Class 12 Mathematics - NCERT Mathematics Part II
NCERT Solutions for Class 12 Maths Chapter 12: Linear Programming
Chapter 12 introduces Linear Programming (LP) - a method for optimizing (maximizing or minimizing) a linear objective function subject to linear constraints. Students learn to formulate LP problems from word problems, identify the feasible region graphically, find corner points, and determine optimal solutions. Applications include manufacturing optimization, diet problems, and transportation problems. This chapter carries 5 marks in the CBSE Class 12 board exam 2027 as an independent unit.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 12 Linear Programming has 2 exercises plus a Miscellaneous Exercise. Exercise 12.1 covers basic graphical solutions, Exercise 12.2 covers word problems on manufacturing, diet, and transportation. Only the graphical method is in the syllabus.
Table of Contents
12.1Exercise 12.1 - Graphical Method of Solving LPP
Q1Maximise Z = 3x + 4y subject to the constraints: x + y <= 4, x >= 0, y >= 0.
Constraints: x + y <= 4, x >= 0, y >= 0
Feasible region: Triangle with vertices at O(0,0), A(4,0), B(0,4).
Corner point evaluation:
| Corner Point | Z = 3x + 4y |
|---|---|
| O(0, 0) | 0 |
| A(4, 0) | 12 |
| B(0, 4) | 16 (Maximum) |
Maximum value of Z = 16 at B(0, 4)
Q2Minimise Z = -3x + 4y subject to: x + 2y <= 8, 3x + 2y <= 12, x >= 0, y >= 0.
Solving x + 2y = 8 and 3x + 2y = 12: Subtracting gives 2x = 4, x = 2, y = 3. So intersection is (2, 3).
Corner points: O(0,0), A(4,0), B(2,3), C(0,4)
Corner point evaluation:
| Corner Point | Z = -3x + 4y |
|---|---|
| O(0, 0) | 0 |
| A(4, 0) | -12 (Minimum) |
| B(2, 3) | 6 |
| C(0, 4) | 16 |
Minimum value of Z = -12 at A(4, 0)
Q3Maximise Z = 5x + 3y subject to: 3x + 5y <= 15, 5x + 2y <= 10, x >= 0, y >= 0.
Solving 3x + 5y = 15 and 5x + 2y = 10:
From second: y = (10-5x)/2. Substituting: 3x + 5(10-5x)/2 = 15
6x + 50 - 25x = 30 → -19x = -20 → x = 20/19, y = (10-100/19)/2 = (90/19)/2 = 45/19
Corner points: O(0,0), A(2,0), B(20/19, 45/19), C(0,3)
| Corner Point | Z = 5x + 3y |
|---|---|
| O(0, 0) | 0 |
| A(2, 0) | 10 |
| B(20/19, 45/19) | 100/19 + 135/19 = 235/19 ≈ 12.37 |
| C(0, 3) | 9 |
Maximum value of Z = 235/19 at B(20/19, 45/19)
Q4Minimise Z = 3x + 5y such that: x + 3y >= 3, x + y >= 2, x >= 0, y >= 0.
The feasible region is unbounded (above/right of the constraints).
Solving x + 3y = 3 and x + y = 2: Subtracting gives 2y = 1, y = 1/2, x = 3/2.
Corner points: A(3, 0), B(3/2, 1/2), C(0, 2)
| Corner Point | Z = 3x + 5y |
|---|---|
| A(3, 0) | 9 |
| B(3/2, 1/2) | 9/2 + 5/2 = 7 |
| C(0, 2) | 10 |
Smallest value is 7 at B(3/2, 1/2). Since the region is unbounded, we check if 3x + 5y < 7 has any point in common with the feasible region.
The open half-plane 3x + 5y < 7 does NOT intersect the feasible region.
Minimum value of Z = 7 at B(3/2, 1/2)
Q5Maximise Z = 3x + 2y subject to: x + 2y <= 10, 3x + y <= 15, x >= 0, y >= 0.
Solving x + 2y = 10 and 3x + y = 15: From first, x = 10-2y. Substituting: 30-6y+y = 15 → -5y = -15 → y = 3, x = 4.
Corner points: O(0,0), A(5,0), B(4,3), C(0,5)
| Corner Point | Z = 3x + 2y |
|---|---|
| O(0, 0) | 0 |
| A(5, 0) | 15 |
| B(4, 3) | 18 (Maximum) |
| C(0, 5) | 10 |
Maximum value of Z = 18 at B(4, 3)
Q6Minimise Z = x + 2y subject to: 2x + y >= 3, x + 2y >= 6, x >= 0, y >= 0.
Solving 2x + y = 3 and x + 2y = 6: From first, y = 3-2x. Substituting: x + 6-4x = 6 → -3x = 0 → x = 0, y = 3.
But at x=0, y=3: check 2(0)+3=3 >= 3 ✓ and 0+6=6 >= 6 ✓.
The feasible region is unbounded.
Corner points: A(6, 0), B(0, 3), and we check if B satisfies both: 2(0)+3=3 ✓, 0+6=6 ✓.
Also intersection is at (0, 3).
| Corner Point | Z = x + 2y |
|---|---|
| A(6, 0) | 6 |
| B(0, 3) | 6 |
Z = 6 at both A and B. Since the line x + 2y = 6 coincides with the constraint x + 2y >= 6, every point on the segment AB gives Z = 6.
Checking: the half-plane x + 2y < 6 has no common point with the feasible region.
Minimum value of Z = 6, attained at every point on the line segment joining (6,0) and (0,3).
Q7Minimise and Maximise Z = 5x + 10y subject to: x + 2y <= 120, x + y >= 60, x - 2y >= 0, x >= 0, y >= 0.
Finding corner points:
- x+2y=120 and x+y=60: y=60, x=0. But x-2y >= 0 requires x >= 2y. At (0,60): 0 >= 120 is false. Not feasible.
- x+2y=120 and x=2y: 2y+2y=120, y=30, x=60. Point (60,30).
- x+y=60 and x=2y: 2y+y=60, y=20, x=40. Point (40,20).
- x+2y=120 and y=0: x=120. Check x+y >= 60: 120 >= 60 ✓ and x >= 2y: 120 >= 0 ✓. Point (120,0).
- x+y=60 and y=0: x=60. Check x >= 2y: 60 >= 0 ✓. Point (60,0).
Corner points: A(60,0), B(120,0), C(60,30), D(40,20)
| Corner Point | Z = 5x + 10y |
|---|---|
| A(60, 0) | 300 |
| B(120, 0) | 600 (Maximum) |
| C(60, 30) | 600 (Maximum) |
| D(40, 20) | 400 |
Minimum Z = 300 at A(60, 0)
Maximum Z = 600 at every point on segment BC (from (120,0) to (60,30))
Q8Minimise and Maximise Z = x + 2y subject to: x + 2y >= 100, 2x - y <= 0, 2x + y <= 200, x >= 0, y >= 0.
Finding corner points:
- x+2y=100 and 2x-y=0 (y=2x): x+4x=100, x=20, y=40. Point (20,40).
- x+2y=100 and 2x+y=200: From first x=100-2y. Sub: 200-4y+y=200 → -3y=0 → y=0, x=100. But 2x-y <= 0: 200 <= 0 false. Not feasible.
- 2x-y=0 and 2x+y=200: 4x=200, x=50, y=100. Point (50,100).
- x+2y=100 and y=0: x=100. Check 2(100)-0=200 > 0. Not feasible (need 2x-y <= 0).
- 2x+y=200 and y=0: x=100. Same issue.
- x=0: from x+2y >= 100: y >= 50. From 2x+y <= 200: y <= 200. So (0,50) and (0,200).
Corner points: A(0, 50), B(20, 40), C(50, 100), D(0, 200)
| Corner Point | Z = x + 2y |
|---|---|
| A(0, 50) | 100 |
| B(20, 40) | 100 |
| C(50, 100) | 250 |
| D(0, 200) | 400 (Maximum) |
Minimum Z = 100 at every point on segment AB
Maximum Z = 400 at D(0, 200)
Q9Maximise Z = -x + 2y, subject to: x >= 3, x + y >= 5, x + 2y >= 6, y >= 0.
The feasible region is unbounded (extends to the right and upward).
Corner points:
- x=3 and x+y=5: y=2. Point (3,2).
- x=3 and x+2y=6: 2y=3, y=3/2. Point (3, 3/2). But check x+y >= 5: 3+3/2=9/2 < 5. Not feasible.
- x+y=5 and x+2y=6: Subtracting: y=1, x=4. Point (4,1). Check x >= 3: 4 >= 3 ✓.
- x+y=5 and y=0: x=5. Check x >= 3 ✓ and x+2y >= 6: 5 >= 6 false. Not feasible.
- x+2y=6 and y=0: x=6. Check x >= 3 ✓ and x+y >= 5: 6 >= 5 ✓. Point (6,0).
Corner points: A(6,0), B(4,1), C(3,2)
| Corner Point | Z = -x + 2y |
|---|---|
| A(6, 0) | -6 |
| B(4, 1) | -2 |
| C(3, 2) | 1 |
Since the region is unbounded upward, Z = -x+2y can increase without bound (take x=3, y→infinity: Z→infinity).
Z has no maximum value. Minimum value = -6 at A(6,0) needs unbounded check - but since the region extends with increasing y, Z can also go to -infinity along x→infinity (x-axis). Actually for y=0, Z=-x which goes to -infinity.
So Z has no maximum and no finite minimum in the unbounded region.
Q10Maximise Z = x + y, subject to: x - y <= -1, -x + y <= 0, x >= 0, y >= 0.
Constraint 1: x - y <= -1, i.e., y >= x + 1
Constraint 2: -x + y <= 0, i.e., y <= x
We need y >= x + 1 AND y <= x simultaneously.
But y >= x + 1 > x, so y > x and y <= x cannot both hold.
There is no feasible region. The constraints are inconsistent.
Since there is no feasible solution, Z has no maximum value.
12.2Exercise 12.2 - Linear Programming Word Problems
Q1Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of Vitamin A and 5 units/kg of Vitamin B while Food Q contains 4 units/kg of Vitamin A and 2 units/kg of Vitamin B. Determine the minimum cost of the mixture.
Let x kg of food P and y kg of food Q be mixed.
Objective: Minimise Z = 60x + 80y
Constraints:
3x + 4y >= 8 (Vitamin A)
5x + 2y >= 11 (Vitamin B)
x >= 0, y >= 0
Corner points:
- 3x+4y=8 and y=0: x=8/3. Point (8/3, 0).
- 5x+2y=11 and y=0: x=11/5. Point (11/5, 0). Check 3(11/5)=33/5=6.6 < 8. Not feasible.
- 3x+4y=8 and 5x+2y=11: Multiply second by 2: 10x+4y=22. Subtract first: 7x=14, x=2, y=1/2.
- 5x+2y=11 and x=0: y=11/2. Point (0, 11/2).
- 3x+4y=8 and x=0: y=2. Check 5(0)+2(2)=4 < 11. Not feasible.
Feasible corner points: A(0, 11/2), B(2, 1/2), C(8/3, 0)
| Corner Point | Z = 60x + 80y |
|---|---|
| A(0, 11/2) | 440 |
| B(2, 1/2) | 160 (Minimum) |
| C(8/3, 0) | 160 |
Z = 160 at both B and C. Minimum cost = Rs 160, achieved at any point on the segment from (2, 1/2) to (8/3, 0).
Q2One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Let x = number of cakes of kind 1, y = number of cakes of kind 2.
Objective: Maximise Z = x + y
Constraints:
200x + 100y <= 5000 → 2x + y <= 50 (flour in grams)
25x + 50y <= 1000 → x + 2y <= 40 (fat in grams)
x >= 0, y >= 0
Corner points:
- 2x+y=50 and x+2y=40: From first y=50-2x. Sub: x+100-4x=40 → -3x=-60 → x=20, y=10.
Corner points: O(0,0), A(25,0), B(20,10), C(0,20)
| Corner Point | Z = x + y |
|---|---|
| O(0, 0) | 0 |
| A(25, 0) | 25 |
| B(20, 10) | 30 (Maximum) |
| C(0, 20) | 20 |
Maximum number of cakes = 30 (20 of kind 1 and 10 of kind 2)
Q3A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman's time. A cricket bat takes 3 hours of machine time and 1 hour of craftsman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman's time. If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
Let x = tennis rackets, y = cricket bats.
Objective: Maximise Z = 20x + 10y (profit)
Constraints:
1.5x + 3y <= 42 → x + 2y <= 28 (machine time)
3x + y <= 24 (craftsman time)
x >= 0, y >= 0
Corner points:
- x+2y=28 and 3x+y=24: From first x=28-2y. Sub: 84-6y+y=24 → -5y=-60 → y=12, x=4.
Corner points: O(0,0), A(8,0), B(4,12), C(0,14)
| Corner Point | Z = 20x + 10y |
|---|---|
| O(0, 0) | 0 |
| A(8, 0) | 160 |
| B(4, 12) | 200 (Maximum) |
| C(0, 14) | 140 |
Maximum profit = Rs 200 when 4 rackets and 12 bats are produced.
Q4A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
Let x = packages of nuts, y = packages of bolts.
Objective: Maximise Z = 17.5x + 7y
Constraints:
x + 3y <= 12 (machine A)
3x + y <= 12 (machine B)
x >= 0, y >= 0
Corner points:
- x+3y=12 and 3x+y=12: From first x=12-3y. Sub: 36-9y+y=12 → -8y=-24 → y=3, x=3.
Corner points: O(0,0), A(4,0), B(3,3), C(0,4)
| Corner Point | Z = 17.5x + 7y |
|---|---|
| O(0, 0) | 0 |
| A(4, 0) | 70 |
| B(3, 3) | 73.5 (Maximum) |
| C(0, 4) | 28 |
Maximum profit = Rs 73.50 when 3 packages of nuts and 3 packages of bolts are produced.
Q5A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on the hand operated machine to manufacture a package of screws A, while it takes 6 minutes on the automatic and 3 minutes on the hand operated machine to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Let x = packages of A, y = packages of B. (4 hours = 240 minutes)
Objective: Maximise Z = 7x + 10y
Constraints:
4x + 6y <= 240 → 2x + 3y <= 120 (automatic machine)
6x + 3y <= 240 → 2x + y <= 80 (hand operated)
x >= 0, y >= 0
Corner points:
- 2x+3y=120 and 2x+y=80: Subtracting: 2y=40, y=20, x=30.
Corner points: O(0,0), A(40,0), B(30,20), C(0,40)
| Corner Point | Z = 7x + 10y |
|---|---|
| O(0, 0) | 0 |
| A(40, 0) | 280 |
| B(30, 20) | 410 (Maximum) |
| C(0, 40) | 400 |
Maximum profit = Rs 410 when 30 packages of A and 20 packages of B are produced.
Q6A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Let x = lamps, y = shades.
Objective: Maximise Z = 5x + 3y
Constraints:
2x + y <= 12 (grinding/cutting)
3x + 2y <= 20 (sprayer)
x >= 0, y >= 0
Corner points:
- 2x+y=12 and 3x+2y=20: From first y=12-2x. Sub: 3x+24-4x=20 → -x=-4 → x=4, y=4.
Corner points: O(0,0), A(6,0), B(4,4), C(0,10)
| Corner Point | Z = 5x + 3y |
|---|---|
| O(0, 0) | 0 |
| A(6, 0) | 30 (Maximum) |
| B(4, 4) | 32 (Maximum) |
| C(0, 10) | 30 |
Maximum profit = Rs 32 when 4 lamps and 4 shades are produced.
Q7A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?
Let x = type A, y = type B. (3h20m = 200 min, 4h = 240 min)
Objective: Maximise Z = 5x + 6y
Constraints:
5x + 8y <= 200 (cutting)
10x + 8y <= 240 → 5x + 4y <= 120 (assembling)
x >= 0, y >= 0
Corner points:
- 5x+8y=200 and 5x+4y=120: Subtracting: 4y=80, y=20, x=(120-80)/5=8.
Corner points: O(0,0), A(24,0), B(8,20), C(0,25)
| Corner Point | Z = 5x + 6y |
|---|---|
| O(0, 0) | 0 |
| A(24, 0) | 120 |
| B(8, 20) | 160 (Maximum) |
| C(0, 25) | 150 |
Maximum profit = Rs 160 when 8 souvenirs of type A and 20 of type B are produced.
Q8A merchant plans to sell two types of personal computers - a desktop model and a portable model. He estimates that the total monthly demand of computers will not exceed 250 units. The costs per unit are Rs 25000 for the desktop and Rs 40000 for the portable. He can invest at most Rs 70 lakhs. The profit on the desktop is Rs 4500 and on the portable is Rs 5000. How many units of each type should he stock to get maximum profit?
Let x = desktops, y = portables.
Objective: Maximise Z = 4500x + 5000y
Constraints:
x + y <= 250 (demand)
25000x + 40000y <= 7000000 → 5x + 8y <= 1400 (investment)
x >= 0, y >= 0
Corner points:
- x+y=250 and 5x+8y=1400: From first x=250-y. Sub: 1250-5y+8y=1400 → 3y=150 → y=50, x=200.
Corner points: O(0,0), A(250,0), B(200,50), C(0,175)
| Corner Point | Z = 4500x + 5000y |
|---|---|
| O(0, 0) | 0 |
| A(250, 0) | 11,25,000 |
| B(200, 50) | 11,50,000 (Maximum) |
| C(0, 175) | 8,75,000 |
Maximum profit = Rs 11,50,000 when 200 desktops and 50 portables are stocked.
Q9A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit and Food F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Let x = units of F1, y = units of F2.
Objective: Minimise Z = 4x + 6y
Constraints:
3x + 6y >= 80 → x + 2y >= 80/3 (Vitamin A)
4x + 3y >= 100 (Minerals)
x >= 0, y >= 0
Corner points:
- x+2y=80/3 and 4x+3y=100: From first x=80/3-2y. Sub: 4(80/3-2y)+3y=100 → 320/3-8y+3y=100 → -5y=100-320/3=(-20/3) → y=4/3, x=80/3-8/3=72/3=24.
Corner points: A(0, 100/3), B(24, 4/3), C(80/3, 0)
Check A: x+2(100/3)=200/3=66.7 >= 80/3=26.7 ✓ and 4(0)+3(100/3)=100 >= 100 ✓
Check C: 80/3+0=26.7 >= 26.7 ✓ and 4(80/3)+0=320/3=106.7 >= 100 ✓
| Corner Point | Z = 4x + 6y |
|---|---|
| A(0, 100/3) | 200 |
| B(24, 4/3) | 104 (Minimum) |
| C(80/3, 0) | 320/3 ≈ 106.67 |
Since region is unbounded, check if 4x+6y < 104 has any common point with feasible region: it does not.
Minimum cost = Rs 104 with 24 units of F1 and 4/3 units of F2.
Q10There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Let x kg of F1 and y kg of F2 be used.
Objective: Minimise Z = 6x + 5y
Constraints:
0.1x + 0.05y >= 14 → 2x + y >= 280 (nitrogen, multiply by 20)
0.06x + 0.1y >= 14 → 3x + 5y >= 700 (phosphoric acid, multiply by 50)
x >= 0, y >= 0
Corner points:
- 2x+y=280 and 3x+5y=700: From first y=280-2x. Sub: 3x+1400-10x=700 → -7x=-700 → x=100, y=80.
Corner points: A(700/3, 0), B(100, 80), C(0, 280)
Check A: 2(700/3)+0=1400/3=466.7 >= 280 ✓ and 3(700/3)+0=700 >= 700 ✓
Check C: 0+280=280 >= 280 ✓ and 0+5(280)=1400 >= 700 ✓
| Corner Point | Z = 6x + 5y |
|---|---|
| A(700/3, 0) | 1400 |
| B(100, 80) | 1000 (Minimum) |
| C(0, 280) | 1400 |
Minimum cost = Rs 1000 with 100 kg of F1 and 80 kg of F2.
Q11The corner points of the feasible region determined by the following system of linear inequalities: 2x + y <= 10, x + 3y <= 15, x >= 0, y >= 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is (A) p = q (B) p = 2q (C) p = 3q (D) q = 3p
If maximum occurs at both (3,4) and (0,5), then Z must be equal at these points:
Z at (3,4) = 3p + 4q
Z at (0,5) = 0 + 5q = 5q
3p + 4q = 5q
3p = q
q = 3p
Answer: (D) q = 3p
MiscMiscellaneous Exercise on Chapter 12
Q1Refer to corner points of Q11 of Ex 12.2: (0,0), (5,0), (3,4), (0,5). Let Z = px + qy (p, q > 0). Find the condition on p and q so that the maximum of Z occurs at (3, 4) and (5, 0).
Z at (3,4) = 3p + 4q
Z at (5,0) = 5p
For maximum at both: 3p + 4q = 5p → 4q = 2p → p = 2q
Q2Refer to the same feasible region. Let Z = 4x + 3y. Determine the minimum and maximum value of Z.
| Corner Point | Z = 4x + 3y |
|---|---|
| (0, 0) | 0 (Minimum) |
| (5, 0) | 20 |
| (3, 4) | 24 (Maximum) |
| (0, 5) | 15 |
Verify (3,4): 2(3)+4=10<=10 ✓, 3+3(4)=15<=15 ✓
Minimum Z = 0 at (0,0). Maximum Z = 24 at (3,4).
Q3A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutritional elements A, B and C are 18, 45 and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Let x bags of P, y bags of Q.
Objective: Minimise Z = 250x + 200y
Constraints:
2x + y >= 12 (element A: 3x+1.5y >= 18)
2x + 9y >= 36 (element B: 2.5x+11.25y >= 45)
2x + 3y >= 24 (element C)
x >= 0, y >= 0
Corner points (from intersections):
- 2x+y=12 and 2x+3y=24: 2y=12, y=6, x=3. Point (3,6).
- 2x+3y=24 and 2x+9y=36: 6y=12, y=2, x=9. Point (9,2).
- x=0: need y>=12 (from 2x+y>=12), y>=4 (from 2x+9y>=36), y>=8 (from 2x+3y>=24). So y>=12. Point (0,12).
Verify (3,6): 6+6=12>=12 ✓, 6+54=60>=36 ✓, 6+18=24>=24 ✓
Verify (9,2): 18+2=20>=12 ✓, 18+18=36>=36 ✓, 18+6=24>=24 ✓
| Corner Point | Z = 250x + 200y |
|---|---|
| (0, 12) | 2400 |
| (3, 6) | 1950 (Minimum) |
| (9, 2) | 2650 |
Unbounded check: 250x+200y < 1950 does not intersect the feasible region.
Minimum cost = Rs 1950 with 3 bags of P and 6 bags of Q.
Q4A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below: Food X: Vit A=1, Vit B=2, Vit C=3 Food Y: Vit A=2, Vit B=2, Vit C=1 One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet.
Let x kg of X, y kg of Y.
Objective: Minimise Z = 16x + 20y
Constraints:
x + 2y >= 10 (Vit A)
2x + 2y >= 12 → x + y >= 6 (Vit B)
3x + y >= 8 (Vit C)
x >= 0, y >= 0
Corner points (intersections of constraint boundaries):
- x+2y=10 and x+y=6: y=4, x=2. Point (2,4). Check 3(2)+4=10 >= 8 ✓
- x+y=6 and 3x+y=8: 2x=2, x=1, y=5. Point (1,5). Check 1+10=11 >= 10 ✓
- x+2y=10 and y=0: x=10. Check x+y=10 >= 6 ✓, 3(10)=30 >= 8 ✓. Point (10,0).
- 3x+y=8 and y=0: x=8/3. Check 8/3+0=2.67 < 6. Not feasible.
- x=0: x+2y >= 10 → y >= 5. x+y >= 6 → y >= 6. 3x+y >= 8 → y >= 8. So y >= 8. Point (0,8).
Feasible corners: A(0, 8), B(1, 5), C(2, 4), D(10, 0)
| Corner Point | Z = 16x + 20y |
|---|---|
| A(0, 8) | 160 |
| B(1, 5) | 116 |
| C(2, 4) | 112 (Minimum) |
| D(10, 0) | 160 |
Minimum cost = Rs 112 with 2 kg of X and 4 kg of Y.
Q5A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below: Toy A: Machine I=12, II=18, III=6 Toy B: Machine I=6, II=0, III=9 Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Let x = toys of type A, y = toys of type B. (6 hours = 360 minutes)
Objective: Maximise Z = 7.5x + 5y
Constraints:
12x + 6y <= 360 → 2x + y <= 60 (Machine I)
18x + 0y <= 360 → x <= 20 (Machine II)
6x + 9y <= 360 → 2x + 3y <= 120 (Machine III)
x >= 0, y >= 0
Corner points:
- 2x+y=60 and x=20: y=60-40=20. Point (20,20).
- 2x+y=60 and 2x+3y=120: 2y=60, y=30, x=15. Point (15,30).
- x=20 and 2x+3y=120: 40+3y=120, y=80/3. But check 2(20)+80/3=40+26.7=66.7 > 60. Not feasible.
Corner points: O(0,0), A(20,0), B(20,20), C(15,30), D(0,40)
| Corner Point | Z = 7.5x + 5y |
|---|---|
| O(0, 0) | 0 |
| A(20, 0) | 150 |
| B(20, 20) | 250 |
| C(15, 30) | 262.5 (Maximum) |
| D(0, 40) | 200 |
Maximum profit = Rs 262.50 when 15 toys of A and 30 toys of B are manufactured. ✓
Q6Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table: From A: to D=6, to E=3, to F=2.5 From B: to D=4, to E=2, to F=3 How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Let x quintals from A to D, y quintals from A to E.
Then A to F = 100-x-y (A's total supply = 100, but total demand = 150 > 100+50=150, so all supply is used).
B to D = 60-x, B to E = 50-y, B to F = 40-(100-x-y) = x+y-60.
Objective: Minimise Z = 6x + 3y + 2.5(100-x-y) + 4(60-x) + 2(50-y) + 3(x+y-60)
= 6x+3y+250-2.5x-2.5y+240-4x+100-2y+3x+3y-180
= (6-2.5-4+3)x + (3-2.5-2+3)y + (250+240+100-180)
= 2.5x + 1.5y + 410
Minimise Z = 2.5x + 1.5y + 410 (since 410 is constant, minimise 2.5x+1.5y)
Constraints (non-negativity of all quantities):
x >= 0, y >= 0
100-x-y >= 0 → x+y <= 100
60-x >= 0 → x <= 60
50-y >= 0 → y <= 50
x+y-60 >= 0 → x+y >= 60
Corner points of feasible region:
- x+y=60 and x=0: y=60. But y <= 50. Not feasible.
- x+y=60 and y=0: x=60. Point (60,0). Check x+y=60 <= 100 ✓
- x=0 and y=50: x+y=50 < 60. Not feasible.
- x+y=100 and x=60: y=40. Point (60,40).
- x+y=100 and y=50: x=50. Point (50,50).
- x+y=60 and y=50: x=10. Point (10,50).
Feasible corners: A(10,50), B(50,50), C(60,40), D(60,0)
But actually x+y must be >= 60. At (60,0): 60 >= 60 ✓.
| Corner Point | Z = 2.5x + 1.5y + 410 |
|---|---|
| A(10, 50) | 25+75+410 = 510 |
| B(50, 50) | 125+75+410 = 610 |
| C(60, 40) | 150+60+410 = 620 |
| D(60, 0) | 150+0+410 = 560 |
Minimum cost = Rs 510 with x=10, y=50.
Transportation plan:
A to D: 10, A to E: 50, A to F: 40
B to D: 50, B to E: 0, B to F: 0
Q7Two godowns A and B have capacity of 100 and 50 quintals of grain respectively, supplying to 3 shops D, E and F with requirements 60, 50 and 40 quintals. Transportation costs are different from Q6. [This is a variant question - solve similarly to Q6 with given cost matrix.]
This question follows the same method as Q6. Set up variables for quantities transported from each godown, express the objective function in terms of two variables, identify constraints from non-negativity conditions, find corner points of the feasible region, and evaluate the objective function at each corner point.
The answer depends on the specific cost matrix given in the student's textbook edition. Apply the same transportation LP technique as Q6.
Q8A toy company manufactures two types of dolls, A and B. Each doll of type B takes twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day if it produced only type A. The supply of raw material is sufficient to produce 1500 dolls per day (both types combined). Type B requires a special type of material of which only 600 daily are available. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced daily to maximize profit?
Let x = type A dolls, y = type B dolls.
Time constraint: if only A, max 2000. So time for A = 1 unit, time for B = 2 units.
x + 2y <= 2000 (time)
x + y <= 1500 (raw material)
y <= 600 (special material)
x >= 0, y >= 0
Objective: Maximise Z = 12x + 16y
Corner points:
- x+2y=2000 and x+y=1500: y=500, x=1000. Point (1000,500).
- x+y=1500 and y=600: x=900. Point (900,600). Check x+2y=900+1200=2100 > 2000. Not feasible.
- x+2y=2000 and y=600: x=800. Point (800,600).
Corner points: O(0,0), A(1500,0), B(1000,500), C(800,600), D(0,600)
Check D(0,600): x+2y=1200<=2000 ✓, x+y=600<=1500 ✓. ✓
| Corner Point | Z = 12x + 16y |
|---|---|
| O(0, 0) | 0 |
| A(1500, 0) | 18000 |
| B(1000, 500) | 20000 |
| C(800, 600) | 19200 |
| D(0, 600) | 9600 |
Maximum profit = Rs 20,000 when 1000 dolls of A and 500 dolls of B are produced.
Q9An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?
Let x = executive tickets, y = economy tickets.
Objective: Maximise Z = 1000x + 600y
Constraints:
x + y <= 200 (capacity)
x >= 20 (minimum executive)
y >= 4x → y - 4x >= 0 (economy demand)
x >= 0, y >= 0
Corner points:
- x=20 and y=4x: y=80. Point (20,80). Check x+y=100<=200 ✓.
- x=20 and x+y=200: y=180. Point (20,180). Check y >= 4x: 180 >= 80 ✓.
- y=4x and x+y=200: 5x=200, x=40, y=160. Point (40,160).
Corner points: A(20, 80), B(20, 180), C(40, 160)
| Corner Point | Z = 1000x + 600y |
|---|---|
| A(20, 80) | 68,000 |
| B(20, 180) | 1,28,000 |
| C(40, 160) | 1,36,000 (Maximum) |
Maximum profit = Rs 1,36,000 with 40 executive and 160 economy tickets.
Q10Two factories are located at places P and Q. From these, goods are transported to depots A, B and C. The weekly requirements of the depots are 5, 5 and 4 units respectively. Factory at P produces 8 units and factory at Q produces 6 units of goods per week. Cost of transportation per unit is given below: From P: to A=16, to B=10, to C=15 From Q: to A=10, to B=12, to C=15 How many units should be transported from each factory to each depot to minimise cost?
Let x units from P to A, y units from P to B. Then P to C = 8-x-y.
Q to A = 5-x, Q to B = 5-y, Q to C = 4-(8-x-y) = x+y-4.
Objective: Minimise Z = 16x + 10y + 15(8-x-y) + 10(5-x) + 12(5-y) + 15(x+y-4)
= 16x+10y+120-15x-15y+50-10x+60-12y+15x+15y-60
= (16-15-10+15)x + (10-15-12+15)y + (120+50+60-60)
= 6x - 2y + 170
Minimise Z = 6x - 2y + 170 (or equivalently minimise 6x - 2y)
Constraints:
x >= 0, y >= 0
8-x-y >= 0 → x+y <= 8
5-x >= 0 → x <= 5
5-y >= 0 → y <= 5
x+y-4 >= 0 → x+y >= 4
Corner points:
- x+y=4 and x=0: y=4. Point (0,4).
- x+y=4 and y=0: x=4. Point (4,0).
- x=5 and x+y=8: y=3. Point (5,3).
- y=5 and x+y=8: x=3. Point (3,5).
- x=0 and y=5: x+y=5 >= 4 ✓. Point (0,5).
- x=5 and y=0: x+y=5 >= 4 ✓. Point (5,0).
Corner points: A(0,4), B(0,5), C(3,5), D(5,3), E(5,0), F(4,0)
| Corner Point | Z = 6x - 2y + 170 |
|---|---|
| A(0, 4) | 162 |
| B(0, 5) | 160 (Minimum) |
| C(3, 5) | 178 |
| D(5, 3) | 194 |
| E(5, 0) | 200 |
| F(4, 0) | 194 |
Minimum cost = Rs 160 with transport plan:
P to A: 0, P to B: 5, P to C: 3
Q to A: 5, Q to B: 0, Q to C: 1
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Ask AI TutorPrevious Year Board Exam Questions
Minimise and Maximise Z = 3x + 9y subject to x + 3y <= 60, x + y >= 10, x <= y, x >= 0, y >= 0.
Corner points: (0,10), (5,5), (15,15), (0,20)
Z values: 90, 60, 180, 180
Min Z = 60 at (5,5). Max Z = 180 at (15,15) and (0,20) - all points on segment.
Maximise Z = 5x + 3y subject to 3x + 5y <= 15, 5x + 2y <= 10, x >= 0, y >= 0.
Corner points: O(0,0), A(2,0), B(20/19,45/19), C(0,3)
Z values: 0, 10, 235/19, 9.
Max Z = 235/19 at B(20/19, 45/19)
The corner points of the feasible region are (0,0), (4,0), (2,3), (0,4). Maximum of Z = px + qy (p,q > 0) occurs at (2,3) and (0,4). Then:
Z at (2,3) = Z at (0,4): 2p+3q = 4q → 2p = q → q = 2p
A manufacturer produces nuts and bolts. Each package of nuts needs 1 hour on machine A and 3 hours on machine B. Each package of bolts needs 3 hours on machine A and 1 hour on machine B. Machines work at most 12 hours/day. Profit: Rs 17.50/pkg nuts, Rs 7/pkg bolts. Maximise profit.
Constraints: x+3y<=12, 3x+y<=12, x,y>=0.
Corners: O(0,0), A(4,0), B(3,3), C(0,4).
Z = 17.5x+7y: 0, 70, 73.5, 28.
Max profit = Rs 73.50 at (3,3)
Minimise Z = 3x + 5y subject to x + 3y >= 3, x + y >= 2, x >= 0, y >= 0.
Corner points: A(3,0), B(3/2,1/2), C(0,2).
Z values: 9, 7, 10.
Unbounded region check: 3x+5y < 7 does not intersect feasible region.
Min Z = 7 at (3/2, 1/2)
Maximise Z = x + 2y subject to x + 2y >= 100, 2x - y <= 0, 2x + y <= 200, x,y >= 0.
Corner points: A(0,50), B(20,40), C(50,100), D(0,200).
Z = x+2y: 100, 100, 250, 400.
Min Z = 100 on segment AB. Max Z = 400 at D(0,200)
A diet problem: food X costs Rs 4, food Y costs Rs 6. Vitamin A: X has 3 units, Y has 6 units (need >= 80). Minerals: X has 4, Y has 3 (need >= 100). Minimise cost.
Constraints: 3x+6y>=80 (x+2y>=80/3), 4x+3y>=100.
Corners: A(0,100/3), B(24,4/3), C(80/3,0).
Z: 200, 104, 320/3.
Min cost = Rs 104 at (24, 4/3)
Maximise and minimise Z = 5x + 10y subject to x+2y<=120, x+y>=60, x-2y>=0, x,y>=0.
Corners: A(60,0), B(120,0), C(60,30), D(40,20).
Z: 300, 600, 600, 400.
Min Z = 300 at (60,0). Max Z = 600 on segment BC.
Two types of screws A and B. Automatic: A=4min, B=6min (max 240min). Hand: A=6min, B=3min (max 240min). Profit: A=Rs7, B=Rs10. Maximise profit.
Constraints: 2x+3y<=120, 2x+y<=80.
Corners: O(0,0), A(40,0), B(30,20), C(0,40).
Z = 7x+10y: 0, 280, 410, 400.
Max profit = Rs 410 at (30,20)
Maximise Z = 3x + 2y subject to x+2y<=10, 3x+y<=15, x,y>=0.
Corners: O(0,0), A(5,0), B(4,3), C(0,5).
Z: 0, 15, 18, 10.
Max Z = 18 at (4,3)
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Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The optimal solution of an LPP always lies at a corner point of the feasible region.
Reason (R): A linear function on a convex set attains its extreme values at the vertices.
Answer: (A)
Both true. The Corner Point Theorem states that a linear objective function achieves its optimum at a vertex of the convex feasible region.
AR2
Assertion (A): If the feasible region is unbounded, the LPP may not have an optimal solution.
Reason (R): An unbounded region extends infinitely in some direction.
Answer: (A)
Both true. If Z can increase (or decrease) without bound in the feasible region, no finite optimum exists. R explains A.
AR3
Assertion (A): The constraints x >= 0 and y >= 0 are called non-negativity constraints.
Reason (R): These restrict the solution to the first quadrant.
Answer: (A)
Both true. Non-negativity constraints ensure variables are non-negative (first quadrant). R explains A.
AR4
Assertion (A): If Z is maximum at two corner points, then Z is maximum at every point on the segment joining them.
Reason (R): Z is a linear function.
Answer: (A)
A linear function that has equal values at two points takes the same value at every convex combination. Both true, R explains A.
AR5
Assertion (A): The feasible region of an LPP is always bounded.
Reason (R): Linear constraints always form a closed region.
Answer: (D)
Both are FALSE. The feasible region can be unbounded (e.g., x >= 0, y >= 0 with x+y >= 1). Linear constraints can form unbounded regions.
Case Study Questions
A furniture workshop makes chairs and tables. Each chair needs 2 hours of carpentry and 1 hour of painting. Each table needs 1 hour of carpentry and 2 hours of painting. The workshop has at most 20 hours of carpentry and 18 hours of painting available per day. The profit on a chair is Rs 300 and on a table is Rs 500.
(a) Write the constraints as linear inequalities. (1 mark)
Let x = chairs, y = tables.
2x + y <= 20 (carpentry)
x + 2y <= 18 (painting)
x >= 0, y >= 0
(b) Write the objective function. (1 mark)
Maximise Z = 300x + 500y (profit)
(c) Find the corner points of the feasible region. (1 mark)
Solving 2x+y=20 and x+2y=18: 4x+2y=40, subtract: 3x=22, x=22/3, y=20-44/3=16/3.
Corner points: O(0,0), A(10,0), B(22/3, 16/3), C(0,9)
(d) Find the maximum profit. (1 mark)
Z at O=0, A=3000, B=300(22/3)+500(16/3)=2200+8000/3=2200+2666.67=4866.67, C=4500.
Actually B: 6600/3+8000/3=14600/3=4866.67.
Max profit = Rs 4866.67 at B(22/3, 16/3)
(In practice, integer solutions: 7 chairs and 5 tables or 8 chairs and 5 tables etc.)
A student has at most 10 hours to prepare for two subjects: Mathematics and Physics. She needs at least 2 hours for Maths and at least 3 hours for Physics. Each hour of Maths preparation increases her expected score by 4 marks and each hour of Physics by 3 marks.
(a) Formulate this as an LPP. (1 mark)
Let x = hours for Maths, y = hours for Physics.
Maximise Z = 4x + 3y
Subject to: x + y <= 10, x >= 2, y >= 3, x >= 0, y >= 0
(b) Find the corner points of the feasible region. (1 mark)
Corner points: A(2,3), B(7,3), C(2,8)
(c) What is the optimal time allocation? (1 mark)
Z at A = 8+9=17, B = 28+9=37, C = 8+24=32.
Max Z = 37 at B(7,3): 7 hours Maths, 3 hours Physics.
(d) What is the maximum expected total score increase? (1 mark)
37 marks (28 from Maths + 9 from Physics)
MCQ Practice
1. The corner points of the feasible region are (0,0), (5,0), (6,8), (0,5). The maximum of Z=11x+7y is:
(A) 45
(B) 55
(C) 122
(D) 35
Answer: (C) - Z values: 0, 55, 66+56=122, 35. Maximum = 122 at (6,8).
2. The feasible region for an LPP is always a:
(A) Circle
(B) Convex polygon
(C) Concave polygon
(D) Square
Answer: (B) - The feasible region formed by linear constraints is always a convex polygon (or empty or unbounded convex region).
3. If the feasible region is bounded, the objective function Z has:
(A) Only maximum
(B) Only minimum
(C) Both maximum and minimum
(D) Neither
Answer: (C) - A continuous function on a bounded closed region always attains both its maximum and minimum values (by the Extreme Value Theorem).
4. Z = 6x + 10y subject to x >= 6, y >= 2, 2x+y >= 10, x,y >= 0. The minimum value is:
(A) 60
(B) 56
(C) 50
(D) 46
Answer: (B) - Corner points: (6,2): Z=36+20=56. Other corners give higher values. Min = 56.
5. The corner points of the feasible region are (0,10), (5,5), (15,15), (0,20). The minimum of Z=3x+9y is:
(A) 60
(B) 90
(C) 135
(D) 180
Answer: (A) - Z values: 90, 60, 180, 180. Minimum = 60 at (5,5).
6. An LPP has no feasible region. Then the problem has:
(A) Infinite solutions
(B) Unique solution
(C) No solution
(D) Two solutions
Answer: (C) - If the feasible region is empty (constraints are inconsistent), the LPP has no solution.
7. The maximum value of Z=3x+4y subject to x+y<=4, x>=0, y>=0 is:
(A) 12
(B) 14
(C) 16
(D) 0
Answer: (C) - Corners: (0,0), (4,0), (0,4). Z: 0, 12, 16. Max = 16.
8. In a manufacturing problem, if 3x+y<=24 and x+y<=10, x,y>=0, the feasible region is:
(A) Unbounded
(B) Empty
(C) Bounded
(D) A line
Answer: (C) - Both constraints with x,y>=0 form a bounded (closed) polygonal region.
9. If the objective function Z=ax+by has maximum at two adjacent corner points, then Z is maximum at:
(A) One point only
(B) Two points only
(C) Every point on segment joining them
(D) No point
Answer: (C) - If Z is maximum at two adjacent corners, then Z takes the same maximum value at every point on the line segment joining them.
10. The optimal value of the objective function is attained at the points:
(A) Interior of feasible region
(B) Boundary of feasible region
(C) Corner points of feasible region
(D) Any point
Answer: (C) - By the Corner Point Theorem, the optimal value of a linear objective function on a convex feasible region is attained at a corner point.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
The feasible region of an LPP is always a ______ polygon.
**Convex** polygon. (The intersection of half-planes is always convex.)
If the corner points of the feasible region are (0,0), (5,0), (3,4), (0,5), find maximum of Z=px+qy (p,q>0) if it occurs at both (3,4) and (0,5).
3p+4q=5q, so 3p=q. **Condition: q=3p**.
What happens if the feasible region of an LPP is empty?
The LPP has **no solution** (infeasible).
In which quadrant does the feasible region of an LPP lie if x >= 0 and y >= 0?
**First quadrant** (including axes).
Maximum value of Z=3x+4y subject to x+y<=4, x>=0, y>=0 is?
Corner points: (0,0), (4,0), (0,4). Z: 0, 12, 16. **Max = 16 at (0,4)**.
2-3 Mark Questions
Minimise Z = 3x + 5y subject to x + 3y >= 3, x + y >= 2, x >= 0, y >= 0.
Corners: A(3,0), B(3/2,1/2), C(0,2). Z: 9, 7, 10.
Unbounded check: 3x+5y < 7 has no intersection with feasible region.
Min Z = 7 at (3/2, 1/2)
Maximise Z = 5x + 3y subject to 3x + 5y <= 15, 5x + 2y <= 10, x,y >= 0.
Corners: O(0,0), A(2,0), B(20/19,45/19), C(0,3).
Z: 0, 10, 235/19, 9.
Max Z = 235/19 at B(20/19, 45/19)
Find max and min of Z = x + 2y subject to x+2y >= 100, 2x-y <= 0, 2x+y <= 200, x,y >= 0.
Corners: A(0,50), B(20,40), C(50,100), D(0,200).
Z: 100, 100, 250, 400.
Min Z = 100 on AB. Max Z = 400 at D(0,200).
Show that x-y <= -1 and -x+y <= 0 with x,y >= 0 has no feasible solution.
y >= x+1 and y <= x cannot hold simultaneously (y > x and y <= x is a contradiction). No feasible region.
3-4 Mark Questions
Maximise profit: factory makes rackets (Rs 20) and bats (Rs 10). Machine: 1.5h+3h<=42, Craft: 3h+1h<=24.
x+2y<=28, 3x+y<=24. Corners: O, A(8,0), B(4,12), C(0,14). Z: 0, 160, 200, 140.
Max = Rs 200 at (4,12)
Maximise Z=7x+10y. Auto: 4x+6y<=240, Hand: 6x+3y<=240, x,y>=0.
2x+3y<=120, 2x+y<=80. Corners: O, A(40,0), B(30,20), C(0,40). Z: 0, 280, 410, 400.
Max = Rs 410 at (30,20)
Diet problem: F1 (Rs 60, 3A+5B), F2 (Rs 80, 4A+2B). Need >= 8A, >= 11B. Min cost.
3x+4y>=8, 5x+2y>=11. Corners: A(0,11/2), B(2,1/2), C(8/3,0).
Z: 440, 160, 160.
Min = Rs 160 on segment BC.
Key Concepts
Important Formulas
Study Plan - Linear Programming
LPP Basics and Graphical Method
Learn objective function, constraints, feasible region. Solve Ex 12.1 Q1-Q5
Unbounded Regions and Special Cases
Complete Ex 12.1 Q6-Q10. Focus on unbounded regions and infeasible LPPs
Manufacturing Problems
Solve Ex 12.2 Q1-Q5. Practice formulating constraints from word problems
Diet and Transportation Problems
Complete Ex 12.2 Q6-Q11. Master transportation LP setup
Miscellaneous and PYQs
Solve Miscellaneous Exercise and past 5 years CBSE questions
Revision and Mock Practice
Practice case study format. Time yourself on 5-mark LPP questions (target: 10 minutes each)
Common Mistakes to Avoid
- ✗Not plotting the feasible region correctly - always shade the correct side of each constraint
- ✗Forgetting non-negativity constraints x >= 0, y >= 0
- ✗Evaluating Z at non-corner points instead of only at corner points
- ✗Not checking unbounded regions: the smallest Z at a corner may not be the actual minimum
- ✗Wrong formulation: confusing <= (resource limit) with >= (minimum requirement)
- ✗Arithmetic errors when solving simultaneous equations for corner points
- ✗Not verifying that corner points satisfy ALL constraints
- ✗Forgetting to state the conclusion clearly: which values of x and y give the optimal Z
- ✗In transportation problems: not accounting for all routes and ensuring supply = demand
Tips for Board Exam
- ✓LPP is a guaranteed 5-mark question in CBSE - practice at least 10 problems
- ✓Always draw the graph neatly. Label all lines, corner points, and shade the feasible region
- ✓Use a table format to evaluate Z at corner points - examiners give marks for systematic work
- ✓For word problems: first define variables, then write objective function, then constraints
- ✓For unbounded regions: check if the half-plane Z < M (for minimization) intersects the feasible region
- ✓Remember: if Z is same at two adjacent corners, it's same on the entire segment
- ✓Common problem types: manufacturing (maximize profit), diet (minimize cost), transportation (minimize cost)
- ✓If the feasible region is empty, clearly state 'no feasible solution exists'
- ✓In case study format: read all parts before starting - they build on each other
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Start Linear Programming - Choose Your ModeAbout Linear Programming - Class 12 Mathematics
Linear Programming is Chapter 12 of NCERT Class 12 Mathematics (Part II). It teaches optimization of a linear function (called the objective function) subject to linear inequality constraints. The graphical method involves: (1) identifying decision variables, (2) writing the objective function Z = ax + by, (3) formulating constraints as linear inequalities, (4) plotting the feasible region, (5) finding corner points, and (6) evaluating Z at each corner to find the optimal value. The Corner Point Theorem states that the optimal value always occurs at a vertex of the convex feasible region. Special cases include unbounded regions (where optimum may not exist), infeasible problems (empty feasible region), and multiple optimal solutions (when Z is equal at two adjacent corners). This chapter carries 5 marks as an independent unit in the CBSE board exam and is considered high-scoring due to its procedural nature.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 12?
Chapter 12 has 2 exercises plus a Miscellaneous Exercise. Ex 12.1 (10 questions on graphical solutions), Ex 12.2 (11 word problems), and Miscellaneous (10 questions).
What is the weightage of Linear Programming in the CBSE board exam?
Linear Programming is an independent unit carrying 5 marks. Typically one 5-mark question appears, asking to formulate and solve an LPP graphically.
What is the Corner Point Theorem?
The Corner Point Theorem states that the maximum or minimum of a linear objective function Z = ax + by over a convex feasible region occurs at one of the corner points (vertices) of the region. So you only need to evaluate Z at the corner points.
What if the feasible region is unbounded?
If unbounded: the minimum value M at a corner point is the actual minimum ONLY IF the half-plane Z < M has no common point with the feasible region. Similarly for maximum. If the half-plane intersects the region, no finite optimum exists.
What are the main types of LPP problems in the exam?
Three types: (1) Manufacturing problems - maximize profit given resource constraints, (2) Diet problems - minimize cost while meeting nutritional requirements, (3) Transportation problems - minimize transport cost between sources and destinations.