NCERT Solutions Class 12 Mathematics Chapter 12
NCERT Solutions Class 12 Mathematics Chapter 12 Linear Programming.
The Class 12 Mathematics textbook has a total of 13 chapters. Chapter 12 is about ‘Linear Programming’ that continues after the chapter ‘ThreeDimensional Geometry’. Linear Programming, also known as ‘Linear Optimisation’, is a method to get the best outcome in a Mathematical model. Linear programming is used for attaining the most optimal solution for a problem with given constraints. In NCERT Solutions Class 12 Mathematics Chapter 12 linear programming, you learn to formulate your actual problem into a Mathematical model. It involves an objective function, linear inequalities subject to constraints. Linear programming has been generally used in optimisation for several reasons. Linear programming is currently utilised in company management, such as Production, Planning, Technology, Transportation, and other related systems.
The majority of companies would like to maximise profit and minimise costs with limited resources. NCERT Solutions Class 12 Mathematics Chapter 12 Linear Programming guides you to learn the techniques of maximising profit with minimum expenses. The characteristics of linear programming are linearity, nonnegativity, finiteness, objective function, and constraints. Solutions provided quickly come in handy for completing the homework and preparing for the exams. The NCERT Solutions Class 12 Mathematics Chapter 12 focuses on helping students to make positive contributions to society by developing costeffective methods. It also suggests how a decisionmaker can employ his productive factors by selecting and allocating the resources. Thus the quality of decisions can be improved by applying Linear programming techniques.
NCERT Solutions Class 12 by Extramarks strengthen your comprehensive capabilities by helping you solve complex linear programming theories in simpler ways. NCERT Solutions Class 12 Mathematics Chapter 12 give students a solid introduction to linear programming Mathematics while studying both advantages and disadvantages of applying linear techniques to different situations under varied conditions.
Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 12
Sometimes there can be other constraints operating outside the problem. This subject teaches you the limitations of Linear Programming as this could be applied only when all the elements related to the problem can be quantified. Studying geometry provides many basic skills and builds logical thinking, analytical reasoning and problemsolving skills. The linear programming technique provides practical solutions to such issues. NCERT Solutions Class 12 Mathematics Chapter 12 teaches you to apply the fundamental plan for any given changing conditions.
The main topics covered in NCERT Solutions for Class 12 Mathematics Chapter 12Linear Programming are:
Exercise  Topic 
12.1  Introduction 
12.2  Linear Programming problems and Mathematical formulation 
12.3  A Mathematical formulation of the problems 
12.4  Graphical method of solving Linear Programming problems 
12.5  Different types of Linear Programming problems 

Introduction
Circumstances that benefit from Linear Programming problems include quality control decisions, product mix decisions, purchasing decisions, material utilisation decisions, production planning and warehousing decisions. NCERT Solutions Class 12 Mathematics Chapter 12 attracts the attention of students to deal with optimising objectives desired in situations that involve limitations. This Mathematics equation assists in creating a beneficial situation, especially in the manufacturing sector. NCERT Solutions Class 12 Mathematics Chapter 12 guides you to find the best obtainable value with the given conditions to be solved. This is quite useful in Economics, Operations research, and Business dealing with logistics. Characteristics of linear programming are :
 Linear programming concludes that all relevant input data and parameters are known with certainty.
 Linear programming’s single objective is to maximise or minimise the quantity for optimisation that caters to variable decisions.
 It helps understand the constraints that limit the ability to achieve the goal.

Linear Programming problems and Mathematical Formulation.
The Linear Programming method assists in achieving the best outcome in a Mathematical model. NCERT Solutions Class 12 Mathematics Chapter 12 educates the students on different forms of results such as maximum profit or the lowest cost, or the best possible price. The description of this model’s requirements is a result of linear relationships.
Types of linear programming problems:
 Diet problems: The objective function is the food intake, and constraints are specified nutritional requirements.
 Manufacturing problems: The objective function is the production rate, and constraints are work hours and the cost of the packaging material.
 Transportation problems: The objective function is transportation, and constraints are specific supply and demand patterns.
 Optimal assignment problems: The objective function is the total assignments done, and constraints are the number of employees and the work hours of each employee.
NCERT Solutions Class 12 Mathematics Chapter 12 helps you analyse various problems and get better results. The main factor of Mathematical formulations is the development of definitions of variables and the functional relationships of the variables required to model the circumstance under study. To formulate a Linear Programming model, recognise the problem data in terms of constants and parameters associated with decision variables.

A Mathematical Formulation of the Problems.
Linear Programming is a Mathematical programming technique in which a Linear function is maximised or minimised when subjected to various constraints. NCERT Solutions Class 12 Mathematics Chapter 12 explains in detail how this technique has been helpful in advising quantitative decisions in Industrial Engineering, Business Planning, Social as well as Physical Sciences. The process for the Mathematical formulation of a Linear Programming problem consists of the following steps.
 Find out the decision variable.
 Recognise the objective function to be maximised or minimised and express it as a Linear programming function of decision variables.
 Spot the set of constraint states and express them as linear inequalities or equations in terms of the decision variables.
NCERT Solutions Class 12 Mathematics Chapter 12, created by Extramarks, assists students through the process of implementing Mathematical formulas in varied industries.

Graphical Method of Solving Linear Programming Problems gives you stepbystep Solutions to all your Queries.
NCERT Solutions Class 12 Mathematics Chapter 12 helps you calculate the Linear Programming problems for which the following steps need to be considered.
 Formulate the linear programming issue – First, you should identify the issue in an industry under the given conditions.
 Build a graph and then plot the various constraints lines – Construct a chart or diagram to show the problem points and variables governing them.
 Discover the valid side of all constraint lines: This is most important in the operations and manufacturing industry.
 Identify the region of a practical solution: Having analysed the problem, apply the best linear programming solution.
 Plot the objective function: It’s vital to know the actual problem amongst the many and develop a solution that yields the best result.
 Ultimately, find out the optimum point: Among all the variables available, it’s imperative to conclude on a perfect point.
NCERT Solutions Class 12 Mathematics Chapter 12 illustrates graphical designs which are used to formulate Mathematical functions. Various plots are used to highlight the objectives and variables.

Different Types of Linear Programming Problems.
The different types of Linear Programming problems are Transportation, Diet, Manufacturing and Optimal Assignment Problems. NCERT Solutions Class 12 Mathematics Chapter 12 identifies the number of decision variables and expresses them in the form of Linear equations or Linear inequations.
Access NCERT Solutions Class 12 Mathematics Chapter 12 Exercises & Answer Solutions
Exercise 12.1 Solutions – 10 Questions (All Questions Require Graphs and Long Questions)
Exercise 12.2 – Linear Programming problems and Mathematical formulation – Questions & Solutions click on the link below.
Exercise 12.2 Solutions – 11 Questions (11 Questions Require Graphs and Long Questions)
Exercise 12.3 A Mathematical formulation of the problems – Questions & Solutions click here to know more.
Exercise 12.4 Graphical method of solving Linear Programming problems guides you through stepbystep solutions to the issues. – Questions & Solutions refer to the link below for further details.
Exercise 12.5 Different types of Linear Programming problems. – Questions & Solutions follow the links
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NCERT Exemplars Class 12 Mathematics Chapter 12 Linear Programming.
NCERT Exemplars are practice books that include additional questions of a higher level and are meant for aiding meticulous learning. They are mainly used for competitive exams. These topics are up to date and are explained thoroughly as per the syllabus of the Class 12 Mathematics. Students can click on the links of these exemplar questions and answers provided by Extramarks for Class 12 Mathematics chapters.
Considering the standard and variety, practising exemplar problems is essential for both school boards and competitive examinations. Students can perfectly evaluate themselves by working on these problems and improving their problemsolving skills for future examinations. NCERT Solutions Class 12 Mathematics Chapter 12 provides a detailed clarification to all questions given in the NCERT Class 12 books.
Key Features of NCERT Solutions Class 12 Mathematics Chapter 12 Linear Programming
To score well in competitive exams, one should have a solid base in Mathematics. NCERT Solutions Class 12 Mathematics Chapter 12 provide you with the solutions to some of the most demanding questions that improve your analytical skills and give you sufficient exposure to any kind of questions that may be asked in board or competitive exams. Our application is extremely beneficial for your preparation for the following reasons.
 NCERT Solutions Class 12 Mathematics Chapter 12 answers to all numerical problems in the textbook and are presented in stepbystep evaluations
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Q.1
$\begin{array}{l}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}=3\mathrm{x}+4\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{constraints}\mathrm{\hspace{0.33em}}:\mathrm{x}+\mathrm{y}\le 4,\mathrm{x}\ge 0,\mathrm{y}\ge 0.\end{array}$
Ans.
The feasible region determined by the constraints,
x + y ≤ 4 … (i)
x ≥ 0, y ≥ 0 …(ii)
is shown in graph:
The corner points of the feasible region are O (0, 0),
A (4, 0), and B (0, 4). The values of Z at these points are as follows.
Corner point  Z = 3x + 4y  
O(0, 0)  0  
A(4, 0)  12  
B(0, 4)  16  →Maximum 
Therefore, the maximum value of Z is 16 at the point B (0, 4).
Q.2
$\begin{array}{l}\mathrm{Minimise}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}=3\mathrm{x}+4\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{x}+\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\le ,\mathrm{x}\ge 0,\mathrm{y}\ge 0.\end{array}$
Ans.
$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{constraints},\\ \mathrm{x}+2\mathrm{y}\le 8...\left(\mathrm{i}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\mathrm{x}+2\mathrm{y}\le 12...\left(\mathrm{ii}\right)\\ \mathrm{x}\ge 0\mathrm{and}\mathrm{y}\ge \mathrm{0}...\left(\mathrm{iii}\right)\mathrm{}\\ \mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{graph}:\end{array}$
The corner points of the feasible region are O (0, 0), A (4, 0), B (2, 3), and C (0, 4).
The values of Z at these corner points are as follows:
Corner point  Z = – 3x + 4 y  
O (0, 0)  0  
A (4, 0)  –12  → Minimum 
B( 2, 3)  6  
C(0, 4)  16 
Therefore, the minimum value of Z is −12 at the point (4, 0).
Q.3
$\begin{array}{l}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}=5\mathrm{x}+3\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}3\mathrm{x}+\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\le ,\mathrm{x}\ge 0,\mathrm{y}\ge 0.\end{array}$
Ans.
The feasible region determined by the system of constraints,
3x + 5y ≤ 15 … (i)
5x + 2y ≤ 10 … (ii)
x ≥ 0, and y ≥ 0 … (iii)
is shown in the given graph:
$\begin{array}{l}\mathrm{The}\mathrm{corner}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{are}\mathrm{O}(0,\; 0),\mathrm{A}(2,\; 0),\\ \mathrm{B}(0,\; 3),\mathrm{and}\mathrm{C}(\frac{20}{19},\hspace{0.17em}\hspace{0.17em}\frac{45}{19}).\\ \mathrm{The}\mathrm{values}\mathrm{of}\mathrm{Z}\mathrm{at}\mathrm{these}\mathrm{corner}\mathrm{points}\mathrm{are}\mathrm{as}\mathrm{follows}:\end{array}$
Corner point  z = 5x + 3y  
O (0, 0)  0  
A (2, 0)  10  
B (0, 3)  9  
C\left(\frac{20}{19},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{45}{19}\right)  \frac{235}{19}  → Maximum 
$\begin{array}{l}\mathrm{Therefore},\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{is}\frac{235}{19}\mathrm{}\mathrm{at}\mathrm{the}\mathrm{point}\\ (\frac{20}{19},\hspace{0.17em}\hspace{0.17em}\frac{45}{19}).\end{array}$
Q.4
$\begin{array}{l}\mathrm{Minimise}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}=3\mathrm{x}+5\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{x}+3\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\ge ,\mathrm{\hspace{0.33em}}\mathrm{x},\mathrm{\hspace{0.33em}}\mathrm{y}\ge 0.\end{array}$
Ans.
\begin{array}{l}\text{The feasible region determined by the system of constraints,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{x}+\text{3y}\ge \text{3}\text{}\text{}\mathrm{\dots}\left(i\right)\\ \text{}\text{}\text{x}+\text{y}\ge \text{2}\text{}\text{}\mathrm{\dots}\left(ii\right)\\ \text{}\text{x,y}\ge \text{0}\text{}\text{}\text{}\mathrm{\dots}\left(iii\right)\\ \text{is shown in the following graph:}\end{array}
$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{graph},\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{unbounded}\mathrm{.}\\ \mathrm{The}\mathrm{corner}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{are}\mathrm{A}(3,\; 0),\mathrm{B}(\frac{3}{2},\mathrm{\hspace{0.17em}}\frac{1}{2}),\\ \mathrm{C}(0,\hspace{0.17em}\hspace{0.17em}2).\mathrm{The}\mathrm{values}\mathrm{of}\mathrm{Z}\mathrm{at}\mathrm{these}\mathrm{corner}\mathrm{points}\mathrm{are}\mathrm{as}\mathrm{follows}:\end{array}$
Corner point  Z = 3x + 5y  
A (3, 0)  9  
B\left(\frac{3}{2},\frac{1}{2}\right)  7  → Smallest 
C (0, 2)  10 
$\begin{array}{l}\mathrm{As}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{is}\mathrm{unbounded},\mathrm{therefore},\; 7\mathrm{may}\mathrm{or}\mathrm{may}\\ \mathrm{not}\mathrm{be}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{.}\\ \mathrm{For}\mathrm{this},\mathrm{we}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{the}\mathrm{inequality},\; 3\mathrm{x}+\; 5\mathrm{y}<\; 7,\mathrm{and}\\ \mathrm{check}\mathrm{whether}\mathrm{the}\mathrm{resulting}\mathrm{half}\mathrm{plane}\mathrm{has}\mathrm{points}\mathrm{in}\mathrm{common}\\ \mathrm{with}\mathrm{the}\mathrm{feasible}\mathrm{region}\mathrm{or}\mathrm{not}.\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{feasible}\\ \mathrm{region}\mathrm{has}\mathrm{no}\mathrm{common}\mathrm{point}\mathrm{with}3\mathrm{x}+\; 5\mathrm{y}<\; 7\mathrm{Therefore},\\ \mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{Z}\mathrm{is}7\mathrm{at}(\frac{3}{2},\frac{1}{2})\mathrm{.}\end{array}$
Q.5
$\begin{array}{l}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}=3\mathrm{x}+2\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{x}+2\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\ge ,\mathrm{\hspace{0.33em}}\mathrm{x},\mathrm{\hspace{0.33em}}\mathrm{y}\ge 0.\end{array}$
Ans.
The feasible region determined by the constraints,
x + 2y ≤ 10 …(i)
3x + y ≤ 15 …(ii)
x ≥ 0, y ≥ 0, …(iii)
is shown in the following graph:
The corner points of the feasible region are A (5, 0), B(4, 3), and C (0, 5).
The values of Z at these corner points are as follows:
Corner point  Z= 3x + 2y  
A(5,0)  15  
B(4, 3)  18  → Maximum 
C(0, 5)  10 
Therefore, the maximum value of Z is 18 at the point (4, 3).
Q.6
$\begin{array}{l}\mathrm{Minimise}\mathrm{\hspace{0.33em}}\mathrm{Z}=\mathrm{x}+2\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}2\mathrm{x}+\mathrm{y}\ge ,\mathrm{x}+2\mathrm{y}\ge ,\mathrm{x},\mathrm{y}\ge 0.\end{array}$
Ans.
$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints},\\ 2\mathrm{x}+\mathrm{y}\ge 3\mathrm{}\dots \left(\mathrm{i}\right)\\ \mathrm{x}+2\mathrm{y}\ge 6\dots \left(\mathrm{ii}\right)\\ \mathrm{}\mathrm{x}\ge 0,\mathrm{y}\ge 0,\mathrm{}\dots \left(\mathrm{iii}\right)\mathrm{}\\ \mathrm{}\mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{graph}:\end{array}$
The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y≥ 0, is as follows:
Corner point  Z = x + 2y 
A(6, 0)  6 
B(0, 3)  6 
Here, the value of Z at points A and B is same. If we take any other point such as (2, 2) on line
x + 2y = 6, then Z = 6.
Thus, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the line, x + 2y = 6.
Show that the minimum of Z occurs at more than two points.
Q.7
$\begin{array}{l}\mathrm{Minimise}\mathrm{\hspace{0.33em}}\mathrm{and}\mathrm{\hspace{0.33em}}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}=5\mathrm{x}+10\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{x}+2\mathrm{y}\le 120,\mathrm{x}+\mathrm{y}\ge ,\mathrm{x}2\mathrm{y}\ge 0;\mathrm{x},\mathrm{y}\ge 0.\end{array}$
Ans.
The feasible region determined by the constraints,
x + 2y ≤ 120 … (i)
x + y ≥ 60 …. (ii)
x−2y≥ 0 ….. (iii)
x, y ≥ 0 …… (iv)
is shown in the following graph:
The corner points of the feasible region are A (60, 0), B(120, 0), C (60, 30), and D (40, 20).
The values of Z at these corner points are as follows:
Corner Point  Z = 5x + 10 y  
A(60, 0)  300  → Minimum 
B(120, 0)  600  → Maximum 
C(60, 30)  600  → Maximum 
D(40, 20)  400 
The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (120, 0) and (60, 30).
Q.8
$\begin{array}{l}\mathrm{Minimise}\mathrm{\hspace{0.33em}}\mathrm{and}\mathrm{\hspace{0.33em}}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}=\mathrm{x}+2\mathrm{y}\\ \mathrm{Subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{x}+2\mathrm{y}\le 100,2\mathrm{x}\mathrm{y}\le ,2\mathrm{x}+\mathrm{y}\le 00;\mathrm{x},\mathrm{y}\ge 0.\end{array}$
Ans.
The feasible region determined by the constraints,
x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0, and y ≥ 0, is as follows:
The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100), and D(0, 200).
The values of Z at these corner points are as follows:
Corner point  Z = x + 2y  
A(0, 50)  100  → Minimum 
B(20, 40)  100  → Minimum 
C(50, 100)  250  
D(0, 200)  400  → Maximum 
The maximum value of Z is 400 at (0, 200) and the minimum value of Z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40)
Q.9
$\begin{array}{l}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.17em}}\mathrm{x}+2\mathrm{y},\mathrm{\hspace{0.33em}}\mathrm{subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{contraints}:\\ \mathrm{x}\ge ,\mathrm{x}+\mathrm{y}\ge ,\mathrm{x}+2\mathrm{y}\ge ;\mathrm{y}\ge 0.\end{array}$
Ans.
$\begin{array}{l}\mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints},\\ \mathrm{\hspace{0.17em}}\mathrm{x}\ge 3...\left(\mathrm{i}\right)\\ \mathrm{}\mathrm{x}+\mathrm{y}\ge 5...\left(\mathrm{ii}\right)\\ \mathrm{}\mathrm{x}+2\mathrm{y}\le \mathrm{6}...\left(\mathrm{iii}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\ge 0\mathrm{}...\left(\mathrm{iv}\right)\\ \mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{graph}:\end{array}$
.
It can be seen that the feasible region is unbounded. The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows:
Corner point  Z = – x + 2y 
A(6, 0)  – 6 
B(4, 1)  – 2 
C(3, 2)  1 
As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, −x + 2y > 1, and check whether the resulting half plane has points in common with the feasible region or not.
The resulting feasible region has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Z has no maximum value.
Q.10
$\begin{array}{l}\mathrm{Maximise}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{y},\mathrm{\hspace{0.33em}}\mathrm{subject}\mathrm{\hspace{0.33em}}\mathrm{to}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{contraints}:\\ \mathrm{x}\mathrm{y}\le ,\mathrm{x}+\mathrm{y}\le ;\mathrm{x},\mathrm{y}\ge 0.\end{array}$
Ans.
$\begin{array}{l}\mathrm{The}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints},\\ \mathrm{x}\mathrm{y}\le 1...\left(\mathrm{i}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}+\mathrm{y}\le 0...\left(\mathrm{ii}\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(\mathrm{iii}\right)\\ \mathrm{is}\mathrm{show}\mathrm{in}\mathrm{graph}:\end{array}$
There is no feasible region and thus, Z has no maximum value.
Q.11 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.12 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.13 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.14 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.15 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.16 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.17 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.18 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.19 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.20 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.21
$\begin{array}{l}\mathrm{The}\mathrm{\hspace{0.33em}}\mathrm{corner}\mathrm{\hspace{0.33em}}\mathrm{points}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{feasible}\mathrm{\hspace{0.33em}}\mathrm{region}\mathrm{\hspace{0.33em}}\mathrm{determined}\mathrm{\hspace{0.33em}}\mathrm{by}\mathrm{\hspace{0.33em}}\mathrm{the}\\ \mathrm{following}\mathrm{\hspace{0.33em}}\mathrm{system}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{linear}\mathrm{\hspace{0.33em}}\mathrm{inequalities}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}+\mathrm{y}\le 10,\mathrm{x}+3\mathrm{y}\le 15,\mathrm{x},\mathrm{y}\ge 0\mathrm{\hspace{0.33em}}\mathrm{are}\mathrm{\hspace{0.33em}}\left(0,0\right),\left(5,0\right),\left(3,4\right)\mathrm{\hspace{0.33em}}\mathrm{and}\\ \left(0,5\right).\mathrm{\hspace{0.33em}}\mathrm{Let}\mathrm{\hspace{0.33em}}\mathrm{Z}=\mathrm{px}+\mathrm{qy},\mathrm{\hspace{0.33em}}\mathrm{where}\mathrm{\hspace{0.33em}}\mathrm{p},\mathrm{q}>0.\mathrm{\hspace{0.33em}}\mathrm{Condition}\mathrm{\hspace{0.33em}}\mathrm{on}\mathrm{\hspace{0.33em}}\mathrm{p}\mathrm{\hspace{0.33em}}\mathrm{and}\mathrm{\hspace{0.33em}}\mathrm{q}\\ \mathrm{so}\mathrm{\hspace{0.33em}}\mathrm{that}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{maximum}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{Z}\mathrm{\hspace{0.33em}}\mathrm{occurs}\mathrm{\hspace{0.33em}}\mathrm{at}\mathrm{\hspace{0.33em}}\mathrm{both}\mathrm{\hspace{0.33em}}\left(3,4\right)\mathrm{\hspace{0.33em}}\mathrm{and}\mathrm{\hspace{0.33em}}\left(0,5\right)\mathrm{\hspace{0.33em}}\mathrm{is}\\ \left(\mathrm{A}\right)\mathrm{p}=\mathrm{q}\left(\mathrm{B}\right)\mathrm{p}=2\mathrm{q}\left(\mathrm{C}\right)\mathrm{p}=3\mathrm{q}\left(\mathrm{D}\right)\mathrm{q}=3\mathrm{p}\end{array}$
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.
Q.22 A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of Vitamin A. The diet requires at least 240 units of calcium, at least 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Ans.
Let x and y be the number of packets of food P and Q respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of the given problem is as follows:
Maximise Z = 6x + 3y (vitamin A)
subject to the constraints
4x + y ≥ 80 … (1)
x + 5y ≥ 115 … (2)
3x + 2y ≤ 150 … (3)
x ≥ 0, y ≥ 0 … (4)
The graph of the inequalities (i) to (iv) is drawn.
The feasible region (shaded) determined by the constraints (i) to (iv) is shown in Figure and it is bounded.
The coordinates of the corner points A, B and C are (2, 72), (15, 20) and (40, 15) respectively. So, the value of Z at these points:
Corner point  Z = 6x + 3y  
(2, 72)  228  
(15, 20)  150  
(40, 15)  285  → Maximum 
From the table, Z is maximum at the point (40, 15). Hence, the amount of vitamin A under the constraints given in the problem will be maximum, if 40 packets of food P and 15 packets of food Q are used in the special diet. The maximum amount of vitamin A will be 285 units.
Q.23 A 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 elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units 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?
Ans.
Let x and y be the number of bags of brand P and Q respectively. Obviously x ≥ 0, y ≥ 0.
The given information can be compiled in a table as follows:
Brand P  Brand Q  Availability
(units/kg) 

Vitamin A (units/kg)  3  1.5  18 
Vitamin B
(units/kg) 
2.5  11.25  45 
Vitamin C
(units/kg) 
2  3  24 
Cost
(Rs/kg) 
250  200 
Mathematical formulation of the given problem is as follows:
Maximise Z = 250x + 200y
subject to the constraints
3x + 1.5y ≥ 18
2x + y ≥ 12 … (1)
2.5x + 11.5y ≥ 45
5x + 23y ≥ 90 … (2)
2x + 3y ≥ 24 … (3)
x ≥ 0, y ≥ 0 … (4)
In the graph the feasible region is unbounded. So, value of Z at the corner points A (18, 0), B(9, 2), C(3, 6) and D(0, 12).
Corner point  Z = 250x + 200y  
A (18, 0)  4500  
B (9, 2)  2650  
C (3, 6)  1950  → Minimum 
D (0, 12)  2400 
In the table, we find that the smallest value of Z is 1950 at the point (3, 6).
Therefore, we have to draw the graph of the inequality 250x + 200y < 1950
Or 5x + 4y < 39.
The resulting open half plane has no point common with the feasible region.Thus, the minimum value of Z is 1950 attained at the point (3, 6). Thus, 3 bags of brand P and 6 bags of brand Q
should be used in the mixture to minimize the cost to Rs 1950.
Q.24 A 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 content of one kg food is given below:
Food  Vitamin A  Vitamin B  Vitamin C 
X  1  2  3 
Y  2  2  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?
Ans.
Let the mixture contains x kg of food X and y kg of food Y. Therefore, x ≥ 0 and y ≥ 0 The given information can be complied in a table as follows.
Food X  Food Y  Availability  
Vitamin A  1  2  10 
Vitamin B  2  2  12 
Vitamin C  3  1  8 
Cost(Rs/kg)  16  20 
The mathematical formulation of the given problem is as follows.
Minimize z = 16x + 20y … (1)
subject to the constraints,
x + 2y ≥ 10 … (2)
x + y ≥ 6 …(3)
3x + y ≥ 8 …(4)
x, y ≥ 0 …(5)
The feasible region determined by the system of constraints is as follows:
In the graph the feasible region is unbounded. So, value of Z at the corner points A (10, 0), B (2, 4), C (1, 5), and D (0, 8).
The values of z at these corner points are as follows:
Corner point  Z = 16x + 20y  
A (10, 0)  160  
B (2, 4)  112  → Minimum 
C (1, 5)  116  
D (0, 8)  160 
Since, the feasible region is unbounded; therefore, 112 may or may not be the minimum value of z.In the table, we find that the smallest value of Z is112 at the point (2, 4). Therefore, we have to draw the graph of the inequality 16x + 20y < 112
Or 4x + 5y < 28.
The resulting open half plane has no point common with the feasible region.
Thus, the minimum value of Z is 112 attained at the point (2, 4). Thus, the mixture should contain 2 kg of food X and 4 kg of food Y. The minimum cost of the mixture is Rs 112.
Q.25 A 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:
Type of toys  Machines  
I  II  III  
A  12  18  6 
B  6  0  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.
Ans.
Let x and y toys of type A and type B respectively be manufactured in a day.
The given information can be complied in a table as follows:
Toys/(time in hrs)  Toys type
A 
Toys type
B 
Availability(hrs) 
Machine, I  12  6  6 
Machine, II  18  0  6 
Machine, III  6  9  6 
If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5,then the given problem can be formulated as follows.
Maximize z = 7.5x + 5y … (1)
subject to the constraints,
2x + y ≤ 60 … (2)
x ≤ 20 … (3)
2x + 3y ≤ 120 … (4)
x, y ≥ 0 … (5)
The feasible region determined by the system of constraints is as follows:
The corner points of the feasible region are A (20, 0), B (20, 20), C (15, 30), and D (0, 40).
The values of z at these corner points can be calculated as follows:
Corner point  Z = 7.5x + 5y  
A(20, 0)  150  
B(20, 20)  250  
C(15, 30)  262.5  → Maximum 
D(0, 40)  200 
The maximum value of Z is 262.5 at (15, 30). Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.
Q.26 An 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?
Ans.
Let x tickets of executive class and y tickets of economy class of aeroplane are sold. The mathematical formulation of the given problem is as follows.
Maximize Z = 1000x + 600y … (1)
subject to the constraints,
x + y ≤ 200 … (2)
x ≥ 20 …(3)
y – 4x ≥ 0 … (4)
x, y ≥ 0 …(5)
The feasible region determined by the constraints is shown in figure:
The corner points of the feasible region are A(20, 80), B(40, 160), and C(20, 180).
The values of z at these corner points can be calculated as follows:
Corner point  Z = 1000x + 600y  
A (20, 80)  68000  
B(40, 160)  136000  →Maximum 
C(20, 180)  128000 
The maximum value of Z is 136000 at (40, 160). Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 136000.
Q.27 Two 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:
Transportation cost per quintal (in Rs)  
From/To  A  B 
D  7  3 
E  6  4 
F  3  2 
How should the supplies be transported in order that the transportation cost is minimum?
What is the minimum cost?
Ans.
Let x quintals and y quintals grain be transported from godown A to the ration shops D and E respectively. Then (150 – x – y) quintals grain will be transported to ration shop F. The requirement at shop D is 60 quintals since x quintals are transported from godown A. Therefore, the remaining (60 −x) quintals will be transported from godown B. Similarly, (50 − y) quintals and 40 − (100 − x − y) = (x + y − 60) quintals will be transported from godown B to shop E and F respectively. The diagram related to this problem is given below:
x ≥ 0, y ≥ 0 and 100 – x – y ≥ 0
⇒ x ≥ 0, y ≥ 0 and x + y ≥ 100
60 – x ≥ 0, 50 – y ≥ 0 and x + y – 60 ≥ 0
⇒ x ≤ 60, y ≤ 50 and x + y ≥ 0
Total transportation cost Z is given by,
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
= 2.5x + 1.5y + 410
The given problem can be formulated as Minimize z = 2.5x + 1.5y + 410 … (1)
subject to the constraints,
x + y ≤ 100 … (2)
x ≤ 60 … (3)
y ≤ 50 … (4)
x + y ≥ 60 … (5)
x, y ≥ 0 … (6)
The feasible region determined by the system of constraints is as follows:
The corner points are A (60, 0), B (60, 40),
C (50, 50), and D (10, 50).
The values of z at these corner points are as follows.
Corner point  Z = 2.5x + 1.5y + 410  
A(60, 0)  560  
B(60, 40)  620  
C(50, 50)  610  
D(10, 50)  510  → Minimum 
The minimum value of Z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
Q.28 An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distance (in km) between the depots and the petrol pumps is given in the following table:
Distance in (km)  
From/To  A  B 
D  7  3 
E  6  4 
F  3  2 
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Ans.
Let x L and y L oil be transported from depot A and B to the petrol pumps, D and E respectively. Then (7000 – x – y) L oil will be transported to petrol pump F. The requirement at petrol pump D is 4500L since x L are transported from depot A. Therefore, the remaining (4500 −x) L will be transported from depot B. Similarly, (3000 − y) L and 3500 − (7000 − x − y) = (x + y − 3500) L will be transported from depot B to petrol pumps E and F respectively.
The diagram related to this problem is given below:
x ≥ 0, y≥ 0 and 7000 – x – y ≥ 0
Þ x ≥ 0, y ≥ 0 and x + y ≤ 7000
4500 – x ≥ 0, 3000 – y ≥ 0 and x + y – 3500 ≥ 0
Þ x ≤ 4500, y ≤ 3000 and x + y ≥ 3500
Cost of transporting 10 L of petrol = Re 1
Cost of transporting 1 L of petrol = Rs (1/10)
Therefore, total transportation cost Z is given by,
$\begin{array}{l}\mathrm{Z}=\frac{7}{10}\mathrm{x}+\frac{6}{10}\mathrm{y}+\frac{3}{10}(7000\mathrm{x}\mathrm{y})+\frac{3}{10}(4500\mathrm{x})+\frac{4}{10}(3000\mathrm{y})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}+\frac{2}{10}(\mathrm{x}+\mathrm{y}3500)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0.7\mathrm{x}+0.1\mathrm{y}+3950\\ \mathrm{The}\mathrm{problem}\mathrm{can}\mathrm{be}\mathrm{formulated}\mathrm{as}\mathrm{follows}\mathrm{.}\\ \mathrm{Minimize}\mathrm{Z}=\mathrm{0}.3\mathrm{x}+\; 0.1\mathrm{y}+\; 3950...(1)\\ \mathrm{subject}\mathrm{to}\mathrm{the}\mathrm{constraints},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{y}\le \mathrm{7000}...\hspace{0.17em}\hspace{0.17em}\left(2\right)\\ \mathrm{x}\le 4500...\hspace{0.17em}\hspace{0.17em}\left(3\right)\\ \mathrm{y}\le 3000...\hspace{0.17em}\hspace{0.17em}\left(4\right)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{y}\le 3500...\hspace{0.17em}\hspace{0.17em}\left(5\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x},\mathrm{\hspace{0.17em}}\mathrm{y}\le 0...\hspace{0.17em}\hspace{0.17em}\left(6\right)\\ \mathrm{The}\mathrm{feasible}\mathrm{region}\mathrm{determined}\mathrm{by}\mathrm{the}\mathrm{constraints}\mathrm{is}:\end{array}$
The corner points of the feasible region are A(3500, 0), B(4500, 0), C(4500, 2500), D(4000, 3000), and E(500, 3000).
The values of Z at these corner points are given in the following table:
Corner point  Z = 0.3x + 0.1y + 3950 

A(3500, 0)  5000  
B(4500, 0)  5300  
C(4500, 2500)  5550  
D(4000, 3000)  5450  
E(500, 3000)  4400  → Minimum 
The minimum value of Z is 4400 at (500, 3000). Therefore, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively. The minimum transportation cost is Rs 4400.
Q.29 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
Kg per bag  
Brand P  Brand Q  
Nitrogen  3  3.5 
Phosphoric acid  1  2 
Potash  3  1.5 
Chlorine  1.5  2 
Ans.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Minimize Z = 3x + 3.5y … (1)
subject to the constraints,
x + 2y ≥ 240 … (2)
3x + 1.5y ≥ 270
or x + 0.5y ≥ 90 … (3)
1.5x + 2y ≥ 310 … (4)
x, y ≥ 0 … (5)
The feasible region determined by the system of constraints is given below:
The corner points are A (140, 50), B (20, 140) and (40, 100).
The values of Z at these corner points are given below:
Corner point  Z = 3x + 3.5y  
A (140, 50)  595  
B (20, 140)  550  
C (40, 100)  470  → Minimum 
The minimum value of Z is 470 at (40, 100).
Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimize the amount of nitrogen.The minimum amount of nitrogen added to the garden is 470 kg.
Q.30 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the maximum amount of nitrogen added in the garden?
Kg per bag  
Brand P  Brand Q  
Nitrogen  3  3.5 
Phosphoric acid  1  2 
Potash  3  1.5 
Chlorine  1.5  2 
Ans.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Maximize Z = 3x + 3.5y … (1)
subject to the constraints,
x + 2y ≥ 240 … (2)
3x + 1.5y ≥ 270
or x + 0.5y ≥ 90 … (3)
1.5x + 2y ≥ 310 … (4)
x, y ≥ 0 … (5)
The feasible region determined by the system of constraints is given below:
The corner points are A (140, 50), B (20, 140) and C (40, 100).
The values of Z at these corner points are given below:
Corner point  Z = 3x + 3.5y  
A (140, 50)  595  → Maximum 
B (20, 140)  550  
C (40, 100)  470 
The maximum value of Z is 595 at (140, 50).
Therefore, 140 bags of brand P and 50 bags of brand Q should be added to the garden to maximize the amount of nitrogen.
The maximum amount of nitrogen added to the garden is 595 kg.
Q.31 A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively on dolls A and B, How many of each should be produced weekly in order to maximize the profit?
Ans.
Let x and y be the number of dolls of type A and B respectively that are produced per week. The given problem can be formulated as follows.
Maximize Z = 12x + 16y … (1)
subject to the constraints,
x + y ≥ 1200 … (2)
y ≤ (x/2)
x ≥ 2y … (3)
x – 3y ≤ 600 … (4)
x, y ≥ 0 … (5)
The feasible region determined by the system of constraints is given below:
The corner points are A (600, 0), B (1050, 150), and C (800, 400).
The values of Z at these corner points are given in the following table:
Corner point  Z = 12x + 16y  
A(600, 0)  7200  
B(1050, 150)  15000  
C (800, 400)  16000  → Maximum 
The maximum value of z is 16000 at (800, 400). Therefore, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of ₹ 16000.
Related Chapters
FAQs (Frequently Asked Questions)
1. What is Linear Programming?
Linear Programming is a process of optimising the problems which are subjected to certain constraints. It’s a process of maximising or minimising the linear functions under linear inequality constraints.
2. What are the requirements of Linear Programming?
The five basic requirements of linear programming are Nonnegativity, Linearity, Objective function, Constraints and Finiteness. These are explained in detail in NCERT Solutions Class 12 Mathematics Chapter 12.