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Install NowNCERT Solutions class 12 Maths Exercise 12.2 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

**Download NCERT solutions for Linear Programming as PDF.**

## NCERT Solutions class 12 Maths Linear Programming

** ****1. Reshma 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.**

**Ans. **Let food P consists of kg and food Q consists of kg.

Minimum Z =

According to question and

and 0

A | B | C | |

0 | 4 | ||

2 | 5 |

Consider

Let

But (0, 0) does not satisfy this inequation, therefore the required half-plane does not contain (0, 0).

Again consider

Let

D | E | F | |

0 | 4 | ||

2 | 5 |

Again (0, 0) does not satisfy this inequation, therefore the required half-plane does not contain (0, 0).

The double shaded region is our solution set. The corners of this region are D (0, 5.5), Q (2, 1.5) and P

Now Z =

AtD(0,5.5)

Z = 60 x 0 + 80 x 5.5 = 440

AtQ (2, 1.5)

Z = 60 x 2 + 80 x 1.5 = 160

AtP

Z = 60 x 80 x 0 = 160

Hence, minimum Z = 160 at means minimum cost Z = Rs. 160 when Reshma mixes food P = 2 kg and food Q = 1.5 kg.

**NCERT Solutions class 12 Maths Exercise 12.2**

**2. One kind of cake requires 200g of flour and 25 g of fat and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cake 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.**

**Ans. **Let number of cakes made of first kind are and that of second kind is

Let to maximize Z =

According to question and

Consider

Let

Here, (0, 0) satisfies this inequation, therefore the required half plane contains (0, 0).

Again consider

Let

Here, again (0, 0) satisfies this inequation, therefore the required half plane contains (0, 0).

The double shaded region is the feasible region which is solution set.

The corner points of this region are O (0, 0), A (25, 0), B (20, 10) and C (0, 20).

Z =

At O (0, 0) Z = 0 + 0 = 0

At A (25, 0) Z = 25 + 0 = 25

At B (20, 10) Z = 20 + 10 = 30

AtC (0, 20) Z = 0 + 20 = 20

Hence, maximum number of cakes Z = 30 when

**NCERT Solutions class 12 Maths Exercise 12.2**

**3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman’s time.**

**(i) What number of rackets and bats must be made if the factory is to work at full capacity?**

**(ii) If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively, find maximum profit of the factory when it works at full capacity.**

**Ans. **Let number of rackets = and number of bats =

A | B | C | |

28 | 20 | 0 | |

0 | 4 | 14 |

**(i)** To maximize Z =

According to question and

Consider

Let

Here (0, 0) satisfies this inequaiton, therefore the required half-plane contains (0, 0).

Again consider

Let

Here (0, 0) satisfies this inequaiton, therefore the required half-plane contains (0, 0).

Now the feasible region is the double-shaded region i.e., the solution set. The corner points of this region are O (0, 0), P (8, 0), Q (4, 12) and R (0, 14).

Now Z =

At O (0, 0) Z = 0 + 0 = 0

At P (8, 0) Z = 8 + 0 = 0

At Q (4, 12) Z = 4 + 12 = 16

At R (0, 14) Z = 0 + 14 = 14

Now maximum Z = 16 at

**(ii)** The profit on a racket = Rs. 20 and the profit on a bat = Rs. 10

Let maximum profit P =

Hence maximum profit P = 20 x 4 + 10 x 12 = Rs.200

**NCERT Solutions class 12 Maths Exercise 12.2**

**4. A 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 maximize his profit, if he operates his machines for at the most 12 hours a day?**

**Ans. **Let number of package of nuts = and number of packages of bolts =

To maximum profit Z =

According to question and

Consider

Let

Therefore, points are A (12, 0), B (0, 4). Also, (0, 0) satisfies this inequation, therefore the required half-plane contains (0, 0).

Again consider

Let

Therefore, points are C (4, 0), D (0, 12).

Again (0, 0) satisfies this inequation, therefore the required half-plane contains (0, 0).

The double shaded region is OBPCO and its corners are O (0, 0), B (0, 4), D (3, 3), C (4, 0).

Now Z =

At O (0, 0) Z = 17.5 x 0 + 7 x 0 = 0

At B (0, 4) Z = 17.5 x 0 + 7 x 4 = 28

At P (3, 3) Z = 17.5 x 3 + 7 x 3 = 73.50

At C (4, 0) Z = 17.5 x 4 + 7 x 0 = 70

maximum profit = Rs. 73.50 at

Hence maximum profit Z = Rs. 73.50 when he produces number of packets of nuts = 3 and number of packets of bolts = 3.

**NCERT Solutions class 12 Maths Exercise 12.2**

**5. A factory manufacturers 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 hand operated machines to manufacture a package of screw A, while it takes 6 minutes on automatic and 3 minutes on hand operated machines 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 a day in order to maximize his profit? Determine the maximum profit.**

**Ans. **Let the manufacturers produces packages of screws A and packages of screws B, then time taken by packages of screws A and packages of screws B on automatic machine = minutes. And time taken by packages of screws A and packages of screws B on hand operated machine = minutes.

Since, each machine is available for at the most 4 hours, i.e., 4 x 60 = 240 minutes.

Therefore, we have and

Profit on selling packages of screws A and packages of screws B = Z =

To find: and such that Z = is maximum subject to , and

Consider

Let

A (60, 0) and B (0, 4)

Here, (0, 0) satisfies this inequation, therefore the required hal-plane contains (0, 0).

Again

Let

C (40, 0) and D (0, 80)

Here also (0, 0) satisfies this inequation, therefore the required hal-plane contains (0, 0).

The feasible portion of the graph satisfying the inequalities and is OABC which is shown shaded in the figure.

Co-ordinates of O, A, B and C are (0, 0), (0, 40), (30, 20) and (40, 0) respectively.

Now Z =

At O (0, 0) Z = 7 x 0 + 10 x 0 = 0

At A (0, 40) Z = 7 x 0 + 10 x 40 = 400

At B (30, 20) Z = 7 x 30 + 10 x 20 = 410

At C (40, 0) Z = 7 x 40 + 10 x 0 = 280

maximum profit Z = Rs. 410 at

Hence, if the manufacturer produces 30 screws of type A and 20 screws of type B, he earn a maximum profit of Rs. 410.

**NCERT Solutions class 12 Maths Exercise 12.2**

**6. A 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 maximize his profit?**

**Ans. **Let the manufacturer produces pedestal lamps and wooden shades, then the time taken by pedestal lamps and wooden shades on grinding/cutting machines = hours and the time taken by pedestal lamps and wooden shades on the sprayer = hours.

Since grinding/cutting machine is available for at the most 12 hours, i.e., and sprayer is available for at the most 20 hours, i.e.,

Profit from the sale of lamps and shades Z =

To find: and such that Z = is maximum subject to constraints , ,

Consider

Let

A | B | C | |

0 | 2 | 4 | |

10 | 7 | 4 |

Now the area represented by is the half-plane containing (0, 0) as (0, 0) satisfies the inequaiton.

Consider

Let

D | E | C | |

0 | 2 | 4 | |

12 | 8 | 4 |

The inequation consists of the half-plane containing (0, ) as (0, 0) satisfies this inequation.

The double shaded region OPCAO is our solution where O (0, 0), P (6, 0), C (4, 4), A (0, 10).

Now Z =

At O (0, 0) Z = 5 x 0 + 3 x 0 = 0

At P (6, 0) Z = 5 x 6 + 3 x 0 = 30

At C (4, 4) Z = 5 x 4 + 3 x 4 = 32

At A (0, 10) Z = 5 x 0 + 3 x 10 = 30

Now maximum Z = 32 at

Hence, maximum profit Z = Rs. 32 when he manufactures 4 pedestal lamps and 4 wooden shades.

**NCERT Solutions class 12 Maths Exercise 12.2**

**7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A requires 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 maximize the profit?**

**Ans. **Let the company manufactures souvenirs of type A and souvenirs of type B, then time taken for cutting souvenirs of type A and souvenirs of type B = minutes and time taken for assembling souvenirs of type A and souvenirs of type B = minutes.

Since 3 hours 29 minutes i.e., 200 minutes are available for cutting, therefore we have

Also since 4 hours i.e., 240 minutes are available for assembling, therefore we have

Thus, our L.P.P. is to maximize profit Z = subject to constraints , ,

Now

Let

40 | 0 | |

0 | 25 |

Again

Let

24 | 20 | 0 | |

0 | 5 | 30 |

In both the equations origin (0, 0) satisfies them and therefore the required half planes are there which contains (0, 0).

The portion of graph satisfying the inequalities and is OABC and is shown in shaded in the figure. Coordinates of the points O, A, B and C are (0, 0), (24, 0), (8, 20) and (0, 25) respectively.

Now Z =

At O (0, 0) Z = 5 x 0 + 6 x 0 = 0

At A (24, 0) Z = 5 x 24 + 6 x 0 = 120

At B (8, 20) Z = 5 x 8 + 6 x 20 = 160

At C (0, 25) Z = 5 x 0 + 6 x 25 = 150

Now maximum Z = 160 at

Hence, maximum profit = Rs. 160 when he manufactures 8 souvenirs of type A and 20 souvenirs of type B.

**NCERT Solutions class 12 Maths Exercise 12.2**

**8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs, 70 lakhs and if his profit on the desktop model is Rs. 4500 and on portable model is Rs. 5000.**

**Ans. **Let number of desktop model computer = and number of a portable model computer =

To maximum profit Z = subject to ,

Consider

Let

Points are A (280, 0), B (0, 175)

Now (0, 0) satisfies the inequation, therefore the required half-plane contains (0, 0).

Again consider

Let

Points are C (250, 0), D (0, 250).

Again (0, 0) satisfies the inequation, therefore the required half-plane contains (0, 0).

The double shaded region OCPBO is the solution set. Its corners are O (0, 0), C (280, 0), P (200, 50), B (0, 200).

Now Z =

AtO (0, 0) Z = 4500 x 0 + 5000 x 0 = 0

At C (280, 0) Z = 4500 x 280 + 5000 x 0 = 12,60,000

(Rejected as )

At P (200, 50) Z = 4500 x 200 + 5000 x 50 = 11,50,000

At B (0, 200) Z = 4500 x 0 + 5000 x 200 = 10,00,000

Now maximum Z = Rs. 11,50,000 at

Hence, maximum profit = Rs. 11,50,000 when he sells 200 desktop models and 50 portable models.

**NCERT Solutions class 12 Maths Exercise 12.2**

**9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F**_{1} and F_{2} are available. Food F_{1} costs Rs. 4 per unit food and F_{2} costs Rs. 6 per unit. One unit of food F_{1} contains 3 units of vitamin A and 4 units of minerals. One unit of food F_{2} 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.**

_{1}and F

_{2}are available. Food F

_{1}costs Rs. 4 per unit food and F

_{2}costs Rs. 6 per unit. One unit of food F

_{1}contains 3 units of vitamin A and 4 units of minerals. One unit of food F

_{2}contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the

**Ans. **Let the number of units of food F_{1} = and number of units of food F_{2} =

To minimize Z = subject to , ,

Consider

Let

A | B | C | |

0 | 5 | 10 | |

Clearly, (0, 0) is not included in the half-plane.

Now again consider

Let

D | E | F | |

25 | 22 | 10 | |

0 | 4 | 20 |

Here, also (0, 0) is not included in the required half-plane.

The double shaded region is our feasible region and its corners are

NCERT Solutions class 12 Maths Exercise 12.2

A P and Q

Now Z =

At A Z = 4 x + 6 x 0 = 106.67

At P Z = 4 x 24 + 6 x = 104

At Q Z = 4 x 0 + 6 x = 200

Hence Z = Rs. 104 is minimum at .

**NCERT Solutions class 12 Maths Exercise 12.2**

**10. There are two types of fertilizers F**_{1} and F_{2}. F_{1} consists of 10% nitrogen and 6% phosphoric acid and F_{2} 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 F_{1} costs Rs. 6/kg and F_{2} costs Rs. 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

_{1}and F

_{2}. F

_{1}consists of 10% nitrogen and 6% phosphoric acid and F

_{2}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 F

_{1}costs Rs. 6/kg and F

_{2}costs Rs. 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

**Ans. **Let the quantity of F_{1} = kg and the quantity of F_{2} = kg

We have to minimize Z = subject to ,

Consider

Let

A | B | C | D | |

0 | 50 | 100 | 140 | |

280 | 180 | 80 | 0 |

Here, (0, 0) is not included in the required half plane and (0, 0) does not satisfy this inequation.

NCERT Solutions class 12 Maths Exercise 12.2

Again consider

Let

E | C | G | |

0 | 100 | 200 | |

140 | 80 | 20 |

Again Here, (0, 0) is not included in the required half plane and (0, 0) does not satisfy this inequation.

NCERT Solutions class 12 Maths Exercise 12.2

The shaded region XPCAY is our feasible region. Its corners are P, C (100, 80) and A (0, 280).

Now Z =

At P Z = 6 x + 5 x 0 = 1400

At C (100, 80) Z = 6 x 100 + 5 x 80 = 1000

At A (0, 280) Z = 6 x 0 + 5 x 280 = 1400

Now minimum cost Z = Rs. 1000 at

**NCERT Solutions class 12 Maths Exercise 12.2**

**11. The corner points of the feasible region determined by the following system of linear inequalities:**

** are (0, 0), (5, 0), (3, 4) and (0, 5). Let where Condition on and so that the maximum of Z occurs at both (3, 4) and (0, 5) is:**

**(A) **

**(B) **

**(C) **

**(D) **

**Ans. **To maximize Z = subject to ,

The corner points are (0, 0), (5, 0), (3, 4) and (0, 5).

Maximum of Z occurs at both (3, 4) and (0, 5)

Z =

At (3, 4) Z =

At (0, 5) Z =

Now maxima occurs at both points,

Hence option (D) is correct.

## NCERT Solutions class 12 Maths Exercise 12.2

NCERT Solutions class 12 Maths Exercise 12.2 PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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