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NCERT Solutions for Class 12 Maths Chapter 12 - Linear Programming Exercise 12.1

Last updated date: 11th Sep 2024
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NCERT Solutions for Maths Class 12 Chapter 12 Exercise 12.1 - FREE PDF Download

NCERT of Exercise 12.1 in Chapter 12 of Class 12 Maths introduces students to the concept of Linear Programming. This exercise focuses on formulating linear inequalities and understanding feasible regions, which are critical for solving optimization problems. Linear programming is widely used in various fields, such as economics, business, and engineering, to maximize or minimize a linear objective function subject to constraints.

Table of Content
1. NCERT Solutions for Maths Class 12 Chapter 12 Exercise 12.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 12 | Vedantu
3. Access NCERT Solutions for Maths Class 12 Chapter 12 - Linear Programming
4. CBSE Class 12 Maths Chapter 12 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Vedantu’s NCERT Solutions of ex 12.1 class 12 provide clear, step-by-step explanations to help students grasp these concepts effectively. By practicing these solutions, students can enhance their problem-solving skills and prepare well for their exams. Download the solutions PDF for detailed guidance.

Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 12 | Vedantu

• This exercise explains the basics of linear programming and its importance in solving optimization problems.

• The linear programming problem and its mathematical formulation involve creating a linear objective function subject to linear constraints.

• The graphical method of solving linear programming problems helps visualize the constraints and identify feasible regions on a graph.

• Feasible solutions are the set of points that satisfy all the constraints of the linear programming problem.

• Infeasible solutions are the set of points that do not meet the constraints and cannot be considered for optimization.

• The optimal (feasible) solution is the point within the feasible region that maximizes or minimizes the objective function, providing the best possible outcome under the given constraints.

• There are ten questions in Maths ex 12.1 class 12 which are fully solved by experts at Vedantu.

Competitive Exams after 12th Science

Access NCERT Solutions for Maths Class 12 Chapter 12 - Linear Programming

Exercise 12.1

1: Maximise $Z=3x+4y$

Subject to the constraints: $x+y\ge 4$, $x\ge 0$, $y\ge 0$

Ans: The constraints: $x+y\ge 4$, $x\ge 0$, and $y\ge 0$, determine the feasible region as shown below:

The values of $Z$at the corner points $O(0,0),A(4,0),B(0,4)$ of the feasible region are:

 Corner point $Z=3x+4y$ $O(0,0)$ $0$ $A(4,0)$ $12$ $B(0,4)$ $16$ Maximum

Hence, the maximum value of $Z$ is  $16$ at the point $B(0,4)$.

2: Minimise $Z=3x+4y$.

Subject to constraints $x+2y\le 8$, $3x+3y\le 12$,$x\ge 0$ and $y\ge 0$.

Ans: The constraints $x+2y\le 8$, $3x+3y\le 12$,$x\ge 0$ and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $O(0,0),A(4,0),B(2,3)$ and $C(0,4)$ of the feasible region are:

 Corner point $Z=3x+4y$ $O(0,0)$ $0$ $A(4,0)$ $-12$ Minimum $B(2,3)$ $6$ $C(0,4)$ $16$

Hence, the minimum value of $Z$ is $-12$ at the point $A(4,0)$.

3: Maximise $Z=5x+3y$

Subject to the constraints: $3x+5y\le 15$, $5x+2y\le 10$, $x\ge 0$, $y\ge 0$

Ans: The constraints: $3x+5y\le 15$, $5x+2y\le 10$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $O(0,0),A(2,0),B(0,3)$ and $C(\frac{20}{19},\frac{45}{19})$ of the feasible region are:

 Corner point $Z=5x+3y$ $O(0,0)$ $0$ $A(2,0)$ $10$ $B(0,3)$ $9$ $C(\frac{20}{19},\frac{45}{19})$ $\frac{235}{19}$ Maximum

Hence, the maximum value of $Z$ is $\frac{235}{19}$ at the point$C(\frac{20}{19},\frac{45}{19})$.

4: Minimise $Z=3x+5y$

Subject to the constraints: $x+3y\ge 3$, $x+y\ge 2$, $x\ge 0$, $y\ge 0$

Ans: The constraints $x+3y\ge 3$, $x+y\ge 2$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

Therefore, the feasible region is unbounded.

The values of $Z$at the corner points $A(3,0),B(\frac{3}{2},\frac{1}{2})$ and $C(0,2)$ of the feasible region are:

 Corner point $Z=3x+5y$ $A(3,0)$ $9$ $B(\frac{3}{2},\frac{1}{2})$ $7$ Smallest $C(0,2)$ $10$

Since, the feasible region is unbounded, $7$ may or may not be the minimum value of  $Z$.

So, we draw the graph of the inequality, $3x+5y<7$, and check if the resulting half-plane has common points with the feasible region.

As the feasible region has no common point with $3x+5y<7$, the minimum value of $Z$ is $7$ at the point $B(\frac{3}{2},\frac{1}{2})$.

5: Maximise $Z=3x+2y$

Subject to the constraints: $x+2y\le 10$, $3x+5y\le 15$, $x\ge 0$, $y\ge 0$

Ans: The constraints, $x+2y\le 10$, $3x+5y\le 15$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $A(5,0),B(4,3)$ and $C(0,5)$ of the feasible region are:

 Corner point $Z=3x+2y$ $A(5,0)$ $15$ $B(4,3)$ $18$ Maximum $C(0,5)$ $10$

Hence, the maximum value of $Z$ is $18$ at the point $B(4,3)$.

6: Minimise $Z=x+2y$

Subject to the constraints: $2x+y\ge 8$, $x+2y\ge 6$, $x\ge 0$, $y\ge 0$

Ans: The constraints $2x+y\ge 8$, $x+2y\ge 6$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $A(6,0)$ and $B(0,3)$ of the feasible region are:

 Corner point $Z=x+2y$ $A(6,0)$ $6$ $B(0,3)$ $6$

As the value of $Z$ at points $A(6,0)$ and $B(0,3)$ is same, we need to take any other point, for example, $(2,2)$ on line $x+2y=6$

Now, $Z=6$ .

Thus, the minimum value of $Z$ occurs at more than $2$ points. Hence, we can say that the value of $Z$ is minimum at every point on the line $x+2y=6$.

7: Minimise and Maximise $Z=5x+10y$

Subject to the constraints: $x+2y\le 120$, $x+y\ge 60$, $x-2y\ge 60$,$x\ge 0$, $y\ge 0$

Ans: The constraints $x+2y\le 120$, $x+y\ge 60$,$x-2y\ge 60$, $x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $A(60,0),B(120,0),C(60,30)$ and $D(40,20)$ of the feasible region are:

 Corner point $Z=5x+10y$ $A(60,0)$ $300$ Minimum $B(120,0)$ $600$ Maximum $C(60,30)$ $600$ Maximum $D(40,20)$ $400$

Hence, the minimum value of $Z$ is $300$ at the point $A(60,0)$and the maximum value of $Z$ is $600$ at all the points lying on the line segment joining the points $B(120,0)$ and $C(60,30)$.

8: Minimise and Maximise $Z=x+2y$

Subject to the constraints: $x+2y\ge 100$, $2x-y\le 0$, $2x+y\le 200$,$x\ge 0$, $y\ge 0$

Ans: The constraints $x+2y\ge 100$, $2x-y\le 0$,$2x+y\le 200$, $x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $A(0,50),B(20,40),C(50,100)$ and $D(0,200)$ of the feasible region are:

 Corner point $Z=x+2y$ $A(0,50)$ $100$ Minimum $B(20,40)$ $100$ Minimum $C(50,100)$ $250$ $D(0,200)$ $400$ Maximum

Hence, the maximum value of $Z$ is $400$ at the point $D(0,200)$and the minimum value of $Z$is $100$ at all the points lying on the line segment joining the points $A(0,50)$ and $B(20,40)$.

9: Maximise $Z=-x+2y$

Subject to the constraints: $x\ge 3$, $x+y\ge 5$, $x+2y\ge 6$ , $y\ge 0$

Ans: The constraints $x\ge 3$, $x+y\ge 5$,$x+2y\ge 6$, and $y\ge 0$ determine the feasible region as shown below:

It is visible that the feasible region is unbounded.

The values of $Z$ at corner points $A(6,0),B(4,1)$ and $C(3,2)$are given as:

 Corner point $Z=-x+2y$ $A(6,0)$ $-6$ $B(4,1)$ $-2$ $C(3,2)$ $1$

Since, the feasible region is unbounded, $Z=1$ may or may not be the maximum value.

So, we will draw the graph of the inequality, $-x+2y>1$, and check if the resulting half-plane has common points with the feasible region.

As the resulting feasible region has common points with the feasible region. Therefore, $Z=1$ is not the maximum value. Also, $Z$ has no maximum value.

10: Maximise $Z=x+y$

Subject to the constraints: $x-y\le -1$, $-x+y\le 0$, $x\ge 0$ , and $y\ge 0$

Ans: The constraints, $x-y\le -1$, $-x+y\le 0$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

It can be seen that there is no feasible region. Thus, $Z$ does not have any maximum value.

Conclusion

Exercise 12.1 of Chapter 12 in Class 12 Maths is essential for mastering linear programming. This exercise focuses on formulating linear programming problems, using the graphical method to solve them, and identifying feasible and optimal solutions. Understanding objective functions, constraints, and the feasible region is crucial. Vedantu's NCERT Solutions offer clear, step-by-step guidance, helping students grasp these concepts effectively. Practicing these solutions will enhance problem-solving skills and exam readiness. Concentrate on the graphical method and evaluating objective functions at corner points for optimal solutions.

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 12 Maths Chapter 12 - Linear Programming Exercise 12.1

1. What are the steps to solve linear programming?

Here are the steps to solve linear programming:

• Determine the factors that influence decisions.

• Compose the objective function.

• Determine the constraints

• Find the optimum point

2. What do you mean by Linear programming?

When a linear function is exposed to various constraints, it is maximised or reduced using the mathematical modelling technique known as "linear programming." In commercial planning and industrial engineering, this approach has worked well for guiding quantitative decisions in the physical and social sciences, though to a lesser extent.

3. How many questions are there in Class 12 Maths Chapter 12 Linear Programming (Ex 12.1) Exercise 12.1?

Class 12 Maths Chapter 12 Linear Programming (Ex 12.1) Exercise 12.1 consists of a total of ten conceptual questions.

4. Why should I practice class 12 maths NCERT chapter 12  Linear Programming (Ex 12.1) Exercise 12.1?

The NCERT book and solutions have been prepared by the best and most highly skilled educators and scholars and have created the content in such a way that all the maths concepts could be understood by each student. Practising exercise 12.1 will help you in clearing your basics about the concept of Linear programming and will help as you go further through the chapter. The NCERT math book has explained the concept of linear programming, on which the first exercise is based, with the help of solved examples that are written in the easiest way.

5. Where can I find the NCERT solutions for Class 12 Maths chapter 12  Linear Programming (Ex 12.1) Exercise 12.1?

NCERT solutions for class 12 maths chapter 12 linear programming (Ex 12.1) exercise 12.1 are free to download in pdf format from Vedantu. Teachers at Vedantu are highly skilled and experts in their subjects, and they have curated these NCERT solutions according to the latest CBSE pattern and guidelines.

6. What is the main focus of Class 12 Maths ex 12.1 solutions?

Class 12 Maths ex 12.1 solutions focuses on the basics of linear programming, including formulating linear programming problems, understanding constraints, and using the graphical method to find feasible and optimal solutions.

7. How does the graphical method help in solving linear programming problems in class 12 maths chapter 12 exercise 12.1?

The graphical method helps by visually representing constraints on a graph, identifying the feasible region, and determining the optimal solution by evaluating the objective function at the corner points of the feasible region in class 12 linear programming exercise 12.1.

8. What are the key components of a linear programming problem in class 12 exercise 12.1?

The key components include the objective function to be optimized, constraints represented by linear inequalities, and the feasible region where these constraints overlap, forming the solution space.

9. Why is understanding constraints important in linear programming in class 12 maths ch 12 ex 12.1?

In linear programming class 12 maths ch 12 ex 12.1 understanding constraints is crucial because they define the feasible region, which is the set of all possible solutions that satisfy the conditions of the problem. Only within this region can the optimal solution be found.

10. How do Vedantu’s NCERT Solutions assist with ex 12.1 class 12?

Vedantu’s NCERT Solutions of linear programming class 12 exercise 12.1 provide detailed, step-by-step explanations and graphical representations, making it easier for students to understand the concepts and effectively solve linear programming problems.