NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming (Ex 12.2) Exercise 12.2
The Calcutta University Commission was responsible for the establishment of Boards of Secondary Education across the nation. The first such board was the U. P. Board of High School and Intermediate Education. However, after a time, the Board was no longer able to carry out this extraterritorial duty. In order to maintain an effective administration, the United Province Board’s jurisdiction was deemed too cumbersome by the United Provinces government, which is why it requested that candidates from regions outside the United Provinces not be allowed to sit for the board examination after 1927–1928. In response to the representation, the Indian government put out two options for the administration of the princely states of Rajputana, Central India, and Gwalior. A unified board for all the impacted areas was one option, while a separate board for each of the afflicted areas was another.
There were many benefits to the Central Board. It was, therefore, decided that a joint board for all the areas should be created. That is how the Central Board of Secondary Education, or the CBSE came into being initially. In present times, the schools that are affiliated with CBSE have to follow the NCERT syllabus. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are created keeping in view the NCERT syllabus.
A selfgoverning organisation called the National Council of Educational Research and Training (NCERT) was created in 1969. It was founded by the Indian government
to support and advise the federal and state governments on policies and programmes aimed at raising educational standards. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 available on the Extramarks website, are solutions to the questions that are given in the NCERT textbook. The main objectives of NCERT and the divisions in the organisation are to conduct, promote, and coordinate research in areas related to school education; to produce and publish model textbooks, supplementary materials, newsletters, journals, and educational kits, among other things, for school students following the NCERT format. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are formulated by the experts keeping this objective in mind as well. For the students of Class 12 and other classes as well, NCERT books have practise questions at the end of each chapter. These questions are based on the topics taught in the previous chapter. Extramarks makes available the NCERT Solutions which are the compilation of all such questions that are present in the NCERT books. Students can make use of the solutions when preparing for their examinations as well as for revising during class tests. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are a part of such solutions compiled by Extramarks’ experts.
For subjects like Mathematics, where problemsolving is an equally important part as the theory part, the questions related to that are divided into multiple exercises. Every chapter has multiple exercises containing different types of questions for the students to understand and practice by continuous revision.
The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are solutions to one of the exercises in the NCERT textbook, which is Mathematics Ex 12.2 Class 12. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 can also be used by the students when they are working on their homework. Mathematics requires a lot of practice and that could be the reason the teachers prefer giving some questions from each exercise as homework for students for to do selfstudy and practice.
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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming (Ex 12.2) Exercise 12.2
Students who want to excel in the Class 12 board exams should begin working on their overall wellrounded education at a young age.The experts at Extramarks understand that to understand complex topics that are covered in higher education, students need to have a good foundation in the simpler concepts that are taught in the lower classes.
Keeping all of this in mind, the experts at Extramarks, along with the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 also have NCERT Solutions for the following classes:
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NCERT Solutions Class 7
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The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are solutions to the topics covered specifically in the 2nd exercise of Chapter 12. Chapter 12 teaches topics that come under the topic of Linear Programming; according to the syllabus issued by the CBSE for the NCERT books.
An Overview of the Topics Covered in Exercise 12.2 of Class 12 Maths NCERT Solutions
Linear programming is a mathematical model whose requirements are represented by linear relationships. Linear programming is also known as Linear Optimisation. It is a method to achieve the best outcome in Mathematical Programming.
The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are the solutions to the questions given in the 2nd exercise of this chapter. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 are not the only solutions available. The solutions to all the other exercises are also available on the Extramarks website.
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Introduction to Linear Programming, related terminology such as Constraints, Objective Function, Optimization, Graphical Method of Solution for Problems in Two Variables, Feasible, and Infeasible Regions (bounded or unbounded), Feasible and Infeasible Solutions, Optimal Feasible Solutions (up to three nontrivial constraints).
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Access NCERT Solutions for Maths Class 12 Chapter 12 – Linear Programming
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NCERT Solution Class 12 Maths of Chapter 12 All Exercises
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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2
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It was established in 1961 by the Indian government as a selfgoverning body that provides advice and assistance to the Central and State Governments on educational policies and programmes. As a result, they have developed and published the NCERT Textbooks, which are designed for all students from Grades 112. Hence, NCERT textbooks are the standard and goto books for all students. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 is one of many such tools available on the Extramarks website. The Mathematics experts at Extramarks understand the importance of all these activities and what role these activities can play in a student’s final result.
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Q.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 the mixture contain x kg of food P and y kg of food Q. Clearly, x ≥ 0 and y ≥ 0.We make the following table from the given data:
Vitamin A
(units/kg) 
Vitamin B
(units/kg) 
Cost
(Rs/kg) 

Food P  3  5  60 
Food Q  4  2  80 
Requirement
(units/kg) 
8  11 
The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B.
Therefore, the constraints are
3x + 4y ≥ 8
5x + 2y ≥ 11
Total cost, Z, of purchasing food is, Z = 60x + 80y
Hence, the mathematical formulation of the problem is:
Minimise Z = 60 x + 80 y …(i)
subject to constraints:
3x + 4y ≥ 8 …(ii)
5x + 2y ≥ 11 …(iii)
x, y ≥ 0 …(iv)
Let us graph the inequalities (ii) to (iv). The feasible region determined by the system to shown in the given graph.
$\begin{array}{l}\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{unbounded}.\\ \mathrm{The}\mathrm{corner}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{are}\\ \mathrm{A}(\frac{8}{3},\mathrm{\hspace{0.17em}}0),\mathrm{\hspace{0.17em}}\mathrm{B}(2,\mathrm{\hspace{0.17em}}\frac{1}{2})\mathrm{and}\mathrm{C}(0,\mathrm{\hspace{0.17em}}\frac{11}{2}).\\ \mathrm{The}\mathrm{values}\mathrm{of}\mathrm{Z}\mathrm{at}\mathrm{these}\mathrm{corner}\mathrm{points}\mathrm{are}\mathrm{as}\mathrm{follows}:\end{array}$
Corner Point  Z = 60x + 80y  
\text{A}\left(\frac{8}{3},\text{\hspace{0.17em}}0\right)  160  Minimum 
\text{B}\left(2,\text{\hspace{0.17em}}\frac{1}{2}\right)  160  
\text{C}\left(0,\text{\hspace{0.17em}}\frac{11}{2}\right)  440 
$\begin{array}{l}\mathrm{As}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{unbounded},\mathrm{therefore},\; 160\mathrm{may}\mathrm{or}\\ \mathrm{may}\mathrm{not}\mathrm{be}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{.}\\ \mathrm{For}\mathrm{this},\mathrm{we}\mathrm{graph}\mathrm{the}\mathrm{inequality},\; 60\mathrm{x}+\; 80\mathrm{y}<\; 160\\ \mathrm{or}3\mathrm{x}+\; 4\mathrm{y}<\; 8,\mathrm{and}\mathrm{check}\mathrm{whether}\mathrm{the}\mathrm{resulting}\mathrm{half}\mathrm{plane}\\ \mathrm{has}\mathrm{points}\mathrm{in}\mathrm{common}\mathrm{with}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{or}\mathrm{not}.\mathrm{It}\mathrm{can}\\ \mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{has}\mathrm{no}\mathrm{common}\mathrm{point}\\ \mathrm{with}3\mathrm{x}+\; 4\mathrm{y}<\; 8\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{minimum}\mathrm{cost}\mathrm{of}\mathrm{the}\mathrm{mixture}\mathrm{will}\mathrm{be}\mathrm{Rs}160\mathrm{at}\mathrm{the}\\ \mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{points}\hspace{0.17em}\hspace{0.17em}(\frac{8}{3},\mathrm{\hspace{0.17em}}0)\mathrm{and}(2,\mathrm{\hspace{0.17em}}\frac{1}{2}).\end{array}$
Q.2 One kind of cake requires 200g flour and 25g of fat, and another kind of cake requires100g 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?
Ans
Let there be x cakes of first kind and y cakes of second kind. Clearly, x ≥ 0 and y ≥ 0.We make the following table from the given data:
Flour (g)  Fat (g)  
Cakes of first kind, x  200  25 
Cakes of second kind, y  100  50 
Availability  5000  1000 
Both cakes are containing at least 5 kg flour and 1kg fat.
Therefore, the constraints are
200 x + 100 y≤ 5000
⇒ 2x + y ≤ 50
25x + 50y ≤ 1000
⇒ x + 2y ≤ 40
Then, maximum number of cakes, Z = x + y
Hence, the mathematical formulation of the problem is:
Maximise Z = x + y …(i)
subject to constraints:
2x + y≥ 50 …(ii)
x + 2y≥ 40 …(iii)
x, y ≥ 0 …(iv)
Let us graph the inequalities (ii) to (iv). The feasible region determined by the system to shown in the given graph.
The corner points are A(25, 0), B(20, 10), O(0, 0), and C(0, 20).
The values of Z at these corner points are given in the following table:
Corner points  Z = x + y  
A (25, 0)  25  
B(20, 10)  30  → Maximum 
C(0, 20)  20  
O(0, 0)  0 
Thus, the maximum numbers of cakes that can be made are 30 (20 of one kind and 10 of the other kind).
Q.3 A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour 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.
(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 the maximum profit of the factory when it works at full capacity.
Ans
(i) Let the number of rackets and the number of bats to be made are x and y respectively.The machine time is not available for more than 42 hours, i.e., 1.5 x + 3y ≤ 42 … (i)
The craft time is not available for more than 24 hours, i.e., 3x + y ≤ 24 … (ii)
When the factory is to work at full capacity,
1.5 x + 3y = 42 and
3x + y = 24
On solving both the equations, we get
x = 4 and y = 12
Thus, 4 rackets and 12 bats must be made in the factory according to given condition.
(ii) Profit on a racket = Rs. 20
Profit on a bat = Rs. 10
Then, Max. Profit, Z = 20 x + 10 y
The given information can be represent as:
Tennis Racket  Cricket Bat  Availability of time  
Machine time(hr)  1.5  3  42 
Craftsman’s Time (hr)  3  1  24 
So, Maximize Z = 10x + 20y … (i)
The constraints are as follows:
1.5 x + 3 y ≤ 42 … (ii)
3x + y ≤ 24 … (iii)
x, y≥ 0 … (iv)
The feasible region determined by the linear inequalities (ii) to (iv) is shown in the figure:
The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0). The values of Z at these corner points are given in the following table:
Corner point  Z=20x+10y  
A (8, 0)  160  
B (4, 12)  200  → Maximum 
C (0, 14)  140  
O (0, 0)  0 
Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.
Q.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 the number of nuts and the number of bolts to be made are x and y respectively. The given information can be written in tabular form as given below:
Nuts  Bolts  Availability(time)  
Machine A  1 hr  3 hrs  12 hrs 
Machine B  3 hrs  1 hr  12 hrs 
Profit  Rs. 17.50  Rs. 7 
The mathematical formulation of the given problem is:
Maximise
Z = 17.5 x + 7 y … (i)
x + 3 y ≤ 12 … (ii)
3x + y ≤ 12 … (iii)
x, y ≥ 0 … (iv)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
The corner points are A(4, 0), B(3, 3), and C(0, 4) and the values of Z at these corner points are given below:
Corner point  Z = 17.5x + 7y  
O(0, 0)  0  
A(4, 0)  70  
B(3, 3)  73.5  → Maximum 
C(0, 4)  28 
The maximum value of Z is 73.5 at the point (3, 3).Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs 73.50.
Q.5 A 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 hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the 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 Rs10. 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 maximize his profit? Determine the maximum profit.
Ans
Let the number of screws, A and B to be made are x and y respectively.
The given information can be written in tabular form as given below:
Screws A  Screws B  Availability  
Automatic Machine  4 min.  6 min.  4 hrs 
Hand operated Machine  6 min.  3 min.  4 hrs 
Profit  Rs. 7  Rs. 10 
The mathematical formulation of the given problem is:
Maximise Z = 7x + 10y … (i)
4x + 6y ≤ 240 … (ii)
6x + 3y ≤ 240 … (iii)
x, y ≥ 0 … (iv)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
The corner points are A(40, 0), B(30, 20), and C(0, 40) and the values of Z at these corner points are given below:
Corner point  Z = 7x + 10y  
O(0, 0)  0  
A(40, 0)  280  
B(30, 20)  410  → Maximum 
C(0, 40)  400 
The maximum value of Z is 410 at the point (30, 20).
Thus, 30 packages of screws A and 20 packages of screws B should be produced each day to get the maximum profit of Rs. 410.
Q.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 number of pedestal lamps and wooden shades to be made are x and y respectively.
The given information can be written in tabular form as given below:
Pedestal lamps  Wooden
lamps 
Availability  
Grinding/Cutting Machine  2 hrs  1 hr  12 hrs 
Sprayer  3 hrs  2 hrs  20 hrs 
Profit  Rs. 5  Rs. 3 
The mathematical formulation of the given problem is:
Maximise Z = 5x + 3y … (i)
2x + y ≤ 12 … (ii)
3x + 2y ≤ 20 … (iii)
x, y ≥ 0 … (iv)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
The corner points are A(6, 0), B(4, 4), and C(0, 10) and the values of Z at these corner points are given below:
Corner point  Z = 5x + 3y  
O(0, 0)  0  
A(6, 0)  30  
B(4, 4)  32  → Maximum 
C(0, 10)  30 
The maximum value of Z is 32 at the point (4, 4).
Thus, 4 pedestal lamps and 4 wooden shades should be produced to maximise his profit.
Q.7 A 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 of 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 number of Souvenirs of type A and Souvenirs of type B to be made are x and y respectively. The given information can be written in tabular form as given below:
Souvenirs A  Souvenirs
B 
Availability  
Time for Cutting  5 min.  8 min.  3 hrs 20 min. 
Time for assembling  10 min.  8 min.  4 hrs 
Profit  Rs.5  Rs.6 
The mathematical formulation of the given problem is:
Maximise Z = 5x + 6y … (i)
5x + 8y ≤ 200 … (ii)
10x + 8y ≤ 240
⇒ 5x + 4y ≤ 120 … (iii)
x, y ≥ 0 … (iv)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
The corner points are A(24, 0), B(8, 20), and C(0, 25) and the values of Z at these corner points are given below:
Corner point  Z = 5x + 6y  
O(0, 0)  0  
A(24, 0)  120  
B(8, 20)  160  → Maximum 
C(0, 25)  150 
The maximum value of Z is 160 at the point (8, 20). Thus, 8 Souvenirs of type A and 20 Souvenirs of type B should be produced to maximise his profit.
Q.8 A merchant plans to sell two types of personal computers − a desktop model and a portable model that will cost ₹ 25,000 and ₹ 40,000 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 ₹ 4500 and on portable model is ₹ 5000.
Ans
Let the number of desktop computer and portable computer to be stock are x and y respectively. The given information can be written in tabular form as given below:
Desktop computer  Portable computer  Availability  
Cost (in Rs.)  25,000  40,000  70,00,000 
No. of units  x  y  250 
Profit (in Rs.)  4500  5000 
The mathematical formulation of the given problem is:
Maximise Z = 4500x + 5000y … (i)
25,000x + 40,000y ≤ 70, 00,000
⇒ 5x + 8y ≤ 1400 … (ii)
x + y ≤ 250 … (iii)
x, y ≥ 0 … (iv)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
The corner points are A(250, 0), B(200, 50), and C(0, 175) and the values of Z at these corner points are given below:
Corner point  Z = 4500x + 5000y  
O(0, 0)  0  
A(250, 0)  11,25,000  
B(200, 50)  1150,000  → Maximum 
C(0, 175)  875,000 
The maximum value of Z is 1150,000 at the point (200, 50).
Therefore, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs. 11,50,000.
Q.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?
Ans
Let the mixture contain x units of food F_{1} and y units of food F_{2}. Clearly, x ≥ 0 and y≥ 0.
The table based on the given data is given below:
Food
F_{1} F_{2} 
Availability  
Vitamin A (units)  3 6  80 
Minerals (units)  4 3  100 
Cost (in Rs./unit)  4 6 
The mathematical formulation of the given problem is:
Minimise Z = 4x + 6y …(i)
3x + 6y ≥ 80 …(ii)
4x + 3y≥ 100 …(iii)
x ≥ 0,≥ 0 …(iv)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
Here, we see that the feasible region is unbounded.
Let us evaluate Z at the corner points A(80/3, 0),
B(24, 4/3) and C(0, 100/3).
Corner point  Z = 4x + 6y  
A(80/3, 0)  106.67  
B(24, 4/3)  104  → Minimum 
C(0, 100/3)  200 
In the table, the smallest value of Z is 104 at the point (24, 4/3). Since, the feasible region is unbounded, so we have to draw the graph of inequality 4x + 6y = 104 i.e., 2x + 3y = 52.Here, feasible region and open half plane has no common point. Therefore, the minimum value of Z is 104 attained at the point (24, 4/3). Hence, the minimum cost of diet that consists of mixture of 24 units of food F_{1} and (4/3) units is Rs.104.
Q.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} cost 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 farmer buy x kg of fertilizer F_{1} and y kg of fertilizer F_{2}. Therefore, x ≥ 0 and y ≥ 0. The given information can be complied in a table as follows.
F_{1}(x)  F_{2}(x)  Requirement  
Nitrogen(%)  10  5  14 
Phosphoric acid (%)  6  10  14 
Cost(Rs/kg)  6  5 
The mathematical formulation of the given problem is
Minimize Z = 6x + 5y …(i)
subject to the constraints,
10%of x + 5% of y ≥ 14
2x + y ≥ 280 … (2)
6% of x + 10% of y ≥ 14
3x + 5y ≥ 700… (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints from (ii) to (iv) is shown as follows:
In the graph the feasible region is unbounded. So, value of Z at the corner points A (700/3, 0), B(100, 80), C(0, 280).
Corner Point  Z = 6x + 5y  
A(700/3, 0)  1400  
B(100, 80)  1000  → Minimum 
C(0, 280)  1400 
In the table, we find that the smallest value of Z is 1000 at the point (100, 80). Therefore, we have to draw the graph of the inequality
6x + 5y < 1000
The resulting open half plane has no point common with the feasible region. Thus, the minimum value of Z is 1000 attained at the point (100, 80). Therefore, 100 kg of fertilizer F_{1} and 80 kg of fertilizer F_{2} should be used to minimize the cost. The minimum cost is Rs 1000.
Q.11
$\begin{array}{l}\mathrm{The}\mathrm{\hspace{0.33em}}\mathrm{corner}\mathrm{\hspace{0.33em}}\mathrm{points}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{feasible}\mathrm{\hspace{0.33em}}\mathrm{region}\mathrm{\hspace{0.33em}}\mathrm{determined}\mathrm{\hspace{0.33em}}\mathrm{by}\mathrm{\hspace{0.33em}}\mathrm{the}\\ \mathrm{following}\mathrm{\hspace{0.33em}}\mathrm{system}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{linear}\mathrm{\hspace{0.33em}}\mathrm{inequalities}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}+\mathrm{y}\le 10,\mathrm{x}+3\mathrm{y}\le 15,\mathrm{x},\mathrm{y}\ge 0\mathrm{\hspace{0.33em}}\mathrm{are}\mathrm{\hspace{0.33em}}\left(0,0\right),\left(5,0\right),\left(3,4\right)\mathrm{\hspace{0.33em}}\mathrm{and}\\ \left(0,5\right).\mathrm{\hspace{0.33em}}\mathrm{Let}\mathrm{\hspace{0.33em}}\mathrm{Z}=\mathrm{px}+\mathrm{qy},\mathrm{\hspace{0.33em}}\mathrm{where}\mathrm{\hspace{0.33em}}\mathrm{p},\mathrm{q}>0.\mathrm{\hspace{0.33em}}\mathrm{Condition}\mathrm{\hspace{0.33em}}\mathrm{on}\mathrm{\hspace{0.33em}}\mathrm{p}\mathrm{\hspace{0.33em}}\mathrm{and}\mathrm{\hspace{0.33em}}\mathrm{q}\\ \mathrm{so}\mathrm{\hspace{0.33em}}\mathrm{that}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{maximum}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}\mathrm{occurs}\mathrm{\hspace{0.33em}}\mathrm{at}\mathrm{\hspace{0.33em}}\mathrm{both}\mathrm{\hspace{0.33em}}\left(3,4\right)\mathrm{\hspace{0.33em}}\mathrm{and}\mathrm{\hspace{0.33em}}\left(0,5\right)\mathrm{\hspace{0.33em}}\mathrm{is}\\ \left(\mathrm{A}\right)\mathrm{p}=\mathrm{q}\left(\mathrm{B}\right)\mathrm{p}=2\mathrm{q}\left(\mathrm{C}\right)\mathrm{p}=3\mathrm{q}\left(\mathrm{D}\right)\mathrm{q}=3\mathrm{p}\end{array}$Ans
Since, Z is maximum at the points (3, 4) and (0, 5).
So, at (3, 4)
Z = 3p + 4q
And at (0, 5)
Z = 0 + 5q
According to given condition,
3p + 4q = 5q ⇒ 3p = q
Therefore, correct option is D.
FAQs (Frequently Asked Questions)
1. What are some tips to perform excellently in the Mathematics Class 12 board examination?
For subjects like Mathematics, practice is the key. The more time students spend with tools like the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2, the higher their chances of conceptually understanding the topics. Students are advised to make use of the many resourceful tools available on the Extramarks website, like the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 and try to solve as many questions as possible for all the topics before the examinations.
2. Should students practice from the solutions given in the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2?
The most important fact about Mathematics is that it is not a subject that is focused only on theory. Certain parts of Mathematics need to be understood in a practical manner by the students and most of it needs to be practised. The students can make use of the solutions available on the Extramarks website for practising, such as the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2.
The students are advised to take help from the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 when they are practising the questions that are given in Exercise 12.2. The NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 can help the students crosscheck their solutions and see if they are on the right track and following the right format for solving the questions.
Trying to remember the solutions given in the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 would not be of much use as the questions are based on the questions in the NCERT book, but they can be framed differently than they are framed in the book and expressed in a unique manner by the students.
If the students try to remember the solutions to the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2 and answer the questions that are asked in the final examinations, and they have different numeric values compared to the questions that were there in the solutions, the students can get very confused.
Instead of trying to remember the stepbystep process to the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2, the students should focus on practising with the help of these solutions. With this, they can conceptually understand the process and solve any question that comes in the examination related to the topic.
3. What are some resources, apart from the NCERT Solutions Class 12 Maths Chapter 12 Exercise 12.2, that are available on the Extramarks website that can be useful for the students for their Mathematics board examination preparation?
At Extramarks, there are different subject experts that are assigned to each subject. The experts have a lot of experience and a deep understanding of the topics that are covered in the NCERT books. These experts are responsible for coming up with resourceful tools for the students to help prepare for their board examinations.
One such resource is the NCERT Solutions for Class 12 Maths, Chapter 12, Exercise 12.2.Many other tools are available on the website. The students can make use of the past years’ papers, extra questions, and important questions when they want to practise different questions related to the same topic. There are also tools available for practising the theory part of Mathematics, such as revision notes, etc.
All these tools and many more are available on the Extramarks website, which can be easily accessed by the students. The students are advised to make use of all these throughout the academic yearStudents can plan their preparation in this manner to avoid stressful situations such as attempting to prepare for every topic and subject at the same time.