Javoblar chiziqsiz programmalashtirish

Yechish. Dastlab masalaning matematik modelini tuzaylik. Idish asosi radiusini x₁ , balandligini x₂ orqali belgilasak, quyidagi analitik masalaga ega bo`lamiz (2-chizma).

В
x2
x
2-chizma
2x₁²  2x₁x₂  min (10.6.2)
x₁²x₂  V₀ (10.6.3)
Bu yerda V₀ - idishning bir barrelga mos sig`imi. (10.6.2)-(10.6.3) masala shartli ekstremum masalasidir:
f x1, x2   2<sup>2</sup>xx12<sub>2</sub>2V<sub>0</sub>x1x20. min, gx<sub>1</sub>, x<sub>2</sub>  x<sub>1</sub>
Masalani normallikka tekshiraylik:
 g 
g x x 1x, 2   xg1   <sub></sub>2x x12 2   0, , (10.6.4)
 x1 
x<sub>2</sub> 
chunki shartga ko`ra, x₁  0, x₂  0.
Demak, Lagranjning normal funksiyasini tuzish mumkin:
F x x ₁, ₂,     2 x₁² 2 x x_{1 2}   x x₁² ₂ V₀ ..
Ko`paytuvchilar qoidasiga ko`ra, agar (x₁, x₂) ekstremal nuqta bo`lsa, u nuqtada
F x x <sub>1</sub>, <sub>2</sub>,
 4x<sub>1</sub>  2x<sub>2</sub>  2x x<sub>1 2</sub>  0,
x<sub>1</sub>
 F x x <sub>1</sub>, <sub>2</sub>,<sup></sup>  2 x1  x12  0,
x<sub>2</sub> F x x <sub>1</sub>, <sub>2</sub>, <sub>2</sub>
x x<sub>1 2</sub> V<sub>0</sub>  0, 
tengliklar o`rinli bo`lishi zarur. Bu tenglamalar sistemasini yechib, x<sub>2</sub><sup>0</sup> <sub></sub> 2x<sub>1</sub><sup>0</sup> ni topamiz, ya’ni eng kam material sarflash uchun bochkaning balandligini asos aylanasi diametriga teng qilib olish zarur ekan.
Agar bochkalar shu usulda yasalsa, iqtisodiy samara eng yuqori bo`lishi shubhasizdir.

10.7. Shartli minimumning yetarlilik sharti

Lagranj ko`paytuvchilari usuli zaruriy shart bo`lib, joiz nuqta qaralayotgan funksiyaning shartli minimum nuqtasi bo`lsa ham, maksimum nuqtasi bo`lsa ham u nuqtada (10.5.3) shart bajarilaveradi. Bu nuqtalarni farqlash uchun yetarlilik shartini keltiramiz.
Ta’rif. Agar shunday m o`lchovli <sub></sub> vektor topilsa va shu nuqtada (10.3.7) tengliklar bajarilsa, qaralayotgan (10.5.1),(10.5.2) masalada X <sup>0</sup> joiz nuqta shartli-statsionar nuqta deyiladi.
Teorema. Qaralayotgan masalada f X , g<sub>i</sub> X , i <sub></sub>1 ,m funksiyalar X <sup>0</sup> nuqta atrofida ikki marta differensiallanuvchi bo`lsin. Shartli-statsionar nuqtaning shartli nisbiy minimum nuqta bo`lishi uchun
 gÒ X 0  y  0 (10.7.1)
x
gipertekislikda
yT  2 F X 20 , y (10.7.2)
x
kvadratik formaning aniq musbat bo`lishi yetarlidir.
Bu teoremani isbotsiz qabul qilib, uning asosiy shartlarini tahlil qilaylik. (10.7.1) va (10.7.2) ifodalardagi T belgi transponirlash belgisi bo`lib, (10.7.1) tenglik g X( ) <sub></sub>0 gipersirtga (bu yerda g X   g<sub>1</sub>  X , ... , g<sub>m</sub>  X  funksiya X <sub></sub> X <sup>0</sup> nuqtada o`tkazilgan urinma gipertekislikni ifodalaydi (3chizma)

3 - chizma
Demak, (10.7.2) shart (10.7.1) gipertekislikda
²F X 2⁰, ^

x
matritsaning musbat aniqlangan bo`lishi, ya’ni (10.7.2) kvadratik formaning musbat aniqlangan bo`lishini anglatadi.
Bu shartni 10.6 mavzuda (bochkalar haqida) gi masalaga qo`llab ko`raylik.
Masalada (10.7.1) gipertekislik
 g g   y₁
     0
x₁ x₂   y₂
tenglikdan iborat bo`lib, uni (10.6.4) yordamida yozsak,
2x x y_{1 2 1} x y₁² ₂  0
yoki
2x₂y₁  x₁y₂  0 (10.7.3)

tenglikka ega bo`lamiz. Shartli-statsionar x<sub>1</sub><sup>0</sup>, x<sub>2</sub><sup>0</sup>   <sup></sup><sub></sub> 3 <sub>2</sub><sup>V</sup><sub></sub><sup>0</sup> ; 23 <sub>2</sub><sup>V</sup><sub></sub><sup>0</sup> <sub></sub><sup></sup> nuqtada esa
<sub></sub>
(10.7.3) tenglik u<sub>1</sub>=-u<sub>2</sub> to`g`ri chiziqdan iborat bo`ladi. Ikkinchi tartibli
2F 2F 
2  2 
matritsa esa ^x^F₂  ^_ ^_^x₂¹_F ^_^x1₂^_F^{x2 }_ korinishda bo`lib, shartli-
 
x2x1 x22 
statsionar nuqtada uning elementlari mos holda quyidagi matritsaning
4 2х200 2 2х10
2 2х<sub>1</sub> 0 
elementlaridan iborat bo`ladi. Bunga mos kvadratik formani (10.7.4)) urinma to`g`ri chiziqda tekshirsak
_{ у}2 _у2 4х2⁰ 0 2 2х1⁰ _ у2_
2 2х₁ 0   у₂ 
ga ega bo`lamiz. Shartli-statsionar nuqtada <sub>х2</sub><sup>0</sup> <sub> 2х1</sub><sup>0</sup> va <sub> </sub> <sup>2</sup><sub>0</sub> ni inobatga
х1 olsak, soddalashtirishlardan so`ng kvadratik formaning qiymati _4у2² ga teng ekanligini ko`ramiz. Bu esa aniq musbat sondir.

Shunday qilib, V₀ barrel neft tashiydigan silindrik bochkaning eng samarali formasini yasash uchun uning o`lchamlarini 10 3 V0 20 3 V0
õ , x  2
2 2
kabi olish zarur va yetarli ekan.