Authors

  • Shodiyev Kamoliddin
  • Mehroj Vahobov
  • Elburs Mingboyev

DOI:

https://doi.org/10.71337/inlibrary.uz.science-research.75389

Keywords:

funksiya ekstrumlari iqtisodiy va qurulish masalalarini yechish usuli matematik usullar iteratsiya koeffitsiyent.

Abstract

Maqolada funksiya ekstrumlarini iqtisodiy va qurulish masalalarini yechishga tadbiqi ko'rib chiqildi va ikki usullarini ko‘rib chiqildi.

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FUNKSIYA EKSTRUMLARINI IQTISODIY VA QURULISH MASALALARINI

YECHISHGA TADBIQI

Kamoliddin Shodiyev

Mirzo Ulug‘bek nomidagi Samarqand davlat arxitektura-qurilish universiteti, i.f.f.d., PhD,

Email:

shodiyevkamoliddin91@gmail.com

Vahobov Mehroj

Mirzo Ulug‘bek nomidagi Samarqand davlat arxitektura-qurilish universiteti ilmiy xodimi,

Email:

vahobovmehroj62@gmail.com

Mingboyev Elburs To'lqin o'g'li

Mizro Ulug'bek nomidagi O'zbekiston Milliy Universiteti, Matematika yo'nalishi talabasi.

https://doi.org/10.5281/zenodo.15089514

Annotatsiya.

Maqolada funksiya ekstrumlarini iqtisodiy va qurulish masalalarini

yechishga tadbiqi ko'rib chiqildi va ikki usullarini ko‘rib chiqildi.

Kalit so'zlar:

funksiya ekstrumlari, iqtisodiy va qurulish masalalarini yechish usuli,

matematik usullar, iteratsiya, koeffitsiyent.


KIRISH

Bizga ma’lumki, biror mahsulot ishlab chiqarish uchun boshlang’ich vaqtda ishlab

chiqarilgan mahsulot soni oz bo’lishiga qaramasdan xarajat ko’payadi, keyinchalik ichki
resurslarni qayta sarflash hisobiga sarflangan xarajat kamayadi. Lekin keyinchalik mahsulot
miqdorini ko’payterish uchun qo’shimcha harajat talab qilinadi natijada xarajat funksiyasi oshib
boradi. Biz bu maqolada iqtisodiy ma‘nolaridan foydalanib harakat funksiyasi foyda funksiyasi,
ishlab chiqarish tushum maksimum bo’lganda uni topish, maksimum foyda nuqtada talab
elastikligini, talab va taklif funksiyalari va hokozalarni iqtisod, qurilish masalalariga tadbiqini
beramiz.

Ishlab chiqarishga sarflangan xarajat

𝑋

ikkiga bo’linadi. O’zgarmas xarajat

𝑂′𝑀𝑋

va

O’zgaruvchi xarajat

𝑂

𝑍𝑋 (

1)

𝑋 = 𝑂

𝑀𝑋 + 𝑂′𝑍𝑋

(1)

ko’rinishda bo’ladi.

𝑂′𝑀𝑋

o’zgarmas xarajat ishlab chiqarish mahsulotlariga bog’liq

bo’lgan holda hamma vaqt mavjud bo’ladi.

𝑂

𝑍𝑋

o’zgaruvchi xarajat esa ishlab chiqariligan

mahsulot sonini

𝑄

desa unga bog’liq bo’ladi.

Xarajat chizig’i

𝑋

ni (1-chizmada) ko’rsatamiz

𝐴

nuqta egilish nuqtasi bo’lgani uchun

𝑋

(𝐶) = 0

𝐶

nuqtadan chapga

𝑋(𝑄)

qavariq

bo’lishi uchun

𝑋

′′

(𝑄) < 0 𝑄 < 𝐶

bo’lganda, lekin


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𝑋

′′

(𝑄) = [𝑋

(𝑄)]

− [𝐿𝑋(𝑄)]

.

Demak,

𝑄 < 𝐶

bo’lganda

𝐿𝑋

o’suvchi bo’ladi. Bu yerda

𝐿𝑋

ishlab chiqarishning ozgina

o’zgarishiga mos kelgan xarajatlar limittik xarjatlar deyladi va u quydagicha topiladi.

𝐿𝑋 = lim

∆𝑄→0

∆𝑋

∆𝑄

= 𝑋

(𝑄)

(2)

Koordinata boshidan

𝑋(𝑄)

grafigiga urinma o’tkazamiz va urinish nuqtasini

𝐵

bilan

bilgilaymiz.

𝑡𝑔𝛼 =

𝐵𝐷
𝑂𝐷

(3)

Bu yerda

𝐵𝐷

kesma

𝑄 = 𝐷

bo’lgandagi xarajatini ifodalaydi,

𝑂𝐷

esa ishlab chiqarish

hajmi

𝐷

ni ifodalaydi. Demak

𝑡𝑔𝛼 =

𝑋(𝐷)

𝐷

(4)

Umumiy xarajatning ishlab chiqarish xarajatiga nisbati o’rtacha xarajatni ifodalaydi va uni

𝑈𝑋

bilan belgilasak

𝑈𝑋 =

𝑋

𝑄

, 𝑡𝑔𝛼 = 𝑈𝑋(𝐷)

(5)

Ikkinchi tarafdan hosilaning geometric ma’nosiga asosan

𝑡𝑔𝛼 = 𝑋

(𝐷) = 𝐿𝑋(𝐷)

(6)

(5) va (6) larni solishtirib

𝑡𝑔𝛼 = 𝑈𝑋(𝐷) = 𝐿𝑋(𝐷)

tenglikni hosil qilamiz, ya’ni

𝐷

nuqta limitik harajat va o’rtacha xarajatlarning kesishish nuqtasining absisasi bo’ladi.

Endi o’rtacha harajatning grafiginiqanday o’zgarishini qaraymiz. Ta’rifga asosan (2), (3)

𝑡𝑔𝛼 =

𝑋(𝑄)

𝑄

bo’lanidan uning stasionar nuqtalarini topamiz:

(𝑈𝑋)

= (

𝑋(𝑄)

𝑄

)

=

𝑋

𝑄 − 𝑋 ∙ 1

𝑄

2

=

𝐿𝑋 ∙ 𝑄 − 𝑋

𝑄

2

= 0

𝐿𝑋 ∙ 𝑄 − 𝑋 = 0 𝐿𝑋 =

𝑋
𝑄

= 𝑈𝑋

Demak,

𝐿𝑋 = 𝑈𝑋

bo’ladigan nuqta stasionar nuqta bo’ladi va unda ekstremum mavjud

bo’lmaydi. Stasionar nuqtadan chapda va o’ngda hosilaning ishorasini tekshiramiz:

(𝑈𝑋)

=

𝐿𝑋 ∙ 𝑄 − 𝑋

𝑄

2

=

𝐿𝑋 ∙ 𝑄 − 𝑈𝑋 ∙ 𝑄

𝑄

2

=

𝐿𝑋 − 𝑈𝑋

𝑄

(7)

1-chizmadan ko’rinib turibdiki

𝑄 < 𝐷

bo’lganda burchak koeffisent urinmaning burchak

koeffisentidan kata

𝑡𝑔𝛽 > 𝑡𝑔𝛼 ,

𝑈𝑋 > 𝐿𝑋 ,

𝐿𝑋 − 𝑈𝑋 = 0 (8)

Natijada

(𝑈𝑋)

< 0

ya’ni

𝐷

dan chap tomonda

𝑈𝑋

ning hosilasi manfiy.

Ko’rsatish mumkinki,

𝑄 > 𝐷 𝑑𝑎 (𝑈𝑋)

> 0 (9)

Shunday qilib

𝐷

nuqta

𝑈𝑋

uchun minimum nuqta bo’ladi. (8) va (9) dan ko’rinadiki,

𝐷

dan

chapdan tomonda

𝐿𝑋

ning grafigi

𝑈𝑋

ning grafigidan yuqorida yotadi.

Agar ishlab chiqarilgan mahsulot soni

𝑄

birlik mahsulot narxi

𝑃

bo’lsa, mahsulotni

sotishdan hosil bo’lgan umumiy tushumini

𝑇

desak

𝑇 = 𝑃𝑄

(10) formula bilan hisoblanadi,

bitta mahsulotni sotishdan kelib chiqqan tushum o’rtacha deyladi va uni

𝑈𝑇

bilan belgilasak

𝑈𝑇 =

𝑇

𝑄

=

𝑃𝑄

𝑄

= 𝑃 (11) 𝑏𝑜

𝑙𝑎𝑑𝑖.

Foyda funsiyaning maksimumini topishni qaraymiz 2-chizmadan umumiy xarajat va

umumiy tushumning grafigi berilgan. Tavarning narxi o’zgarmas bo’lgan

(𝑃 − 𝑜

𝑧𝑔𝑎𝑟𝑚𝑎𝑠)


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holini qaraymiz. Bu holda tushum

𝑇

va harajat

𝑋

lar ayermasiga teng bo’lgani uchun

𝑀 𝑣𝑎 𝑁

nuqtalarning foyda nol bo’lgan holda to’g’ri keladi.

𝑄 < 𝑀 𝑣𝑎 𝑄 > 𝑁

harajat chizig’i tushim chizig’idan yuqorida joylashadi, yani bu holatda

ishlab chiqarish korxona zarar kurishi bilan ishlaydi

𝑀 < 𝑄 < 𝑁

oralig’ida esa korxona foyda

oladi, uni

𝐹

bilan belgilasak

𝐹

= (𝑇 − 𝑋)

= 0, 𝑇

= 𝑋

𝑙𝑒𝑘𝑖𝑛

𝑇(𝑄) = (𝑃 ∙ 𝑄)

𝑃 𝑦𝑎𝑛𝑖 𝑋

(𝑄) = 𝑃

Bundan kurinadiki, xajat grafigiga o’tkazilgan urinmaning burchak koeffsinti, tushum

chizig’ining burchak koeffsintiga teng bo’lishi kerak. Bu nuqta

𝐾

bo’ladi va maksimum foyda

𝑄

qiymatga mos keladi.

Biror qurlish fermasi ishlab chiqarayotgan mahsulotga bo’lgan talab funksiyasi

𝑃 =

300 − 4𝑄

formula bilan berilgan bo’lsin o’rtacha xarajat esa

𝑈𝑋 = 𝑄

2

− 76𝑄1120 +

7200

𝑄

formula bilan berilgan bo’lsa ( bunda

𝑃

birlik tavarning narxi,

𝑄

tavarning bir oylik soni).

a)

Tushum maksimum bo’ladigan ishlab chiqarish hajmi

b)

Limit xarajat minimum bo’ladigan ishlab chiqarish hajmi

c)

Foyda maksimum bo’ladigan ishlab chiqarish hajmi
1) Maksimum foyda nuqtasi talab elastikligini. Masalani yechimini topamiz:
a) Tushum

𝑇(𝑄) = 𝑃𝑄

ko’rinishida deb olsak, bu holda

𝑇(𝑄) = (300 − 4𝑄)𝑄 = 300𝑄 − 4𝑄

2

Stasionar nuqtalarini toppish uchun hosila olib, nolga tenglashtirib yechamiz.

𝑇

(𝑄) = 300 − 8𝑄 = 0, 𝑄

300

8

= 37,5

Tavar ishlab chiqarish hajmi

𝑄 = 37

bo’lganda tushum funksiyasi ekstremum bo’ladi,

tushum funksiyasi maksimum yoki minmumga ega bo’lishini topish uchun ikkinchi hosilani
olamiz.

𝑇

′′

(𝑄) = −8 < 0

Ikkinchi hosila manfiy bo’lgani uchun

𝑄 = 37,5

da tushum funksiyasi maksimumga

erishadi.

Demak,

𝑄 = 37,5

bo’lganda ferma tushum maksimumi bo’ladi.


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b) Limit harajat minimum bo’ladigan

𝑄

qning qiymatini topish uchun umumiy harajatni

topamiz

𝑋(𝑄) = 𝑈𝑋 ∙ 𝑄 = 𝑄

3

− 76𝑄

2

1256𝑄 + 7200

limit harajatni toppish uchun hosila olamiz

𝐿𝑋 = 𝑋

(𝑄) = 3𝑄

2

− 152𝑄 + 1256

Stasionar nuqtani toppish uchun buyndan hosila olamiz va uni nolga tenglashtirib

yechamiz

𝐿

= (3𝑄

2

− 152𝑄 + 1256)

= 6𝑄 − 152 = 0, 𝑄 = 25

Demak

𝐿𝑋

ekstirimum qiymati

𝑄 = 25

da qabul qiladi. Bu nuqtada maksimum yoki

minmumga erishishni toppish uchun yana bir marta hosila olamiz

(𝐿𝑋)

′′

= 6 > 0

Shunday qilib

𝑄 = 25

bo’lganda limitik harajat minmum qiymatini qabul qiladi.

c)

Foyda funksiyasini topamiz.

𝐹(𝑄) = 𝑇(𝑄) − 𝑋(𝑄) = 300𝑄 − 4𝑄

2

− 𝑄

3

+ 76𝑄

2

− 1260𝑄 − 7200

yoki

𝐹(𝑄) = −𝑄

3

− 72𝑄

2

− 960𝑄 − 7200

birinchi hosilasini olamiz

𝐹

(𝑄) = −3𝑄

2

+ 144𝑄 − 960

Stasionar nuqtalarini topamiz, buning uchun nolga tenglashtirib yechamiz

3𝑄

2

− 144𝑄 + 960 = 0, 𝑄 = 40 𝑣𝑎 𝑄 = 8

Foyda funksiyasi ikkita

𝑄 = 40 𝑣𝑎 𝑄 = 8

statsionar nuqtalarga ega, qaysida

maksimumga, qaysida minmumga ega bo’lishini toppish uchun ikkinchi tartibli hosilasini olamiz
va unga kritik nuqtalarni quyib tekshiramiz

𝐹

′′

(𝑄) = −6𝑄 + 144

𝐹

′′

(40) = −6 ∙ 40 + 144 = −240 + 144 = −96 < 0

𝐹

′′

(8) = −6 ∙ 8 + 144 = −48 + 144 = 96 > 0

Demak, foyda funksiyasi

𝑄 = 40

da maksimum qiymat qabul qiladi.

𝑃 = 300 − 4 ∙ 40 = 300 − 160 = 140 𝑠𝑜′𝑚

Shunday chiqib, 40 ta tavar 140 so’mdan sotilsa 5600 so’m foyda olinadi.
2) Talab elastikligini narxga nisbatan topamiz

𝑄

mahsulot

𝑃

esa uning narxi bo’lsa

mahsulot narxiga nisbatan talab elastikligi

𝐸(𝑄, 𝑃)

ni quydagiga topar edik

𝐸(𝑄, 𝑃) = 𝑄

(𝑃)

𝑃

𝑄

masalani shartidan

𝑃

= (300 − 4𝑄)

1 = −4𝑄

(𝑃) 𝑄

(𝑃) =

1
4

Demak talab elastikliga

𝐸(𝑃, 𝑄) = −

1
4

𝑃

𝑄

= −

300 − 4𝑄

𝑄

= 2 −

150

𝑄

Endi, qurilish funksiyasi maksimal foyda olgan

𝑄 = 40

nuqtada elastikligini topamiz

𝐸(𝑃, 40) = 2 −

150

40

= 2 − 3,25 = −1,25

da iborat bo’ladi.

Xulosa qilib aytganda funksiyaning hosilasi tushunchasi matematikaning eng ko’p

tadbiqlanadigan sohalardan biri. Bu yerda uning iqtisodiy masalalarga qisman tadbiqini ko’rib
o’tdik.

Jumladan ishlab chiqarishdagi tushum, o’rtacha tushum, limitik tushum ishlab

chiqarishdagi o’zgarmas xarajat, o’zgaruvchi xarajat, limitik xarajatga talab elastikliga


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foyda(daromad), maksimum foyda va boshqa shunga o’xshash masalalarni yechish mumkin. Bu
esa hozirgi zamonning dolzarb masalalaridan hisoblanadi.

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20.

Axmadovich, M. B. (2020). Sfera sirtida joylashgan uchburchaklarni yechishning ba'zi
usullari.

Science and Education

,

1

(2), 23-27.

21.

Usarov, S., Zikiryaev, S., Mardonov, B., & Namazov, G. (2024, May). Numerical analysis
of the process of heat transfer in inhomogeneous media. In

AIP Conference

Proceedings

(Vol. 3147, No. 1). AIP Publishing.

22.

Aхмадович М. Б. . (2024). Интерактивные Веб-Технологии Для Развития
Логического Мышления Инженеров Будущего В Условиях Цифровой
Трансформации Образования.

Miasto Przyszłości

,

52

, 755–761. Retrieved from

https://miastoprzyszlosci.com.pl/index.php/mp/article/view/4713

23.

Mardonov Baxodir Axmadovich. (2024). KELAJAKDAGI MUHANDISLARNI
RAQAMLI TA’LIM ASOSIDA O‘QITISH, SAMARALI VEB-KONTENT YARATISH
METODOLOGIYASI.

IJTIMOIY

FANLARDA

INNOVATSIYA

ONLAYN

ILMIY

JURNALI

,

4

(9),

42–45.

Retrieved

from

https://sciencebox.uz/index.php/jis/article/view/11916





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