Lagranj ko’paytuvchilari usuli

6.3. Lagranj ko’paytuvchilari usuli
n
o
„
zgaruvchili funksiyalar uchun Lagranjning ko
„
paytuvchilar usuli
.
Aytaylik
n
o„zgaruvchili
1
2
,
, ...,
n
z
f x x
x
funksiya va
m
ta chegaraviy shartlar
berilgan bo„lsin
1
2
1
1
2
2
1
2
1
2
,
, ...,
,
, ...,
0,
,
, ...,
0,
.............................
,
, ...,
0.
n
n
n
m
n
z
f x x
x
x x
x
x x
x
x x
x
.
Lagranj funksiyani tuzamiz
80
1
2
1
2
1 1
2
2
,
, ...,
, ,
, ...,
...
n
m
m
m
F x x
x
f
.
Lagranj funksiyasining barcha noma`lumlari bo„yicha xususiy hosila olib
ularni nolga tenglashtirib, stasionar nuqtalarni aniqlaymiz
0,
1,
0,
1,
i
j
F
i
n
x
j
m
.
Yuqoridagi kabi to„la differensial
2
d F
ning ishorasini aniqlaymiz. Agar
aniqlangan stasionar nuqtada
2
0
d F
shart bajarilsa, bu nuqta shartli maksimum,
aksincha esa, ya`ni
2
0
d F
bo`lsa, shartli minimum bo„ladi. Buni boshqacha
usulda, ya`ni quyidagi matrisadan foydalanib ham aniqlashimiz mumkin.
1
1
1
1
1
2
3
2
2
2
2
1
2
3
1
2
3
1
0
0
...
0
...
0
0
...
0
...
................................................................................................
0
0
...
0
...
n
n
m
m
m
m
n
x
x
x
x
x
x
x
x
x
x
x
x
A
1
2
3
2
2
2
2
2
2
1
1
1
1
2
1
3
1
2
2
2
2
1
2
2
2
2
2
2
1
2
3
2
2
2
2
2
1
2
2
3
3
3
3
1
3
2
3
...
...
...
...
...
...
.............................
m
n
m
n
m
n
F
F
F
F
x
x
x
x
x x
x x
x x
F
F
F
F
x
x
x
x x
x
x x
x x
F
F
F
F
x
x
x
x x
x x
x
x x
2
2
2
2
1
2
2
1
2
3
.....................................................................
...
...
n
m
n
n
n
n
n
n
F
F
F
F
x
x
x
x x
x x
x x
x
A matrisadagi
81
1
2
3
2
2
2
2
2
1
2
1
3
1
2
2
2
2
2
2
1
2
3
2
2
2
2
2
2
3
1
3
2
3
2
2
2
1
2
...
...
...
..............................................................
n
n
n
n
n
n
F
F
F
F
x
x x
x x
x x
F
F
F
F
x x
x
x x
x x
F
F
F
F
H X
x x
x x
x
x x
F
F
F
x x
x x
x
2
2
3
...
n
F
x
x
matrisa, Gesse matrisasidir.
Quyidagi qoidani kiritamiz:
-
Agar
A
matrisaning burchak minorlari
2
1
2
2
,
,...,
m
m
m n
H
H
H
ishoralari
1
m
ning ishoralari bilan mos bo„lsa, u holda bu stasionar nuqta
1
2
,
, ...,
n
z
f x x
x
funksiyaning minimum nuqtasi bo„ladi.
-
Agar
A
matrisaning burchak minorlari
2
1
2
2
,
,...,
m
m
m n
H
H
H
ishoralari
almashinib tursa va
2
1
m
H
minorning ishorasi
1
1
m
ning ishoralari bilan mos
tushsa, u holda bu stasionar nuqta
1
2
,
, ...,
n
z
f x x
x
funksiyaning maksimum
nuqtasi bo„ladi.
Misol.
Funksiya shartli ekstremum nuqtasini toping
3
L x
x
y
chegaraviy shartda:
2
2
10
x
y
Yechish
. Lagranj funksiyasi tuziladi
2
2
, ,
3
10
F x y
x
y
x
y
.
Lagranj funksiyasining xususiy hosilalarini aniqlaymiz
1
2
2
2
2
2
1
1 2
0,
,
2
1
3
2
3 2
0,
,
1
2
2
1
3
10
0.
10
0.
2
2
F
x
x
x
F
y
y
y
F
x
y
1
1
1
1
1
1
2
2
2
2
2
2
1
1
3
1
;
1;
3
2
2
2
2
1
1
1
3
;
1;
3
2
2
2
2
x
y
x
y
Demak, sistema ikkita yechimga ega ekan
82
1
1
2
2
1
,
1, 3
2
1
,
1,
3 .
2
M
M
Bu nuqtalarning har birida Gesse matrisasini hisoblaymiz.
2
2
2
2
10
2 ,
10
2 .
x
y
x
y
x
x
x
y
y
y
2
2
2
2
2
1 2
2 ,
3 2
2 ,
1 2
0.
x
y
y
F
x
x
F
y
y
F
x
x y
1
2
2
2
2
1
1
1
2
2
2
2
2
1
2
2
0
0
2
2
2
2
0
2
0
2
x
x
x
y
F
F
A
x
x
x
x x
y
F
F
x
x x
x
.
1
1, 3
M
nuqtada
0
2
2
0
1
3
2
2
0
8 1
1/ 2
0
40
0
2
0
2
3
0
1/ 2
x
y
A
x
y
.
Demak,
1
1, 3
M
nuqtada
1
2
3
L x
x
x
funksiya shartli minimumga ega
ekan
min
1 3 3 10
L x
.
Shu kabi
2
1,
3
M
nuqtada quyidagini aniqlaymiz
0
2
2
0
1
3
2
2
0
8
1 1/ 2
0
40
0
2
0
2
3
0
1/ 2
x
y
A
x
y
.
Demak,
2
1,
3
M
nuqtada
1
2
3
L x
x
x
funksiya shartli minimumga ega
bo„lar ekan
min
1 3
3
10
L x
.
Misol.
Funksiya shartli ekstremum nuqtasini toping
2
2
2
1
2
3
min max
L x
x
x
x
chegaraviy shartlarda:
83
1
2
3
1
2
4
2
3
12
x
x
x
x
x
Yechish
. Lagranj funksiyasi tuziladi
2
2
2
1
2
3
1
1
2
3
2
1
2
4
2
3
12
F
x
x
x
x
x
x
x
x
Lagranj funksiyasi
F
dan
1
2
3
,
,
x x
x
noma`lum o„zgaruvchilar va
1
2
,
parametrlar bo„yicha xususiy hosilalar olib, ularni nolga tenglashtirib, quyidagi
tenglamalar sistemasi hosil qilinadi:
1
1
2
1
2
1
2
2
3
1
3
1
2
3
1
1
2
2
2
2
0
2
3
0
2
0
4
0
2
3
12
0
F
x
x
F
x
x
F
x
x
F
x
x
x
F
x
x
Bundan
1
2
3
72
28 32
,
,
,
,
, min
19,37
19
19 19
x x x
L x
.
Misol.
Funksiya shartli ekstremum nuqtasini toping
1 2
2 3
L x
x x
x x
chegaraviy shartlarda:
1
2
2
3
2
2
4.
x
x
x
x
Yechish
. I-
usul
. Lagranj funksiyasi tuziladi
1
2
3
1
2
1 2
2 3
1
1
2
2
2
3
,
, , ,
2
2
4
F x x x
x x
x x
x
x
x
x
Lagranj funksiyasi
F
dan
1
2
3
,
,
x x
x
noma`lum o„zgaruvchilar va
1
2
,
parametrlar bo„yicha xususiy hosilalar olib, ularni nolga tenglashtirib, quyidagi
tenglamalar sistemasi hosil qilinadi:
84
2
1
1
1
3
1
2
2
2
2
3
1
2
1
2
3
1
0
0
2
0
2
0
2
4
0
F
x
x
F
x
x
x
F
x
x
F
x
x
F
x
x
Bundan
1
2
3
,
,
2,
4, 4 , min
8
х x x
L x
.
II-
usul
. Noma`lumlarni ketma-ket yo„qotish usuli. Buning uchun chegaraviy
shartdagi ikkita tenglamadan birini yo„qotib, bitta tenglamadan iborat chegaraviy
shart tuzamiz. Chegaraviy shartdagi birinchi tenglamadan
2
x
noma`lumni topib,
ikkinchi tenglamaga va maqsad funksiyasiga qo„yib, quyidagi chiziqli
programmalashtirish masalasini hosil qilamiz
1
2
1 2
2 3
1
1
1
3
1 3
1
3
2
1
2
1
2
1
2
3
1
3
1
3
2
2
2
2
2
2
2
2
4.
2
6.
2
2
4.
L x
x x
x x
L x
x x
x
x
L x
x
x x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Lagranj funksiyasini tuzamiz
1
2
1
3
1 3
1
3
1
3
, ,
2
2
2
6
F x x
x
x x
x
x
x
x
Lagranj funksiyasi
F
dan
1
3
,
x
x
o„zgaruvchilar va
parametr bo„yicha
xususiy hosilalar olib, ularni nolga tenglashtirib, quyidagi tenglamalar sistemasi
hosil qilinadi:
1
3
1
1
3
1
1
1
3
1
3
3
1
3
2
2
0
2
2
0
2
2 2
0
2 2
0
2
2
6
0
4.
2
6
0
F
x
x
x
x
x
F
x
x
x
x
x
x
x
F
x
x
Bundan
2, 4
M
stasionar nuqtani aniqlaymiz. Bu nuqtani ekstremumga
tekshiramiz. Buning uchun ikkinchi tartibli xususiy hosilani aniqlaymiz
1
3
3
2
1
3
2
1
2
1
2
3
2
1
3
1
3
2
2
2,
2 2
0,
2
2
1.
x
x
x
F
x
x
x
F
x
x
F
x
x
x x
85
2
2
2
2
2
2
2
2
2
1
3
1
3
1
3
1
3
1
1
3
2
2
1
3
1
3
2
2
0
2
2
2
F
F
F
d F
dx
dx
dx dx
dx
dx
dx dx
dx
dx dx
x
x
x x
.
Endi
2
d F
ning ishorasini aniqlash uchun stasionar nuqtalarning xossasidan
foydalanamiz,
1
3
2
6
x
x
belgilash olib
1
3
1
3
1
3
1
1
3
3
1
3
1
3
2
6
2
6
2
0
x
x
d
dx
dx
x
x
dx
x
x
dx
dx
dx
x
x
.
Demak,
1
3
2
dx
dx
, bundan esa
1
2
2
2
2
2
1
1
1
1
1
2
2
2
0
2
dx
d F
dx
dx
dx
dx
dx
.
Demak, stasionar
2, 4
M
nuqta maqsad funksiyasining minimum nuqtasi
bo„lar ekan
1
2
2
1 3
1
3
min
2
2
2
2 4 2
2
2 4
8
L x
x
x x
x
x
.
Bu optimal yechimlardan foydalanib, boshlang„ich masalaning yechimini
aniqlaymiz. Bundan esa berilgan masalaning ekstremal qiymati quyidagicha
bo`ladi
1 2
2 3
1
2
3
2
1
2
2
3
,
,
2, 4, 4
2
4
4 4
2
min
8
2 2.
2
4.
L x
x x
x x
x x x
L x
x
x
L x
x
x
x
III usul. Endi masalaning optimal yechimini tuzilgan matrisaning burchak
minorlaridan foydalanib aniqlaymiz. Buning uchun, quyidagi xususiy hosilalarni
aniqlaymiz
1
2
3
1
1
2
1
1
1
2
2
1
1
2
3
2
1,
2
1
2
0.
x
x
x
x
x
x
x
x
x
x
x
x
1
2
3
2
2
3
1
2
2
3
2
2
2
3
3
2
4
0,
2
4
1
2
4
2.
x
x
x
x
x
x
x
x
x
x
x
x
1
2
2
3
2
3
3
3
2
2
2
1
2
1
2
2
1
1
1
2
1
2
2
1
3
1
2
1
3
1
2
2
1
2
2
1
3
2
2
2
1
3
1
2
2
2
2
3
2
3
0
1
0
0
0
0
2
0.
1.
2
0.
x
x
x
x
x
x
F
F
F
x
x
x
x
x x
x
F
F
F
x
x
x
x
x
x
x
x x
F
F
F
x
x
x
x
x
x x
x
86
1
2
3
1
1
1
1
2
3
2
2
2
1
2
3
2
2
2
1
2
2
1
1
1
2
1
3
2
2
2
1
2
2
2
2
2
1
2
3
2
2
2
1
2
2
3
3
3
1
3
2
0
0
0
0
0
0
1
1
0
0
0
0
1
2
1
0
0
1
0
1 1
1
0
1
0
2
0
1
0
x
x
x
x
x
x
F
F
F
A
x
x
x
x x
x x
F
F
F
x
x
x x
x
x x
F
F
F
x
x
x x
x x
x
A
matrisaning
burchak
minorlarining
2
1
2
2
,
,...,
m
m
m n
H
H
H
ishoralarini
aniqlashimiz kerak. Bunda
m
parametr masalaning chegaraviy shartlar soni,
n
parametr esa noma`lumlar soni. Bu masalada ikkita chegaraviy shart va uchta
noma`lum qatnashmoqda, shuning uchun
2,
3
m
n
.
2
1
5,
5
m
n
m
bo„lgani
uchun, bunda faqat
5
H
minor qaraladi.
A
matrisa 5-tartibli bo„lgani uchun
5
H
minor bu matrisa bilan ustma –ust
tushadi.
5
H
minorning tartibini pasaytirish orqali uning ishorasini aniqlaymiz.
4 2
0
0
1
1
0
0
0
1
1
0
0
0
0
1
2
0
0
0
1
2
1
0
0
1
0
1
0
0
1
0
1 1
1
0
1
1 1
1
0
1
0
2
0
1
0
2
0
2
1
2
0
1
1
0
0
0
1
2
1
2
1
1
4
0
1
0
1
0
3
2
2
2
1
2
H
Ta`rifga asosan
2
1
1
1
m
bo„lgani uchun,
2
1
m
H
ning ishorasi,
1
m
ning ishorasi bilan mos tushadi. Bundan kelib chiqib, topilgan stasionar
1
2
3
,
,
2,
4, 4
х x x
nuqta, berilgan funksiyaning minimal nuqtasi bo„lar ekan
min
8
L x
.
|