Linear Programming: Class 12 Maths Chapter 12 Hinglish mein

Seedhi lines wali limits ke andar sabse best plan

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

60 min
1 simulation
14 practice questions
15 MCQs

Is chapter mein aap kya seekhenge

Optimisation problem pehchaano, aur woh features bhi jo use linear programming problem banate hain
Word problem ko decision variables, ek linear objective function aur linear constraints (non-negativity samet) mein badlo
Do variables wale linear constraints ka graph banao, uska feasible region shade karo aur uske corner points nikaalo
Bounded feasible region par linear objective ka maximum ya minimum nikaalne ke liye corner point method use karo
Feasible region unbounded ho to open half-plane test se tay karo ki optimum exist karta hai ya nahi
Aise problems pehchaano jinke infinitely many optimal solutions hote hain, aur aise jinka koi feasible solution nahi hota

NCERT Mathematics Class 12, Chapter 12, pp. 394–405 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

12.1 Introduction

Har asli plan ki kuch limits hoti hain. Dukaan ke paas utna hi paisa hota hai, kitchen ke paas utne hi ghante, tempo mein utni hi jagah. In limits ke andar bhi hum best plan chahte hain: sabse zyada profit, sabse chhota bill, sabse kam delivery time. Yeh chapter aise sawaalon ki ek important family ke liye ek saaf-suthra method deta hai.

Yeh raha woh problem jo hum poore chapter mein saath le chalenge. Diwali se pehle Anjali Jaipur mein ghar par do products banati hai: painted diyas ke boxes aur jar candles. Diyas ke ek box ke liye ½ kg wax aur 2 ghante ka kaam lagta hai; ek jar candle ke liye 1 kg wax aur 1 ghanta. Tyohaar se pehle uske paas 20 kg wax aur 50 working hours bache hain. Har box par woh ₹30 kamaati hai aur har jar par ₹40, aur jo bhi banati hai sab bik jaata hai. Use har ek ke kitne banane chahiye?

Ek samajhdaar pehla kadam hai kuch plans try karke unhe compare karna, jaise Fig. 12.1 karta hai. Sirf boxes: 25 boxes ke baad 50 ghante khatam ho jaate hain, jabki shelf par 40 boxes ka wax pada hai, to woh ₹750 kamaati hai. Sirf jars: 20 jars ke baad wax khatam aur woh ₹800 kamaati hai. 10 boxes aur 15 jars ka mix saara wax use karta hai par sirf 35 ghante, aur ₹900 kamaata hai, jo dono pure plans se zyada hai. Kaagaz par 24 boxes aur 6 jars ₹960 kamaate, lekin us plan ko 54 ghante chahiye, jo uske paas nahi hain.

3/410 boxes aur 15 jars: ₹900, dono se behtar.

Fig. 12.1Chaar plans dono limits ke against check kiye gaye. Mix karne se fayda hota hai, par ek plan time limit tod deta hai.

Do horizontal bars, ek wax ke liye aur ek time ke liye, dono ek dashed limit line ke against naape gaye hain, aur upar-neeche plan aur uska profit likha hai. Step 1, 25 boxes aur koi jar nahi: wax 20 mein se 12.5 kg, time 50 mein se 50 ghante, profit 750 rupees, time khatam ho jaata hai. Step 2, koi box nahi aur 20 jars: wax 20 mein se 20 kg, time 50 mein se 20 ghante, profit 800 rupees, wax khatam ho jaata hai. Step 3, 10 boxes aur 15 jars: wax 20 mein se 20 kg, time 50 mein se 35 ghante, profit 900 rupees, yaani mix zyada kamaata hai. Step 4, 24 boxes aur 6 jars: wax 20 mein se 18 kg, lekin time 50 mein se 54 ghante, isliye time bar limit ke paar ek hatched zone mein chala jaata hai aur plan possible nahi hai.

Isse teen baatein seekhne ko milti hain. Profit plan par depend karta hai; kuch plans koi limit tod dete hain aur woh available hi nahi hote; aur dono products ko mix karna sirf ek banane se behtar ho sakta hai. Phir bhi guess karna achhi strategy nahi hai. Sirf whole numbers gine to bhi Anjali ke paas 356 possible plans hain, aur lucky guess ke saath yeh proof nahi milta ki usse behtar kuch nahi hai.

Jo sawaal kuch limits ke andar kisi quantity ki sabse badi ya sabse chhoti value poochte hain, unhe optimisation problems kehte hain. Jab quantity aur saari limits unknowns mein linear hon, jaise Anjali ke case mein, to sawaal ek linear programming problem (रैखिक प्रोग्रामन) hota hai, short mein LPP. Aise problems wahan aate hain jahan bhi koi cheez kam ho: sabse kam cost mein ingredients blend karna, machines schedule karna, delivery vans ke routes banana, hospital shifts mein staff lagana.

Asli problems mein hazaaron unknowns ho sakte hain aur unhe computer solve karta hai. Is chapter mein hum do unknowns tak rahenge, jahan ek graph sab kuch ek saath dikha deta hai, aur har problem graphical method se solve karenge.

Quick check

3 mein se 1

Diyas ke 24 boxes aur 6 jar candles banana Anjali ka ek possible plan hai.

SahiGalat

Answer dekho

Answer Galat

Wax theek hai (12 kg + 6 kg = 18 kg), lekin time

48 + 6 = 54

ghante hai, jo uske 50 se zyada hai. Plan ko har limit maanni padti hai, isliye yeh plan bahar hai, chahe iska profit kitna bhi attractive lage.

12.2 Linear Programming Problem aur uska Mathematical Formulation

Jo plans humne try kiye, unhe phir se dekho. Do features saaf dikhte hain. Anjali ke paas choice hai: numbers ka har pair (boxes, jars) ek plan hai, aur alag plans alag amount kamaate hain. Aur yeh choice wax aur ghanton se ghiri hui hai, jinhe koi plan cross nahi kar sakta. Linear programming problem ki shape bilkul yahi hai: ek quantity jise jitna ho sake upar (ya neeche) le jaana hai, aur choices par seedhi lines wali fences ki ek list.

12.2.1 Problem ka mathematical formulation

Maan lo $x$ diyas ke boxes ki sankhya hai aur $y$ jar candles ki sankhya. Yahi quantities Anjali choose karti hai, isliye inhe decision variables kehte hain. Inmein se koi bhi negative nahi ho sakta:

non-negativity

x \geq 0, y \geq 0

(12.1)

Wax. Ek plan $\frac{1}{2} x + y$ kg wax use karta hai, aur yeh 20 kg se zyada nahi ho sakta. Dono sides ko double karne se fraction hat jaata hai:

wax

x + 2 y \leq 40

(12.2)

Time. Plan mein $2 x + y$ ghante lagte hain, aur sirf 50 available hain:

time

2 x + y \leq 50

(12.3)

Goal. ₹30 per box aur ₹40 per jar par, rupees mein profit hai

profit

Z = 30 x + 40 y

(12.4)

Saare pieces ikattha karo to poori kahaani chaar lines mein aa jaati hai. $Z = 30 x + 40 y$ ko maximise karo, in conditions ke saath

\begin{aligned} x + 2 y & \leq 40 \\ 2 x + y & \leq 50 \\ x, y & \geq 0 \end{aligned}

Definition

Linear programming problem (LPP)

Aisa problem jo $x \geq 0$ , $y \geq 0$ ke un saare choices mein, jo linear inequalities (ya equations) ki di gayi list ko satisfy karte hain, ek linear expression $Z = a x + b y$ ki sabse badi ya sabse chhoti value maangta hai.

ExampleAnjali ka problem: (12.1)–(12.3) ke saath $Z = 30 x + 40 y$ ko maximise karo.

Is description ke har hisse ka ek naam hai. Jis expression ko jitna ho sake bada ya chhota banana hai, woh objective function (उद्देश्य फलन) hai. Jo inequalities choices ko gherti hain, woh constraints (व्यवरोध) hain; khaas constraints $x \geq 0$ , $y \geq 0$ non-negativity restrictions hain.

Definition

Objective function

Woh quantity jise optimise karna hai, decision variables mein ek linear expression $Z = a x + b y$ ke roop mein likhi hui, jahan $a$ aur $b$ fixed numbers hain.

ExampleAnjali ka profit $Z = 30 x + 40 y$ , rupees mein.

Definition

Constraints

Decision variables par conditions, har ek linear inequality ya equation, jo har acceptable plan ko poori karni hoti hain.

ExampleWax $x + 2 y \leq 40$ , time $2 x + y \leq 50$ , aur $x \geq 0$ , $y \geq 0$ .

Har LPP ek optimisation problem hai, lekin ulta sach nahi hai. Fencing ki fixed length se sabse bada area $x y$ gherne ke liye rectangular pen ki sides chunna ek optimisation problem hai, par uska objective linear nahi hai, isliye is chapter ke methods us par apply nahi hote.

Step by step

Chaar kadam mein LPP formulate karna

Step 1 / 4

Unknowns ka naam rakho

Tay karo ki tumhare control mein kya hai aur har quantity ko uski unit ke saath ek letter do: " $x$ = banaye gaye boxes ki sankhya".

1 / 4

Example

Canteen ka budget, symbols mein

Chennai ke ek school ki canteen mid-morning snack mein sundal ki trays aur upma ki trays deti hai. Sundal ki ek tray 2 units protein deti hai aur ₹200 ki padti hai; upma ki ek tray 1 unit deti hai aur ₹50 ki padti hai. Snack mein kam se kam 12 units protein hona chahiye, aur stove subah zyada se zyada 10 trays paka sakta hai. Ise ek LPP ki tarah likho jiska aim sabse kam cost hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Surat ke ek tailor ke paas zyada se zyada 30 m kapda hai. Ek kurta mein 2 m lagta hai aur ek dupatta mein 1 m.

x

kurtas aur

y

dupattas ke saath, cloth constraint hai

x + 2 y \leq 30

2 x + y \geq 30

2 x + y \leq 30

2 x + y = 30

Answer dekho

Answer (C) $2 x + y \leq 30$

Laga kapda

2 x + y

metres hai aur "at most 30" ka matlab

\leq 30

hai.

x + 2 y

likhne se dono kapde aapas mein badal jaate hain,

\geq

kam se kam 30 m maangta, aur

=

har metre use karne par majboor karta.

12.2.2 Linear programming problems solve karne ka graphical method

Do decision variables ke saath har plan plane ka ek point $(x, y)$ hai. Class XI mein tumne dekha tha ki linear inequality straight line ki ek side rakhti hai, yaani ek half-plane. $x + 2 y \leq 40$ ka graph banane ke liye line $x + 2 y = 40$ ko uske intercepts (40, 0) aur (0, 20) se draw karo. Phir line se hatkar koi bhi point test karo, aam taur par origin: $0 \leq 40$ sach hai, isliye origin wali side rakhi jaati hai.

Har constraint ke liye yahi karke results ko overlap karne par Fig. 12.2 milta hai. Jo plans saare tests paas karte hain, woh milkar feasible region (सुसंगत क्षेत्र) banate hain. Anjali ke liye yeh quadrilateral $O P Q R$ hai. Teen corners axes par hain; chautha, $Q$ , woh jagah hai jahan dono lines katti hain. $x + 2 y = 40$ aur $2 x + y = 50$ ko saath mein solve karne par $x = 20$ , $y = 10$ milta hai.

4/4Overlap $O P Q R$ hi feasible region hai.

Fig. 12.2Har constraint ek half-plane rakhta hai; feasible region woh hai jo sabmein common hai.NCERT Fig 12.1 · p. 397

x aur y ke liye 0 se 50 tak axes. Step 1 poora first quadrant shade karta hai, jahan x aur y kam se kam 0 hain. Step 2 solid wax line x + 2y = 40 ko (40, 0) se (0, 20) tak draw karta hai aur origin wali side ko hatch karta hai. Step 3 dashed time line 2x + y = 50 ko (25, 0) se (0, 50) tak jodta hai aur uski origin wali side ko doosri direction mein hatch karta hai. Step 4 sirf overlap rakhta hai, yaani quadrilateral OPQR jiske corners O(0, 0), P(25, 0), Q(20, 10) aur R(0, 20) hain, feasible region ke roop mein shaded; Q woh jagah hai jahan dono lines milti hain.

Definition

Feasible region

LPP ke woh saare points jo har constraint test paas karte hain, $x \geq 0$ aur $y \geq 0$ samet. Plane ka baaki hissa infeasible region hai.

ExampleAnjali ka quadrilateral $O P Q R$ .

Definition

Feasible solution

Aisa plan jo har constraint poora kare, yaani feasible region ka ek point; boundary points bhi gine jaate hain. Jo plan ek bhi constraint tod de, woh infeasible solution hai.

Example(10, 15) edge $Q R$ par hai aur feasible hai; (24, 6) time limit todta hai aur infeasible hai.

Feasible solutions mein se hum aisa chahte hain jo objective ko uski best value de: ek optimal solution. Use dhoondhne se pehle feasible aur infeasible plans ko alag pehchaanne ki practice karo.

Try karke dekho

Possible hai ya nahi?

Tumhe chahiye

pencil aur paper

Pehle guess karo

Calculate karne se pehle: tumhe kaun sa plan koi limit todta lagta hai?

22 boxes aur 5 jars12 boxes aur 15 jars8 boxes aur 16 jars

Har plan ke liye $x + 2 y$ nikaalo aur use 40 se compare karo.
$2 x + y$ nikaalo aur use 50 se compare karo.
Agar dono tests paas hon, to plan possible hai: uska profit $30 x + 40 y$ nikaalo.

Table 12.1

Plan $(x, y)$	Possible?
(22, 5)
(12, 15)
(8, 16)
(21, 9)

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 12.3

SimulationAnjali ke corners ke against ek plan test karo

Dono axes par minus 5 se 30 tak ek coordinate plane, jismein chaar fixed points O(0, 0), P(25, 0), Q(20, 10) aur R(0, 20) hain, jo Anjali ke feasible region ke corners hain, aur ek draggable point X hai jo (10, 15) se shuru hota hai aur whole numbers par snap hota hai. X ke chalne par uske coordinates dikhte hain, taaki learner wax aur time constraints check kar sake aur dekh sake ki X quadrilateral OPQR ke andar hai ya nahi.

Fig. 12.3O, P, Q aur R feasible region ke fixed corners hain. Plan X ko drag karo aur uske coordinates padho; jab woh

O P Q R

ke andar ya uske edges par ho, tab woh feasible hai.

Try karo

X ko (21, 9) par drag karo. Yeh edge $P Q$ ke paas lagta hai: kya $2 x + y \leq 50$ sach hai?
Edge $Q R$ par, jahan $x + 2 y = 40$ hai, teen whole-number points dhoondho.
X ko (15, 15) par rakho. Kaun sa constraint fail hota hai, aur kitne se?

Best plan kahan rehta hai

Feasible region mein infinitely many points hain, isliye hum unhe ek-ek karke test nahi kar sakte. Ek picture batati hai ki kahan dekhna hai. Ek profit fix karo, maan lo ₹800. Jo plans exactly itna kamaate hain, woh $30 x + 40 y = 800$ satisfy karte hain, jo ek straight line hai. Target badlo to line apne hi parallel khisakti hai, kyunki uska slope, $- \frac{3}{4}$ , target par depend nahi karta. Barabar profit wali lines ko iso-profit lines kehte hain (jab $Z$ cost ho to iso-cost lines).

3/4 $Z = 1000$ : line region ko sirf $Q (20, 10)$ par chhooti hai.

Fig. 12.4Profit badhaane se line bahar ki taraf khisakti hai; aakhri contact point corner

Q

hai.

Anjali ka feasible region OPQR jiske corners O(0, 0), P(25, 0), Q(20, 10) aur R(0, 20) hain, aur ek arrow jo upar aur right ki taraf point karta hai, yaani woh direction jismein profit badhta hai. Ek dashed profit line apne hi parallel khisakti hai aur peeche halki copies chhodti hai. Step 1: 400 rupee wali line region ko cross karti hai. Step 2: 800 rupee wali line abhi bhi use cross karti hai aur R se guzarti hai. Step 3: 1000 rupee wali line region ko sirf corner Q(20, 10) par chhooti hai, jo maximum ke roop mein highlighted hai. Step 4: 1200 rupee wali line region ko poori tarah miss kar deti hai, isliye koi plan itna nahi kamaata.

Line ko bahar ki taraf dhakelne se profit badhta hai. Yeh region ko tab tak kaatti rehti hai jab tak $Z = 1000$ na ho jaaye, jahan yeh region ko sirf ek point $Q (20, 10)$ par chhooti hai; usse oonchi koi bhi line region ko poori tarah miss karti hai. To Anjali ka best plan hai 20 boxes aur 10 jars, jo ₹1000 kamaata hai. Khisakti line jis aakhri point ko chhooti hai, woh hamesha ek corner hota hai, ya agar line kisi edge ke saath pade, to uske do corners samet woh poora edge. Yahi picture neeche ke do theorems ka idea hai, jinhe hum bina proof ke use karte hain.

InteractiveFeasible-region explorer

Is chapter ke chhe problems ka ek interactive graph: Anjali ki candles, canteen, tie example, unbounded example, recharge vouchers aur tiffin service. Feasible region shaded hai, uske corner points apne coordinates ke saath marked hain, aur har constraint line drawn hai. Ek dashed objective line ax + by = Z ko gol handle se drag kiya ja sakta hai ya slider se khisakaya ja sakta hai; ek status panel batata hai ki line region ko cross karti hai, use sirf ek corner par ya ek edge ke saath chhooti hai (ek optimum), ya use miss karti hai. Do aur sliders a aur b badalte hain, jisse line tilt hoti hai aur best corner khisakta hai. Ek button har corner par Z ki value list karta hai. Unbounded regions ke liye panel samjhaata hai ki Z kab bina limit ke badh ya ghat sakta hai, aur tiffin problem ke liye batata hai ki koi feasible region nahi hai. Jab tak learner kuch na kare, kuch nahi hilta.

Fig. 12.5Is chapter ka koi problem chuno, phir objective line ko drag karo (ya slider use karo) aur dekho ki woh region ko aakhri baar kahan chhooti hai.

a

aur

b

sliders se objective ko tilt karo aur best corner badalta dekho.

Definition

Corner point

Feasible region ka ek vertex: region ka woh point jahan uski do boundary lines katti hain.

ExampleAnjali ke region ke liye $O$ , $P$ , $Q$ aur $R$ .

Theorem 1Optimum ek corner par hota hai

Maan lo objective $Z = a x + b y$ wale ek LPP ka feasible region $R$ ek convex polygon hai. Jab bhi $Z$ , $R$ par sabse badi ya sabse chhoti value tak pahunchta hai, woh yeh value $R$ ke kisi corner point par zaroor leta hai (shayad doosre points par bhi).

Proof ka idea. Maan lo best value sirf $R$ ke andar ke kisi point par mile. Us point se $Z$ badhaane wali direction mein ek chhota kadam $R$ ke andar hi rehta hai aur badi value deta hai, jo impossible hai. Isliye best value boundary par milti hai.

Kisi bhi edge ke saath $Z$ constant rate se badalta hai, kyunki edge ek straight segment hai aur $Z$ linear hai. To edge ke saath $Z$ ek end par sabse bada, doosre par sabse chhota, ya constant hota hai. Dono hi tarah koi na koi corner edge ke kisi bhi point jitna achha ya usse behtar hota hai.

Definition

Bounded region

Aisa feasible region jo kisi circle ke andar fit ho jaaye. Jo region kisi direction mein bina ant ke failta jaaye, woh unbounded hai.

ExampleAnjali ka region bounded hai; region $x + y \geq 3$ , $x, y \geq 0$ unbounded hai.

3/3Bada circle bhi kaam nahi aata: region unbounded hai.

Fig. 12.6Bounded ka matlab "kisi circle ke andar fit ho jaaye"; unbounded region har circle se bahar nikal jaata hai.

Do panels. Left: Anjali ka quadrilateral feasible region, uske chaaron taraf ek dashed circle, jis par bounded, circle mein fit likha hai. Step 1 mein sirf yeh panel poora hai. Step 2 mein right panel ek unbounded wedge-shaped region dikhata hai jo do lines ke beech upar-right ki taraf khulta jaata hai, ek arrow dikhata hai ki yeh aage chalta rehta hai, aur ek dashed circle sirf uska kuch hissa cover karta hai, jis par unbounded, circle mein nahi likha hai. Step 3 mein circle bada hota hai lekin region phir bhi uske bahar nikla rehta hai.

Theorem 2Bounded regions mein dono hamesha hote hain

Jab feasible region $R$ bounded ho, to objective $Z = a x + b y$ , $R$ par sabse badi aur sabse chhoti value zaroor leta hai, aur dono mein se har ek $R$ ke kisi corner point par milti hai.

Proof ka idea. Bounded convex polygon ka har point uske corners ka weighted average hota hai, jahan non-negative weights ka sum 1 hai. Kyunki $Z$ linear hai, us point par uski value corners par uski values ka wahi weighted average hoti hai.

Average kabhi bhi average kiye gaye numbers mein sabse bade se zyada nahi ho sakta, na sabse chhote se kam. Isliye sabse badi corner value $R$ par maximum hai, aur sabse chhoti corner value minimum.

Corner point method

Yeh theorems ek kabhi na khatam hone wali search ko ek chhoti si table mein badal dete hain: corners list karo, har ek par $Z$ nikaalo, aur answer padh lo. Anjali ke liye:

Table 12.2Har corner par Anjali ka profitNCERT corner values ki table · p. 398

Corner	$Z = 30 x + 40 y$ (₹)
$O (0, 0)$	0
$P (25, 0)$	750
$Q (20, 10)$	1000 (maximum)
$R (0, 20)$	800

Region bounded hai, isliye Theorem 2 se sabse badi entry maximum hai. Anjali ko diyas ke 20 boxes aur 10 jar candles banane chahiye, ₹1000 ke profit ke liye. Dhyaan do ki yeh plan saara wax ( $20 + 2 \times 10 = 40$ ) aur saare ghante ( $2 \times 20 + 10 = 50$ ) use karta hai. Sabse chhoti entry, $O$ par ₹0, minimum hai, ek aise sawaal ka sahi jawaab jo kisi ne poocha hi nahi: kuch mat banao, kuch mat kamao.

Step 4 kaam kyun karta hai? Open half-plane $a x + b y > M$ bilkul wahi jagah hai jahan $Z$ , $M$ ko peeche chhod deta hai. Agar usmein koi feasible point nahi hai, to corner value $M$ ko koi nahi harata, isliye $M$ maximum hai. Agar koi feasible point usmein hai, to koi maximum exist nahi kar sakta: Theorem 1 se maximum kisi corner par hota aur isliye zyada se zyada $M$ hota, jabki woh point pehle hi usse behtar hai. $m$ ka test bhi aise hi kaam karta hai.

Example 1NCERT Example 1 · p. 399

Sivakasi ki ek print shop

Ek chhoti press greeting cards ke batches aur wall calendars ke batches print karti hai. Dono mein se kisi bhi ek batch mein paper ka ek roll lagta hai, aur stock mein 8 rolls hain. Card batch ko press par 3 ghante chahiye aur calendar batch ko 1 ghanta; press 18 ghante free hai. Card batch ₹2500 kamaata hai aur calendar batch ₹600. $x$ card batches aur $y$ calendar batches ke saath, sabse zyada kamaane wala plan nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Profit line $Z = 15 000$ region ko sirf $L (6, 0)$ par chhooti hai.

Fig. 12.7Best corner kisi axis par bhi ho sakta hai: yahan shop ek bhi calendar print nahi karti.NCERT Fig 12.2 · p. 400

0 se 10 tak axes. Solid paper line x + y = 8 aur dashed press line 3x + y = 18 shaded region OLMN ko gherti hain, jiske corners O(0, 0), L(6, 0), M(5, 3) aur N(0, 8) hain. Step 1 sirf region aur uske corners dikhata hai. Step 2 har corner par rupees mein profit list karta hai: L 15 000, M 14 300, N 4800, O 0. Step 3 steep dashed profit line 2500x + 600y = 15 000 draw karta hai jo region ko sirf L par chhooti hai, jo highlighted hai, saath mein note best: L.

Example 2NCERT Example 2 · p. 400

Sabse kam cost par canteen

Formulation example mein banaya gaya canteen problem solve karo: $Z = 200 x + 50 y$ ko minimise karo, in conditions ke saath

\begin{aligned} 2 x + y & \geq 12 \\ x + y & \leq 10 \\ x, y & \geq 0 \end{aligned}

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Triangle $U V W$ ; cost line $Z = 800$ ise $W (2, 8)$ par chhooti hai.

Fig. 12.8Aisa region jismein origin nahi hai: sabse kam cost

W

par hai.NCERT Fig 12.3 · p. 400

0 se 12 tak axes. Step 1 solid protein line 2x + y = 12 draw karta hai aur origin se door wali side hatch karta hai, saath mein note ki O test mein fail hota hai. Step 2 dashed stove line x + y = 10 jodta hai aur uski origin wali side hatch karta hai. Step 3 sirf overlap dikhata hai, yaani triangle jiske corners U(6, 0), V(10, 0) aur W(2, 8) hain, costs 1200, 2000 aur 800 rupees list karta hai, aur dashed cost line 200x + 50y = 800 draw karta hai jo triangle ko W par chhooti hai, jo least: W ke saath highlighted hai.

Try karke dekho

Ruler ko profit line banao

Tumhe chahiye

graph paper
ek ruler aur ek set square
Scale par draw kiya gaya Anjali ka region $O P Q R$

Pehle guess karo

Agar har box ₹50 kamaata aur har jar sirf ₹20, to kaun sa corner sabse zyada profit deta?

P (25, 0)

Q (20, 10)

R (0, 20)

Rates ke pehle pair ke liye koi ek iso-profit line draw karo, jaise $30 x + 40 y = 600$ jo (20, 0) aur (0, 15) se guzarti hai.
Set square ko uske saath lagakar rakho aur ruler ko parallel rakhte hue $O$ se door khisakao.
$O P Q R$ ka woh aakhri corner note karo jise ruler chhoota hai, aur wahan profit nikaalo.
Rates ke baaki do pairs ke liye dohrao.

Table 12.3

Rates (box, jar)	Aakhri corner
₹30, ₹40
₹50, ₹20
₹10, ₹40

Tumhari entries apne aap save hoti hain.

Quick check

3 mein se 1

Corner point method ke steps sahi order mein lagao.

Sabse badi ya sabse chhoti value chuno
Feasible region draw karo
Har corner point nikaalo
Har corner par Z nikaalo
Unbounded ho to half-plane test karo

Answer dekho

Answer

Feasible region draw karo
Har corner point nikaalo
Har corner par Z nikaalo
Sabse badi ya sabse chhoti value chuno
Unbounded ho to half-plane test karo

Pehle region, phir uske corners, phir values ki table. Half-plane test sabse aakhir mein aata hai, aur sirf tab jab region unbounded ho, kyunki woh table se mili candidate value ko check karta hai.

Ties: kai optimal solutions

Example 3NCERT Example 3 · p. 401

Do corners mein shared minimum

$Z = 4 x + 2 y$ ka minimum aur maximum nikaalo, in conditions ke saath

\begin{aligned} 2 x + y & \geq 8 \\ x + y & \leq 10 \\ x & \leq 6 \\ x, y & \geq 0 \end{aligned}

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/3Line $Z = 16$ edge $E A$ ke saath padi hai: $E A$ ka har point optimal hai.

Fig. 12.9Jab objective kisi edge ke parallel ho, to poora edge optimal ho sakta hai.NCERT Fig 12.4 · p. 401

0 se 12 tak axes. Solid line 2x + y = 8, dashed line x + y = 10 aur dotted vertical line x = 6 ek shaded paanch-sided region ko gherti hain, jiske corners A(4, 0), B(6, 0), C(6, 4), D(0, 10) aur E(0, 8) hain. Ek panel corners par Z = 4x + 2y list karta hai: 16, 24, 32, 20 aur 16. Step 1 region aur values dikhata hai. Step 2 poore edge EA ko ek moti line se highlight karta hai aur note karta hai ki objective line 2x + y = 8 ke saath padi hai, isliye minimum 16 poore EA par milta hai. Step 3 dashed objective line 4x + 2y = 32 draw karta hai jo region ko sirf C par chhooti hai, jo C pe max 32 ke saath highlighted hai.

Unbounded feasible regions

Jab region infinity tak chala jaata hai, to corner values ki table kahaani ka ant nahi hai. Method ka step 4 tay karta hai ki sabse achhi dikhne wali corner value sach mein optimal hai ya nahi.

Example 4NCERT Example 4 · p. 402

Koi minimum hi nahi

Graph se $Z = 2 x - 3 y$ ki minimum value nikaalo, agar koi ho, in conditions ke saath

\begin{aligned} x + 2 y & \geq 6 \\ x & \leq 6 \\ x - y & \leq 3 \\ x, y & \geq 0 \end{aligned}

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Half-plane $2 x - 3 y < - 9$ ke kuch points region mein bhi hain: koi minimum nahi.

Fig. 12.10Half-plane

Z < - 9

abhi bhi region se milta hai, isliye koi minimum nahi hai.NCERT Fig 12.5 · p. 402

0 se 12 tak axes. Dotted line x + 2y = 6, solid line x − y = 3 aur dashed vertical line x = 6 ek shaded strip ko gherti hain jo bina ant ke upar chalti jaati hai, saath mein upar ki taraf ek arrow jis par aage likha hai. Iske corners (0, 3), (4, 1) aur (6, 3) hain. Step 1 region dikhata hai. Step 2 corners par Z = 2x − 3y list karta hai: −9, 5 aur 3. Step 3 dashed line 2x − 3y = −9 ko (0, 3) se draw karta hai aur uske upar wala half-plane hatch karta hai; strip ka ek bada hissa us half-plane ke andar hai, outline kiya hua, aur feasible point (2, 6) Z = −14 ke saath marked hai. Isliye −9 minimum nahi hai.

Test hamesha fail nahi hota. Agle problem mein region phir se unbounded hai, phir bhi minimum exist karta hai, jaisa sensible prices wale cost problem mein aam taur par hota hai.

Example

Jab half-plane test paas ho jaaye

Pune ke ek office ko is mahine apne field staff ke liye kam se kam 12 GB extra data aur kam se kam 1200 minutes extra calls chahiye. Voucher A ₹50 mein 3 GB aur 100 minutes deta hai; voucher B ₹40 mein 1 GB aur 300 minutes deta hai. Sabse kam kharch ke liye use har ek ke kitne kharidne chahiye?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Half-plane $50 x + 40 y < 270$ ka region ke saath koi common point nahi hai: ₹270 minimum hai.

Fig. 12.11Half-plane

Z < 270

region ko miss karta hai, isliye

E

par ₹270 sabse kam cost hai.

0 se 14 tak axes. Solid line 3x + y = 12 aur dashed line x + 3y = 12 ek shaded region ko gherti hain jo origin se door hai aur upar-right ki taraf chala jaata hai. Iske corners (0, 12), E(3, 3) aur (12, 0) hain. Step 1 region dikhata hai. Step 2 costs 480, 270 aur 600 rupees list karta hai. Step 3 dashed cost line 50x + 40y = 270 ko E se draw karta hai aur uske neeche, origin ki side wala half-plane hatch karta hai; yeh region ko sirf E par chhoota hai, isliye koi common point nahi hai, aur panel par least: 270 rupees likha hai.

Jab koi plan kaam na kare

Example 5NCERT Example 5 · p. 403

Ek promise jo boxes nibha nahi sakte

Mumbai ki ek tiffin service ne ek client se roz kam se kam 50 tiffins ka promise kiya hai, $x$ regular aur $y$ large. Regular tiffins steel boxes mein jaate hain, aur roz sirf 20 saaf boxes taiyaar hote hain; large tiffins three-tier carriers mein jaate hain, aur woh 25 hain. ₹40 per regular aur ₹60 per large tiffin par sabse kam cost wala plan nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Dono ko koi satisfy nahi karta: problem ka koi feasible solution nahi hai.

Fig. 12.12Dono constraints kabhi overlap nahi karte, isliye koi feasible region nahi hai.NCERT Fig 12.6 · p. 403

0 se 60 tak axes. Step 1 solid line x + y = 50 draw karta hai aur origin se door wali side hatch karta hai, jis par orders likha hai. Step 2 dashed edges x = 20 aur y = 25 jodta hai aur axes ke saath bane rectangle ko hatch karta hai, jis par boxes likha hai; uska door wala corner (20, 25) marked hai. Step 3 rectangle aur line ke beech ki khaali band ko shade karta hai, us par gap likhta hai, aur note jodta hai ki dono nahi milte, max 45 tiffin hi aate hain.

In saare examples mein do patterns baar-baar dikhe. Pehla, humne jo bhi feasible region draw kiya woh convex (उत्तल) tha: apne kisi bhi do points ke saath usmein unhe jodne wala poora segment bhi hota hai, jaise Fig. 12.13 dikhata hai. Aisa hamesha hota hai, kyunki har half-plane convex hai aur convex sets ka overlap bhi convex hota hai. Doosra, optimum, agar exist kare, to ek corner par hota hai; jab do corners mein tie ho, to woh unke beech ke edge par fail jaata hai.

2/3Ek notch segment $P^{'} Q^{'}$ ko region se bahar jaane deta hai.

Fig. 12.13Feasible regions half-planes ke overlaps hain, isliye woh hamesha convex hote hain.

Step 1: ek paanch-sided convex region jiske andar do points P aur Q ek segment se jude hain jo andar hi rehta hai, us par convex likha hai. Step 2: uske bagal mein ek region jiski left side mein ek triangular notch kata hai; uske do points P′ aur Q′ ko jodne wala segment notch ko cross karta hai, aur region ke bahar wala hissa dashed draw kiya gaya hai, us par non-convex likha hai. Step 3: notch wali shape ki jagah note aata hai ki 5 half-planes ka overlap convex hi hai, aur dashed lines convex region ke paanchon edges ko aage badhaati hain, jo dikhata hai ki yeh paanch half-planes ka overlap hai, isiliye har feasible region convex hota hai.

Answers ki kismein

3 mein se 1

Har situation ko usse milne wali baat se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

Region bounded hai : Max aur min dono exist karte hain
Best ke liye do corners mein tie : Unke beech ka har point optimal hai
Constraints kabhi overlap nahi karte : Koi feasible solution nahi
Unbounded; ek feasible point $M$ ko peeche chhodta hai : Koi maximum nahi

Yeh hain Theorem 2, ties wala remark, infeasible case (Example 5) aur corner point method ka step 4 (Example 4).

Practice · 14 questions

Practice questions answers ke saath (NCERT “Practice 12.1”)

Short answerEasyNCERT Exercise 12.1, Q1
$Z = 2 x + 5 y$ ko maximise karo, jahan $2 x + y \leq 6$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMaximum 30, (0, 6) par.
Line $2 x + y = 6$ axes ko (3, 0) aur (0, 6) par milti hai, aur origin wali side rakhi jaati hai. Region woh triangle hai jiske corners (0, 0), (3, 0) aur (0, 6) hain, jo bounded hai.
Table 12.4
Corner $Z$
(0, 0) 0
(3, 0) 6
(0, 6) 30
Maximum 30 hai, (0, 6) par.
Short answerMediumNCERT Exercise 12.1, Q2
$Z = - 2 x + 5 y$ ko minimise karo, jahan $x + y \leq 7$ , $2 x + y \leq 10$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMinimum −10, (5, 0) par.
$x + y = 7$ axes ko (7, 0) aur (0, 7) par milti hai; $2 x + y = 10$ (5, 0) aur (0, 10) par. Dono ki origin wali side rakhi jaati hai.
Lines wahan katti hain jahan $x = 3$ aur $y = 4$ (doosri mein se pehli ghataao). Bounded region ke corners (0, 0), (5, 0), (3, 4) aur (0, 7) hain.
Table 12.5
Corner $Z$
(0, 0) 0
(5, 0) −10
(3, 4) 14
(0, 7) 35
Minimum −10 hai, (5, 0) par.
Short answerMediumNCERT Exercise 12.1, Q3
$Z = 3 x + 4 y$ ko maximise karo, jahan $2 x + 3 y \leq 12$ , $3 x + y \leq 9$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMaximum $\frac{117}{7}$ , $(\frac{15}{7}, \frac{18}{7})$ par.
$2 x + 3 y = 12$ axes ko (6, 0) aur (0, 4) par milti hai; $3 x + y = 9$ (3, 0) aur (0, 9) par. Dono ki origin wali side rakhi jaati hai.
$3 x + y = 9$ se $y = 9 - 3 x$ . Phir $2 x + 27 - 9 x = 12$ se $x = \frac{15}{7}$ aur $y = \frac{18}{7}$ milta hai.
Table 12.6
Corner $Z$
(0, 0) 0
(3, 0) 9
$(\frac{15}{7}, \frac{18}{7})$ $\frac{𝟏𝟏𝟕}{𝟕}$
(0, 4) 16
Region bounded hai. Maximum $\frac{117}{7} \approx 16.7$ hai, jo (0, 4) wale 16 se bas thoda zyada hai, isiliye exact corner nikaalna zaroori hai.
Short answerHardNCERT Exercise 12.1, Q4
$Z = 2 x + 3 y$ ko minimise karo, jahan $x + 2 y \geq 4$ , $3 x + y \geq 6$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMinimum $\frac{34}{5}$ , $(\frac{8}{5}, \frac{6}{5})$ par.
Dono constraints $O$ se door wali side rakhte hain, isliye region unbounded hai. $x + 2 y = 4$ axes ko (4, 0) aur (0, 2) par milti hai; $3 x + y = 6$ (2, 0) aur (0, 6) par.
Corners: (0, 6), (4, 0), aur crossing point. $3 x + y = 6$ se $y = 6 - 3 x$ , isliye $x + 12 - 6 x = 4$ aur $x = \frac{8}{5}$ , $y = \frac{6}{5}$ .
Table 12.7
Corner $Z$
(0, 6) 18
$(\frac{8}{5}, \frac{6}{5})$ $\frac{𝟑𝟒}{𝟓}$
(4, 0) 8
Open half-plane $2 x + 3 y < \frac{34}{5}$ test karo. Iska edge crossing corner se guzarta hai, slope $- \frac{2}{3}$ ke saath, jo dono constraint lines ke slopes $- 3$ aur $- \frac{1}{2}$ ke beech hai, isliye yeh region ko sirf wahin chhoota hai. Half-plane ka region ke saath koi common point nahi hai, aur minimum $\frac{34}{5}$ hai.
Short answerMediumNCERT Exercise 12.1, Q5
$Z = 4 x + 5 y$ ko maximise karo, jahan $2 x + y \leq 14$ , $x + 3 y \leq 12$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMaximum 34, (6, 2) par.
$2 x + y = 14$ axes ko (7, 0) aur (0, 14) par milti hai; $x + 3 y = 12$ (12, 0) aur (0, 4) par. Dono ki origin wali side rakhi jaati hai.
$x = 12 - 3 y$ se: $24 - 6 y + y = 14$ , isliye $y = 2$ aur $x = 6$ . Bounded region ke corners (0, 0), (7, 0), (6, 2) aur (0, 4) hain.
Table 12.8
Corner $Z$
(0, 0) 0
(7, 0) 28
(6, 2) 34
(0, 4) 20
Maximum 34 hai, (6, 2) par.
Short answerMediumNCERT Exercise 12.1, Q6
$Z = 2 x + 3 y$ ko minimise karo, jahan $2 x + 3 y \geq 12$ , $2 x + y \geq 8$ , $x \geq 0$ , $y \geq 0$ . Dikhao ki infinitely many points minimum dete hain.
Solution dikhao
AnswerMinimum 12, (3, 2) aur (6, 0) ko jodne wale segment ke har point par.
Dono constraints $O$ se door wali side rakhte hain; region unbounded hai. $2 x + 3 y = 12$ axes ko (6, 0) aur (0, 4) par milti hai; $2 x + y = 8$ (4, 0) aur (0, 8) par. Yeh (3, 2) par katti hain, kyunki ghataane par $2 y = 4$ milta hai.
Table 12.9
Corner $Z$
(0, 8) 24
(3, 2) 12
(6, 0) 12
Open half-plane $2 x + 3 y < 12$ pehla constraint todta hai, isliye koi feasible point usmein nahi hai: 12 minimum hai.
Yeh (3, 2) aur (6, 0) par milta hai, aur isliye unhe jodne wale edge ke har point par, jaise (4.5, 1). Yaani infinitely many points.
Short answerHardNCERT Exercise 12.1, Q7
$Z = 2 x + 6 y$ ka minimum aur maximum nikaalo, jahan $x + 3 y \leq 30$ , $x + y \geq 12$ , $x - 3 y \geq 0$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMinimum 24, (12, 0) par; maximum 60, (15, 5) aur (30, 0) ko jodne wale segment ke har point par.
$x + 3 y = 30$ axes ko (30, 0) aur (0, 10) par milti hai, origin wali side rakhi. $x + y = 12$ unhe (12, 0) aur (0, 12) par milti hai, $O$ se door wali side rakhi. Line $x = 3 y$ $O$ se guzarti hai; (3, 0) test karne se pata chalta hai ki region iske neeche hai.
Corners: $x = 3 y$ , $x + y = 12$ se (9, 3) par aur $x + 3 y = 30$ se (15, 5) par milti hai; $x$ -axis se (12, 0) aur (30, 0) milte hain. Region bounded hai.
Table 12.10
Corner $Z$
(9, 3) 36
(12, 0) 24
(30, 0) 60
(15, 5) 60
Minimum 24, (12, 0) par. Maximum 60 (30, 0) aur (15, 5) par aata hai, isliye unhe jodne wale segment ke har point par: $x + 3 y = 30$ ke saath $Z = 2 (x + 3 y)$ constant hai.
Short answerHardNCERT Exercise 12.1, Q8
$Z = 2 x + y$ ka minimum aur maximum nikaalo, jahan $2 x + y \geq 20$ , $3 x - y \leq 0$ , $x + y \leq 60$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerMinimum 20, (0, 20) aur (4, 12) ko jodne wale segment ke har point par; maximum 75, (15, 45) par.
$2 x + y = 20$ : $O$ se door wali side. $x + y = 60$ : origin wali side. $y = 3 x$ $O$ se guzarti hai; (0, 1) test karne se pata chalta hai ki region iske upar hai.
Corners: $y$ -axis par (0, 20) aur (0, 60); $y = 3 x$ , $2 x + y = 20$ se (4, 12) par aur $x + y = 60$ se (15, 45) par milti hai. Region bounded hai.
Table 12.11
Corner $Z$
(0, 20) 20
(4, 12) 20
(15, 45) 75
(0, 60) 60
Maximum 75, (15, 45) par. Minimum 20, (0, 20) aur (4, 12) par, isliye unhe jodne wale poore segment ke saath, jo $2 x + y = 20$ par hai.
Short answerHardNCERT Exercise 12.1, Q9
$Z = - x + 3 y$ ko maximise karo, jahan $x \geq 1$ , $2 x + y \geq 8$ , $x + 2 y \geq 10$ , $y \geq 0$ .
Solution dikhao
AnswerKoi maximum nahi hai.
Region unbounded hai. Iske corners hain line $x = 1$ par (1, 6), (2, 4) jahan $2 x + y = 8$ , $x + 2 y = 10$ se milti hai, aur (10, 0).
Table 12.12
Corner $Z$
(1, 6) 17
(2, 4) 10
(10, 0) −10
Sabse badi corner value $M = 17$ hai. Open half-plane $- x + 3 y > 17$ test karo: point (1, 10) feasible hai ( $x = 1$ , $12 \geq 8$ , $21 \geq 10$ ) aur $Z = 29 > 17$ deta hai.
Half-plane region se milta hai; line $x = 1$ par upar jaate hue $Z = 3 y - 1$ bina limit ke badhta hai. Isliye $Z$ ka koi maximum nahi hai.
Short answerMediumNCERT Exercise 12.1, Q10
$Z = x + 3 y$ ko maximise karo, jahan $- 2 x + y \geq 3$ , $2 x - y \geq 1$ , $x \geq 0$ , $y \geq 0$ .
Solution dikhao
AnswerKoi feasible region nahi, isliye koi maximum nahi.
Pehla constraint kehta hai $y - 2 x \geq 3$ aur doosra kehta hai $y - 2 x \leq - 1$ .
Koi bhi point dono ko satisfy nahi kar sakta, kyunki uske liye $3 \leq y - 2 x \leq - 1$ chahiye hoga. Graph par dono half-planes parallel lines $y = 2 x + 3$ aur $y = 2 x - 1$ ki door wali sides par hain.
Koi feasible region nahi hai, isliye problem ka koi feasible solution nahi hai aur $Z$ ka koi maximum nahi hai.
11
Long answerHard
Alappuzha ka ek self-help group coir doormats aur coir baskets bunta hai. Ek doormat mein 2 kg fibre aur 1 ghanta bunai lagti hai; ek basket mein 1 kg aur 2 ghante. Har hafte group ke paas 36 kg fibre aur 48 ghante hote hain, aur uska buyer zyada se zyada 15 doormats leta hai. Ek doormat ₹60 kamaata hai aur ek basket ₹50.
1. Problem ko ek LPP ki tarah formulate karo.
  Solution dikhao
  Answer $Z = 60 x + 50 y$ ko maximise karo, jahan $2 x + y \leq 36$ , $x + 2 y \leq 48$ , $x \leq 15$ , $x, y \geq 0$ .
  Maan lo $x$ doormats ki sankhya hai aur $y$ baskets ki sankhya. Rupees mein profit $Z = 60 x + 50 y$ hai, jise maximise karna hai.
  Fibre se $2 x + y \leq 36$ milta hai aur bunai ke ghanton se $x + 2 y \leq 48$ . Buyer $x \leq 15$ jodta hai, aur counts se $x, y \geq 0$ bhi milta hai.
2. Ise corner point method se solve karo.
  Solution dikhao
  Answer8 doormats aur 20 baskets; maximum ₹1480.
  Corners: (0, 0), (15, 0), (15, 6) jahan $x = 15$ , $2 x + y = 36$ se milti hai, (8, 20) jahan $2 x + y = 36$ , $x + 2 y = 48$ se milti hai, aur (0, 24).
  Table 12.13
  Corner $Z$ (₹)
  (0, 0) 0
  (15, 0) 900
  (15, 6) 1200
  (8, 20) 1480
  (0, 24) 1200
  Region bounded hai, isliye maximum ₹1480 hai, 8 doormats aur 20 baskets ke saath. 15 doormats wali buyer limit yahan binding nahi hai.
Quick answerEasy
Constraints $x + y \leq 6$ , $x \leq 4$ , $x, y \geq 0$ ke liye, kaun sa point feasible region ka corner hai?
(2, 2)(6, 0)(4, 6)(4, 2)
Answer dekho
Answer (D) (4, 2)
(4, 2) dono $x = 4$ aur $x + y = 6$ par hai. (6, 0) $x \leq 4$ todta hai, (4, 6) $x + y \leq 6$ todta hai, aur (2, 2) feasible hai lekin region ke andar hai, corner nahi.
Quick answerEasy
Agar kisi LPP ka feasible region unbounded ho, to objective function ka maximum nahi ho sakta.
SahiGalat
Answer dekho
Answer Galat
Unbounded region mein optimum sirf shayad na ho. Jaise, voucher region par $Z = - x - y$ ka maximum $- 6$ hai, (3, 3) par: half-plane $- x - y > - 6$ ka matlab $x + y < 6$ hai, jo region ko miss karta hai.
Quick answerMedium
Maan lo Anjali ke profits badal kar ₹45 per box aur ₹30 per jar ho jaate hain, wax aur time limits wahi rehti hain. Uska naya maximum profit kitna hai, rupees mein?
rupees
Answer dekho
Answer 1200 rupees
$Z = 45 x + 30 y$ $O$ par 0, $P$ par 1125, $Q$ par $900 + 300 = 1200$ aur $R$ par 600 deta hai. Maximum abhi bhi $Q (20, 10)$ par hai: ₹1200.

Corner	$Z$
(0, 0)	0
(3, 0)	6
(0, 6)	30

Corner	$Z$
(0, 0)	0
(5, 0)	−10
(3, 4)	14
(0, 7)	35

Corner	$Z$
(0, 0)	0
(3, 0)	9
$(\frac{15}{7}, \frac{18}{7})$	$\frac{𝟏𝟏𝟕}{𝟕}$
(0, 4)	16

Corner	$Z$
(0, 6)	18
$(\frac{8}{5}, \frac{6}{5})$	$\frac{𝟑𝟒}{𝟓}$
(4, 0)	8

Corner	$Z$
(0, 0)	0
(7, 0)	28
(6, 2)	34
(0, 4)	20

Corner	$Z$
(0, 8)	24
(3, 2)	12
(6, 0)	12

Corner	$Z$
(9, 3)	36
(12, 0)	24
(30, 0)	60
(15, 5)	60

Corner	$Z$
(0, 20)	20
(4, 12)	20
(15, 45)	75
(0, 60)	60

Corner	$Z$
(1, 6)	17
(2, 4)	10
(10, 0)	−10

Corner	$Z$ (₹)
(0, 0)	0
(15, 0)	900
(15, 6)	1200
(8, 20)	1480
(0, 24)	1200

Recap

Summary aur key points

Optimisation problem limits ke andar kisi quantity ki sabse badi ya sabse chhoti value poochta hai; LPP aisa problem hai jismein quantity aur har limit linear hoti hai (Section 12.1).
LPP formulate karne ke liye decision variables ka naam rakho, objective $Z = a x + b y$ likho, har limit ko linear constraint mein translate karo, aur $x, y \geq 0$ jodo (Section 12.2.1).
Har constraint ek half-plane rakhta hai; jo points sabko satisfy karte hain woh feasible region banate hain, boundary samet (Section 12.2.2).
Iso-profit (ya iso-cost) lines $a x + b y = k$ parallel hoti hain; unhe khisakane se dikhta hai ki optimum ek corner par milta hai (Best plan kahan rehta hai).
Theorem 1: agar feasible region par $Z$ ki koi optimal value hai, to woh ek corner point par milti hai (Theorem 1).
Theorem 2: bounded feasible region par $Z$ hamesha maximum aur minimum leta hai, dono kisi corner par (Theorem 2).
Corner point method: corners nikaalo, har ek par $Z$ nikaalo, sabse bada $M$ aur sabse chhota $m$ lo (method).
Unbounded region par $M$ maximum tabhi hai jab $a x + b y > M$ region ko miss kare, aur $m$ minimum tabhi jab $a x + b y < m$ use miss kare (Example 4).
Agar optimum ke liye do corners mein tie ho, to unhe jodne wale segment ka har point optimal hai (Example 3).
Jo constraints kabhi overlap nahi karte, unse na koi feasible region bachta hai aur na koi solution (Example 5).
Feasible regions hamesha convex hote hain, kyunki woh half-planes ke overlaps hain (Jab koi plan kaam na kare).

History seNCERT Historical Note · p. 404

Wartime planning se har delivery van tak

Linear programming bahut practical chintaon se nikla. 1939 mein Soviet mathematician Leonid Kantorovich ne dikhaya ki plywood factories mein production ko linear models se kaise plan karein, aur 1941 mein American mathematician Frank Hitchcock ne kai sources se kai destinations tak saamaan bhejne ke sabse saste tareeke ko study kiya, yaani transportation problem.

Second World War ke dauraan aur uske thik baad, military planners ke saamne supplies, transport aur costs ke bade-bade sawaal the, aur yeh subject tezi se aage badha. 1945 mein economist George Stigler ne nutritional needs ki ek list poori karne wali sabse sasti diet ke baare mein poocha. 1947 mein George Dantzig ne simplex method invent kiya, jo feasible region ke ek corner se doosre corner tak chalta hai, har baar $Z$ ko behtar karta hai, jab tak koi behtar corner na bache: yeh is chapter ka hi corner point idea hai, hazaaron variables ke liye efficient banaya hua.

Is kaam ke liye Kantorovich ne Tjalling Koopmans ke saath 1975 ka Nobel Memorial Prize in Economic Sciences share kiya. 1984 mein Indian mathematician Narendra Karmarkar ne, jo tab Bell Labs mein the, ek fast interior-point method publish kiya jo region ke edges ke saath chalne ki jagah uske andar se hokar jaata hai. Aaj airlines, railways, refineries aur online shops roz linear programmes solve karte hain.

Flash cards · 7

Key ideas

1 / 7

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 12.1, Q1	Practice Q1
NCERT Exercise 12.1, Q2	Practice Q2
NCERT Exercise 12.1, Q3	Practice Q3
NCERT Exercise 12.1, Q4	Practice Q4
NCERT Exercise 12.1, Q5	Practice Q5
NCERT Exercise 12.1, Q6	Practice Q6
NCERT Exercise 12.1, Q7	Practice Q7
NCERT Exercise 12.1, Q8	Practice Q8
NCERT Exercise 12.1, Q9	Practice Q9
NCERT Exercise 12.1, Q10	Practice Q10

Quick check: MCQs

12.1 Introduction3 questions
12.2.1 Problem ka mathematical formulation3 questions
12.2.2 Linear programming problems solve karne ka graphical method3 questions
Corner point method3 questions
Answers ki kismein3 questions

Glossary · 14

Key terms (English + Hindi)

bounded regionपरिबद्ध क्षेत्रparibaddh kshetra: Aisa feasible region jo kisi circle ke andar fit ho jaaye; jo bina ant ke chalta jaaye woh unbounded hai.

constraintव्यवरोधvyavarodh: Ek linear inequality ya equation jo LPP mein decision variables ki choices ko limit karti hai.Chapter mein pehli baar yahan aaya →
convex regionउत्तल क्षेत्रuttal kshetra: Aisa region jismein uske kisi bhi do points ko jodne wala poora segment ho.Chapter mein pehli baar yahan aaya →
corner pointकोणीय बिंदुkoniya bindu: Feasible region ka woh point jahan uski do boundary lines milti hain.

decision variablesनिर्णायक चरnirnayak char: LPP mein hum jo unknowns choose karte hain, jaise do products ke banaye jaane wale numbers $x$ aur $y$ .Chapter mein pehli baar yahan aaya →

feasible regionसुसंगत क्षेत्रsusangat kshetra: LPP ke saare constraints ko satisfy karne wale points ka set, non-negativity samet.Chapter mein pehli baar yahan aaya →
feasible solutionसुसंगत हलsusangat hal: Feasible region ka koi bhi point, uske andar ya uski boundary par.

half-planeअर्धतलardhatal: Kisi straight line ki ek side par plane ka hissa; $x$ aur $y$ mein ek linear inequality aisa hi hissa describe karti hai.Chapter mein pehli baar yahan aaya →

iso-profit line: Ek line $a x + b y = k$ jiske saath objective ki value constant $k$ rehti hai; aisi lines parallel hoti hain.Chapter mein pehli baar yahan aaya →

linear programming problemरैखिक प्रोग्रामन समस्याrekhik programan samasya: Aisa optimisation problem jiska objective function aur saare constraints linear hon, aur decision variables non-negative hon.Chapter mein pehli baar yahan aaya →

non-negativity restrictionsऋणेतर व्यवरोधrinetar vyavarodh: Constraints $x \geq 0$ , $y \geq 0$ .Chapter mein pehli baar yahan aaya →

objective functionउद्देश्य फलनuddeshya phalan: Linear function $Z = a x + b y$ jise maximise ya minimise karna hai.Chapter mein pehli baar yahan aaya →
optimal solutionइष्टतम हलishtatam hal: Aisa feasible solution jahan objective function apni maximum ya minimum value leta hai.Chapter mein pehli baar yahan aaya →
optimisation problemइष्टतमीकरण समस्याishtatamikaran samasya: Aisa problem jo limits ke andar kisi quantity, jaise profit ya cost, ki sabse badi ya sabse chhoti value poochta hai.Chapter mein pehli baar yahan aaya →

12.1 Introduction

12.2 Linear Programming Problem aur uska Mathematical Formulation

12.2.1 Problem ka mathematical formulation

12.2.2 Linear programming problems solve karne ka graphical method

Best plan kahan rehta hai

Corner point method

Ties: kai optimal solutions

Unbounded feasible regions

Jab koi plan kaam na kare

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye