  # !   # µ œ

!  #

D#$ #  D#$

#

B #

C

$ $

$

$

- [ - - [

-

 #D#B#$  #-C#$  #-  D#$  #- #B#C#$  #- #C#B#$  #-œ

 #B#C#$  #-D#$  #- #C#B#$  #-D#$  #- #D#B#$  #-C#$  #- . Second order conditions:

[µ [µ

ÐT Ñ œ  "#  ! ÐT Ñ œ  &%  ! T

$ " ; . % " " minimum.

[µ [µ

ÐT Ñ œ  "#  ! ÐT Ñ œ  &%  ! T

$ # ; . % # # maximum.

The proposed problem has minimum equal on point $ Ð"ß "ß "Ñ and maximum equal

 $ on point Ð  "ß  "ß  "Ñ.

%) The condition is satisfied on point , T " † / œ " † /$ $; rewritten the condition as B /#C D# # C /B #D# # œ ! we calculate the gradient of

0 ÐBß Cß DÑ œ B /#C D# # C /B #D# #.

f0 œ Ð/#C D# # #BC/B #D# #ß %BC/#C D# # /B #D# #ß #BD/#C D# # %CD /B #D# #Ñ and f0 ÐT Ñ œ Ð  / ß $ / ß  # / Ñ$ $ $ , but any element of f0 ÐT Ñ is different from zero thus we can have in implicy form a function of kind ÐBß CÑ È DÐBß CÑ or

ÐBß DÑ È CÐBß DÑ or ÐCß DÑ È BÐCß DÑ. If we choose the first kind,

fDÐ"ß "Ñ œ  0 ÐT Ñ0 ÐT ÑDBww ß  0 ÐT Ñ0 ÐT ÑCDww  œ  # //$$ß  # /$ /$$ œ  ß# #" $. The equation of tangent plane will be

D  " œ fDÐ"ß "Ñ † B  " Í D  " œ  ÐB  "Ñ  ÐC  "Ñ Í C  "

  "# $#

B  $ C  # D œ !.

(3)

& fD œ Ð$B  Cß B  $C Ñ D œ 'B D Ð"ß !Ñ œ D Ð!ß "Ñ œ '

) , # α α # [ 'Cα . BBww CCww ,

 α 

D Ð"ß !Ñ œ D Ð!ß "Ñ œwwBC CBww α; the condition is true when $' œα# Êαœ „'.

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