Class 12 Maths · Chapter 12
Short answer:
Is chapter me Linear Programming Problem (LPP) formulate karna, graphical method se solve karna (bounded aur unbounded feasible region dono cases), aur Corner Point Method seekhte hain. Neeche 14 questions hain — LPP formulation word problems (manufacturing, diet, transportation-type), feasible region describe karna, corner points nikalna, aur objective function evaluate karke optimal (max/min) value find karna. Ek unbounded region example bhi hai jisme multiple optimal solutions milte hain (poore edge par).
Linear Programming ek technique hai jisse limited resources (time, material, money) ko is tarah allocate karte hain ki profit maximum ho ya cost minimum ho — bina kisi constraint ko todhe. Real life me factory production planning, diet planning, aur transportation cost minimize karne jaise problems isi tarah solve hote hain. Is chapter ka core skill hai: word problem ko equations me convert karna (formulation), phir un equations se bana feasible region graphically samajhna, aur uske corner points (vertices) par objective function check karke best answer nikalna.
Chapter 12 Summary — 5 Minute Revision
1. Linear Programming Problem (LPP) kya hai
LPP ek aisa problem hai jisme ek linear objective function (jaise profit ya cost) ko linear constraints (inequalities) ke under maximize ya minimize karna hota hai.
- Decision variables: unknowns jo hum decide karte hain — usually x aur y (jaise kitni units product A aur B banayein)
- Objective function: Z = ax + by — jise maximize ya minimize karna hai
- Constraints: linear inequalities jo resources ki limit batati hain (time, material, budget)
- Non-negativity restrictions: x ≥ 0, y ≥ 0 (kyunki quantity negative nahi ho sakti)
2. Mathematical Formulation — LPP banane ke 4 steps
- Decision variables identify karo (jise find karna hai — usually x, y)
- Objective function likho — Z = ax + by (maximize ya minimize)
- Har resource/condition ko ek inequality me convert karo (constraints)
- Non-negativity add karo: x ≥ 0, y ≥ 0
Word problem me keywords dekho: "at most", "not more than", "cannot exceed" → ≤ ; "at least", "not less than", "minimum" → ≥.
3. Graphical Method
- Har constraint ki boundary line plot karo (equality maan kar)
- Inequality ke direction ke hisaab se shade karo (origin test: (0,0) ko constraint me daal kar check karo satisfy hota hai ya nahi)
- Sab constraints ka common shaded region = feasible region
- Feasible region ke corners (vertices) = corner points
4. Bounded vs Unbounded Feasible Region
| Type | Meaning | Optimal solution |
|---|---|---|
| Bounded | Region chaaron taraf se closed hai (finite area) | Max aur Min dono guaranteed exist karte hain |
| Unbounded | Region kisi direction me infinite tak extend hoti hai | Min exist kar sakta hai (agar objective ke coefficients positive hain), lekin Max ke liye extra check zaroori hai — kabhi max exist hi nahi karta |
↔ Table ko side me swipe karein
5. Corner Point Theorem (Corner Point Method)
Corner Point Theorem kehta hai ki agar optimal solution exist karta hai, to wo feasible region ke kisi corner point (vertex) par hi milega — isliye har corner point par objective function evaluate karke compare karte hain.
- Feasible region ke saare corner points nikalo (constraint lines ko pairs me solve karke, simultaneous equations)
- Har corner point par Z = ax + by ka value calculate karo
- Bounded region: sabse bada value = Maximum, sabse chhota = Minimum
- Unbounded region: candidate max/min corner se milta hai, lekin verify karna padta hai ki open half-plane (jaise ax+by > M ya < m) feasible region se koi common point rakhta hai ya nahi — agar nahi rakhta, wahi actual optimal hai
6. Types of LPP word problems (common categories)
- Manufacturing/Production problem: limited raw material/machine time me profit maximize karna (kitne units A aur B banayein)
- Diet problem: minimum cost me required nutrients (vitamins, calories) fulfil karna
- Transportation-type problem: limited supply/demand ke saath cost minimize karna ya allocation optimize karna
In sab me formulation ka process same hai — sirf context alag hota hai.
7. Multiple Optimal Solutions
Kabhi kabhi objective function ki line, feasible region ke ek poore edge (side) ke parallel ho jaati hai — tab us edge ke saare points par Z ka same optimal value milta hai, sirf ek point par nahi.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q14)
Q1. Ek furniture manufacturer table aur chair banata hai. Ek table banane me 4 hours carpentry aur 2 hours painting lagti hai; ek chair banane me 2 hours carpentry aur 1 hour painting lagti hai. Total available time: carpentry 60 hours, painting 30 hours. Table se profit ₹30 aur chair se ₹20 hai. LPP formulate karo (solve mat karo).
Let x = number of tables, y = number of chairs banayi jayein.
Maximize Z = 30x + 20y
Subject to constraints:
4x + 2y \le 60 \quad \text{(carpentry)}
2x + y \le 30 \quad \text{(painting)}
x \ge 0,\ y \ge 0 \text{ (non-negativity)}
Note: dono constraints actually same line 2x+y ≤ 30 ban jaate hain (4x+2y≤60 ko 2 se divide karo), lekin formulation stage me dono alag likhna sahi hai jab tak simplify na karo.
Q2. Feasible region x + y ≤ 4, x ≥ 0, y ≥ 0 ke corner points nikalo aur Z = 3x + 4y ka maximum value find karo.
Constraint lines: x + y = 4, x-axis (y=0), y-axis (x=0).
\text{Corner points: } (0,0),\ (4,0),\ (0,4)
| Corner point | Z = 3x+4y |
|---|---|
| (0,0) | 0 |
| (4,0) | 12 |
| (0,4) | 16 |
↔ Table ko side me swipe karein
\text{Maximum } Z = 16 \text{ at } (0,4)
Region bounded hai (triangle), isliye maximum guaranteed exist karta hai.
Q3. Solve graphically: Minimize Z = 3x + 5y, subject to x + 2y ≥ 10, x + y ≥ 6, x ≥ 0, y ≥ 0.
Constraint lines ki intersection nikalo — x + 2y = 10 aur x + y = 6 ko simultaneously solve karo:
(x+2y) - (x+y) = 10 - 6 \Rightarrow y = 4,\ x = 2
Feasible region unbounded hai (dono constraints ≥ type). Corner points check karo — har constraint line ka woh intercept jo dusre constraint ko bhi satisfy kare:
\text{Corner points: } (10,0),\ (2,4),\ (0,6)
| Corner point | Z = 3x+5y |
|---|---|
| (10,0) | 30 |
| (2,4) | 6+20=26 |
| (0,6) | 30 |
↔ Table ko side me swipe karein
\text{Minimum } Z = 26 \text{ at } (2,4)
Verify: open half-plane 3x+5y < 26 ka feasible region se koi common point nahi hai, isliye 26 hi actual minimum hai.
Q4. Solve graphically: Maximize Z = 5x + 3y, subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
Dono lines ki intersection nikalo: 3x+5y=15 aur 5x+2y=10.
6x + 10y = 30 \ \ (\times 2) \quad;\quad 25x + 10y = 50\ \ (\times 5)
19x = 20 \Rightarrow x = \tfrac{20}{19},\quad y = \tfrac{45}{19}
Corner points (bounded region):
| Corner point | Z = 5x+3y |
|---|---|
| (0,0) | 0 |
| (2,0) | 10 |
| (0,3) | 9 |
| (20/19, 45/19) | (100+135)/19 = 235/19 ≈ 12.37 |
↔ Table ko side me swipe karein
\text{Maximum } Z = \tfrac{235}{19} \text{ at } \left(\tfrac{20}{19}, \tfrac{45}{19}\right)
Q5. (Diet problem) Ek diet me kam se kam 8 units Vitamin A aur 10 units Vitamin B chahiye. Food P ke ek unit me 2 units A aur 1 unit B hai, cost ₹6/unit. Food Q ke ek unit me 1 unit A aur 2 units B hai, cost ₹3/unit. LPP formulate karo minimum cost ke liye.
Let x = units of Food P, y = units of Food Q.
\text{Minimize } Z = 6x + 3y \text{ (total cost)}
Constraints:
2x + y \ge 8 \quad \text{(Vitamin A)}
x + 2y \ge 10 \quad \text{(Vitamin B)}
x \ge 0,\ y \ge 0
Yeh classic diet problem type LPP hai — objective ≥ constraints ke saath minimize karna.
Q6. Solve graphically: Minimize Z = x + 2y, subject to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0. Feasible region unbounded hai — dikhao.
Intersection of 2x+y=3 aur x+2y=6:
2(2x+y) - (x+2y) = 6 - 6 \Rightarrow 3x = 0 \Rightarrow x = 0,\ y = 3
Corner points: line x+2y=6 ka x-intercept (6,0) [check 2x+y≥3: 12≥3 ✓], aur (0,3) [dono lines yahan intersect karti hain].
| Corner point | Z = x+2y |
|---|---|
| (6,0) | 6 |
| (0,3) | 6 |
↔ Table ko side me swipe karein
\text{Minimum } Z = 6 \text{ — dono corner points par SAME value}
Isका matlab: segment (6,0) se (0,3) tak (line x+2y=6 ka poora edge) har point par Z=6 minimum milta hai — yeh multiple optimal solutions ka example hai. Verify: open half-plane x+2y < 6 feasible region ko touch nahi karta, isliye 6 hi actual minimum hai. Maximum yahan exist NAHI karta kyunki region unbounded hai aur x, y dono badhne par Z bhi infinitely badhta jayega.
Q7. (Manufacturing problem) Do machines M1 aur M2 available hain. Product x ko M1 par 3 hrs aur M2 par 1 hr chahiye. Product y ko M1 par 1 hr aur M2 par 2 hrs chahiye. M1 total 12 hrs, M2 total 10 hrs available hai. Profit x se ₹4, y se ₹5. LPP formulate karo.
Let x, y = units of product X aur Y.
\text{Maximize } Z = 4x + 5y
3x + y \le 12 \quad \text{(M1)}
x + 2y \le 10 \quad \text{(M2)}
x \ge 0,\ y \ge 0
Q8. Q7 ke LPP ko graphical method se solve karo — corner points aur maximum Z nikalo.
Intersection of 3x+y=12 aur x+2y=10:
y = 12 - 3x \text{ ko dusre me daalo: } x + 2(12-3x) = 10
x + 24 - 6x = 10 \Rightarrow -5x = -14 \Rightarrow x = \tfrac{14}{5} = 2.8
y = 12 - 3(2.8) = 12 - 8.4 = 3.6
Corner points (bounded region):
| Corner point | Z = 4x+5y |
|---|---|
| (0,0) | 0 |
| (4,0) | 16 |
| (0,5) | 25 |
| (2.8, 3.6) | 11.2+18=29.2 |
↔ Table ko side me swipe karein
\text{Maximum } Z = 29.2 \text{ at } (2.8, 3.6)
Check (0,5) valid hai kyunki 3(0)+5=5≤12 ✓, aur (4,0) valid kyunki 4+0=4≤10 ✓.
Q9. Corner points solve karne ka tareeka batao — constraint lines 2x + 3y = 12 aur x + y = 5 ka intersection point nikalo.
x + y = 5 se x = 5 - y. Isse pehli equation me substitute karo:
2(5-y) + 3y = 12 \Rightarrow 10 - 2y + 3y = 12 \Rightarrow y = 2
x = 5 - 2 = 3
Intersection point = (3, 2). Isko verify karo: 2(3)+3(2)=6+6=12 ✓ aur 3+2=5 ✓.
Q10. (Transportation-type problem) Ek company do godowns A aur B se do shops P aur Q tak saman bhejti hai. Godown A me 60 units aur B me 40 units stock hai. Shop P ko 40 units aur Q ko 60 units chahiye. A se P tak transport cost ₹6/unit, A se Q tak ₹3/unit. LPP formulate karo (allocation minimize cost ke liye) — sirf x units A se P tak bheje jaayein, is basis par formulate karo.
Let x = units A se P ko bheje jaayein. Baaki A se Q ko = 60−x. P ko total 40 chahiye, isliye B se P ko (40−x) units, aur B se Q ko 40−(40−x) = x units bhejni hongi (B ka total 40 units use hoga).
\text{Cost} = 6x + 3(60-x) + c_2(40-x) + c_3 x
(Yahan c₂, c₃ = B se P aur B se Q ki per-unit cost, jo problem me di gayi honi chahiye.) General form:
\text{Minimize } Z = ax + b(60-x) + c(40-x) + d\,x
0 \le x \le 40 \quad \text{(non-negativity + supply/demand limits)}
Is tarah transportation problem ek single-variable LPP me convert ho jaata hai jab supply-demand equal ho.
Q11. Determine karo ki following region bounded hai ya unbounded: x + y ≤ 10, x ≥ 2, y ≥ 2.
x ≥ 2 aur y ≥ 2 lower bounds dete hain, x+y ≤ 10 upper bound deta hai.
\text{Corner points: } (2,2),\ (8,2),\ (2,8)
Yeh region ek closed triangle hai (teeno sides finite lines se bane hain) — isliye bounded region hai.
Q12. (Land allocation problem) Ek farmer ke paas 50 acres zameen hai. Wheat me per acre profit ₹4000, rice me per acre profit ₹3000. Wheat ko per acre 2 labour-days, rice ko per acre 1 labour-day chahiye — total 80 labour-days available. LPP formulate karo.
Let x = acres wheat, y = acres rice.
\text{Maximize } Z = 4000x + 3000y
x + y \le 50 \quad \text{(land)}
2x + y \le 80 \quad \text{(labour)}
x \ge 0,\ y \ge 0
Q13. Q12 ke LPP ko solve karo.
Intersection of x+y=50 aur 2x+y=80:
(2x+y)-(x+y) = 80-50 \Rightarrow x = 30,\ y = 20
| Corner point | Z=4000x+3000y |
|---|---|
| (0,0) | 0 |
| (40,0) | 160000 |
| (0,50) | 150000 |
| (30,20) | 120000+60000=180000 |
↔ Table ko side me swipe karein
Check (40,0): 2(40)+0=80≤80 ✓, x+y=40≤50 ✓. Check (0,50): x+y=50≤50 ✓, labour=50≤80 ✓.
\text{Maximum profit } Z = ₹1{,}80{,}000 \text{ at } x=30,\ y=20
Q14. Feasible region ke corner points (0,0), (5,0), (3,4), (0,6) diye hain. Z = 2x + 3y ka maximum aur minimum dono nikalo.
| Corner point | Z=2x+3y |
|---|---|
| (0,0) | 0 |
| (5,0) | 10 |
| (3,4) | 6+12=18 |
| (0,6) | 18 |
↔ Table ko side me swipe karein
\text{Maximum } Z = 18 \text{ (at (3,4) aur (0,6) dono — multiple optimal solutions)}
\text{Minimum } Z = 0 \text{ at } (0,0)
Yaad rakho: sabhi CHAAR corner points check karna zaroori hai — sirf kuch points check karke jaldi answer mat do, warna galat max/min mil sakta hai.
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Standard LPP form | Maximize/Minimize Z = ax + by, subject to constraints (inequalities in x, y), and x ≥ 0, y ≥ 0 |
| Decision variables | Unknown quantities jo determine karni hain — usually x aur y |
| Objective function | Z = ax + by — jo maximize ya minimize karni hai |
| Constraints | Linear inequalities (≤, ≥, =) jo resource limits ya requirements represent karti hain |
| Non-negativity restriction | x ≥ 0, y ≥ 0 (quantities negative nahi ho sakti) |
| Feasible region | Sab constraints ka common (intersection) region — sab feasible solutions yahi hote hain |
| Feasible solution | Koi bhi (x,y) jo saari constraints satisfy kare |
| Optimal solution | Feasible solution jispe Z ka max ya min value ho |
| Corner Point Method — Step 1 | Feasible region ke saare corner points (vertices) nikalo — constraint lines ko pairs me simultaneously solve karke |
| Corner Point Method — Step 2 | Har corner point par Z = ax + by evaluate karo |
| Corner Point Method — Step 3 (bounded) | Sabse bada value = Maximum; sabse chhota = Minimum — dono guaranteed exist karte hain |
| Corner Point Method — Step 3 (unbounded) | Candidate max/min corner points se milta hai, par verify karo ki open half-plane ax+by > M (max ke liye) ya ax+by < m (min ke liye) ka feasible region se koi common point nahi hai |
| Unbounded region — Maximum | Kabhi exist nahi bhi kar sakta agar region ax+by ki direction me infinitely extend karta ho |
| Multiple optimal solutions | Agar do adjacent corner points par Z ka same value aaye, to unke beech ka poora edge optimal hota hai |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Non-negativity constraints (x ≥ 0, y ≥ 0) likhna bhool jaana. Har LPP formulation me yeh do conditions hamesha add karo — inke bina feasible region galat ban jaata hai aur negative quantities allowed ho jaati hain jo real life me possible nahi.
- Inequality ki direction galat le lena jab feasible region shade karo. Origin test use karo — (0,0) ko constraint me substitute karke check karo satisfy hota hai ya nahi, aur us according shade karo. 'At least' ka matlab ≥ hota hai, 'at most' ka matlab ≤ — inhe swap mat karo.
- Sirf kuch corner points check karke answer de dena, sab nahi. Corner Point Method me feasible region ke HAR corner point par Z evaluate karna zaroori hai — agar ek bhi corner point miss ho gaya to max/min galat aa sakta hai.
- Unbounded feasible region me directly bada value ko maximum maan lena bina verify kiye. Unbounded region me maximum kabhi exist hi nahi karta agar objective function ki value region me infinitely badh sakti hai — pehle check karo ki open half-plane feasible region ko intersect karta hai ya nahi.
- Simultaneous equations solve karte waqt arithmetic error karna. Do constraint lines ka intersection nikalte waqt elimination/substitution me sign galti ya calculation mistake corner point ko hi galat bana degi — jisse poora answer wrong ho jayega. Hamesha final point ko dono equations me daal kar verify karo.
- Word problem me decision variables galat define kar dena. x aur y ko clearly define karo (jaise 'x = number of tables', na ki 'x = carpentry hours') — galat variable definition se poori objective function aur constraints galat ban jaati hain.
Board-Style Important Questions
- 1 mark: LPP me 'feasible region' ki definition do.
- 2 marks: Corner Point Theorem ka statement likho.
- 3 marks: Ek LPP ka feasible region unbounded diya gaya hai — explain karo ki iske liye maximum aur minimum kab exist karte hain.
- 4 marks: Ek word problem diya jaayega (manufacturing/diet type) — usko LPP ke form me formulate karo (objective function + constraints + non-negativity).
- 5-6 marks: Graphical method se ek LPP solve karo — feasible region describe karo, corner points nikalo, aur optimal (max/min) value find karo.
Aksar Poochhe Jaane Wale Sawaal
Linear Programming Problem kya hota hai?
LPP ek aisa mathematical problem hai jisme kisi linear objective function (jaise profit ya cost) ko kuch linear constraints (inequalities) ke under maximize ya minimize karte hain, saath me x ≥ 0, y ≥ 0 ki non-negativity restriction bhi hoti hai.
Feasible region aur corner points me kya farak hai?
Feasible region woh poora area hai jahan saari constraints ek saath satisfy hoti hain. Corner points us region ke vertices (kone) hain — jaha do boundary lines aapas me milti hain. Corner Point Theorem ke according, optimal solution hamesha kisi corner point par hi milta hai.
Bounded aur unbounded feasible region me optimal solution kaise differ karta hai?
Bounded region me maximum aur minimum dono guaranteed exist karte hain. Unbounded region me minimum exist kar sakta hai (agar coefficients positive hain), lekin maximum kabhi exist nahi bhi kar sakta — isko verify karne ke liye open half-plane check karna padta hai.
Agar do corner points par Z ka value same aaye to kya hota hai?
Iska matlab optimal solution ek point par nahi balki un do corner points ko jodne waale poore edge (line segment) par milta hai — isko 'multiple optimal solutions' kehte hain.
LPP formulate karte waqt kaunse steps follow karne chahiye?
Pehle decision variables define karo (x, y), phir objective function likho (Z = ax+by), phir har condition ko inequality (constraint) me convert karo, aur last me non-negativity restriction x≥0, y≥0 add karo.
Diet problem aur manufacturing problem ke formulation me kya farak hai?
Dono ka process same hota hai — sirf context alag hai. Manufacturing problem me usually resources (time, material) ≤ limited hote hain aur profit maximize karte hain. Diet problem me nutrients ki requirement ≥ minimum hoti hai aur cost minimize karte hain.
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