Linear Programming - Exercise-wise Questions and Answers For Class 12 Maths - Free PDF Download
In NCERT Solutions Class 12 Maths Chapter 12, you’ll explore the basics of Linear Programming—a cool chapter that shows you how to solve real-life problems by making the best possible choices within given limits. You’ll learn how to use graphs and simple equations to find solutions to situations where you need to maximize or minimize something, like profit or cost.
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With the step-by-step NCERT Solutions from Vedantu, tricky concepts like forming inequalities and finding the best answer become much easier. You can also grab a free, easy-to-understand PDF to help with your revision. For more help with your Class 12 syllabus, check out the Class 12 Maths syllabus.
If you want to boost your exam practice, don’t forget to review all the NCERT Solutions for Class 12 Maths. This chapter carries 5 marks in your CBSE exam.
Chapter 12 of NCERT Class 12 Maths deals with Linear Programming. Linear Programming is a technique used to find the optimal solution to a problem where we need to maximize or minimize a linear function subject to a set of constraints that are also linear in nature.
Exercise 12.1: This exercise consists of 10 questions that are based on formulating linear programming problems and solving them graphically. These questions help students understand the basic concepts and steps involved in solving linear programming problems.
Access NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming
Exercise 12.1
1. Maximize \[\mathbf{Z=3x+4y}\]
Subject to the constraints: \[\mathbf{x+y\le 4}\], \[\mathbf{x\ge 0}\], \[\mathbf{y\ge 0}\]
Ans: The given constraints are, \[x+y\le 4\], \[x\ge 0\], \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points O (0, 0), A (4, 0), and B (0, 4) at the corners
The points at the corners in the feasible region are O (0, 0), A (4, 0), and B (0, 4). Z assumes the following values on these points.
Table of values
Thus, the maximum value of Z is 16 at the point B (0,4).
2. Minimize Z=-3x+4y subject to \[\mathbf{x+2y\le 8}\], \[\mathbf{3x+2y\le 12}\], \[\mathbf{x\ge 0}\], \[\mathbf{y\ge 0}\]
Ans: The given constraints are, \[x+2y\le 8\], \[3x+2y\le 12\], \[x\ge 0\]and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points O (0, 0), A (4, 0), B (2, 3), and C (0, 4) at the corners
The points at the corners in the feasible region are O (0, 0), A (4, 0), B (2, 3), and C (0, 4). Z assumes the following values on these points.
Thus, the minimum value of Z is -12 at the point (4, 0).
3.Maximize \[\mathbf{Z=5x+3y}\] subject to \[\mathbf{3x+5y\le 15}\], \[\mathbf{5x+2y\le 10}\], \[\mathbf{x\ge 0}\], \[\mathbf{y\ge 0}\].
Ans: The given constraints are,\[3x+5y\le 15\], \[5x+2y\le 10\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points O (0, 0), A (2, 0), B (0, 3), at the corners
The points at the corners in the feasible region are O (0, 0), A (2, 0), B (0, 3), and \[C\left( \frac{20}{19},\frac{45}{19} \right)\]. Z assumes the following values on these points.
Thus, the maximum value of Z is \[\frac{235}{19}\] at the point \[\left( \frac{20}{19},\frac{45}{19} \right)\].
4.Minimize \[Z=3x+5y\]such that \[\mathbf{x+3y\ge 3}\], \[\mathbf{x+y\ge 2}\], \[\mathbf{x,y\ge 0}\]
Ans: The given constraints are, \[x+3y\ge 3\], \[x+y\ge 2\], and \[x,y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points A (0, 2), B (3, 0), at the corners
We can see that the feasible region is not bounded.
The points at the corners in the feasible region are A (3, 0), \[B\left( \frac{3}{2},\frac{1}{2} \right)\] , and C (0, 2). Z assumes the following values on these points.
Since the feasible region is unbounded, we cannot be sure that 7 is the minimum value of Z. To confirm this, we need to sketch the graph of the inequality, \[3x+5y<7\], and see if the resulting plane has any point in common with the feasible region.
From the graph that we sketched, we can see that there is no common point between feasible regions and the sketched inequality \[3x+5y<7\].
Z achieves minimum value 7 at \[\left( \frac{3}{2},\frac{1}{2} \right)\].
5.Maximize Z=3x+2y subject to \[\mathbf{x+2y\le 10}\], \[\mathbf{3x+y\le 15}\], \[\mathbf{x,y\ge 0}\].
Ans: The given constraints are, \[x+2y\le 10\], \[3x+y\le 15\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points A (5, 0), B (4, 3), and C (0, 5) at the corners
The points at the corners in the feasible region are A (5, 0), B (4, 3), and C (0, 5). Z assumes the following values on these points.
Z achieves maximum value 18 at (4,3).
6.Minimize \[Z=x+2y\]subject to \[2x+y\ge 3\], \[x+2y\ge 6\], \[x,y\ge 0\].
Ans:The given constraints are, \[2x+y\ge 3\], \[x+2y\ge 6\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points A (6, 0) and B (0, 3) at the corners
The points at the corners in the feasible region are A (6, 0) and B (0, 3). Z assumes the following values on these points.
We can see that the value of Z is the same on both A and B hence, we will need to check on the other point on the line \[x+2y=6\]as well. The value of Z is 6 at point (2,2) also. Hence, the minimum value of Z Occurs at more than two points
Thus, the value of Z is minimum at every point on the line, \[x+2y=6\]
7. Minimize and Maximize Z=5x+10y subject to\[\mathbf{x+2y\le 120}\], \[\mathbf{x+y\ge 60}\], \[\mathbf{x-2y\ge 0}\], \[\mathbf{x,y\ge 0}\].
Ans:The given constraints are, \[x+2y\le 120\], \[x+y\ge 60\], \[x-2y\ge 0\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points A (60, 0), B (120, 0), C (60, 30), and D (40, 20) at the corners
The points at the corners in the feasible region are A (60, 0), B (120, 0), C (60, 30), and D (40, 20). Z assumes the following values on these points.
Z achieves maximum and minimum values as 600 and 300 respectively. The point of maximum value is all the points on the line segment joining (120, 0) and (60, 30) and minimum value is (60,0).
8.Minimize and Maximize Z=x+2ysubject to \[\mathbf{x+2y\ge 100}\], \[\mathbf{2x-y\le 0}\], \[\mathbf{2x+y\le 200}\], \[\mathbf{x,y\ge 0}\].
Ans:The given constraints are, \[x+2y\ge 100\], \[2x-y\le 0\], \[2x+y\le 200\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is
Feasible Region having points A(0, 50), B(20, 40), C(50, 100), and D(0, 200) at the corners
The points at the corners in the feasible region are A(0, 50), B(20, 40), C(50, 100), and D(0, 200). Z assumes the following values on these points.
Z achieves maximum and minimum values as 400 and 100 respectively. The point of maximum value is (0,200) and minimum value is all points on the line joining the points (0, 50) and (20, 40).
9. Maximize Z=-x+2y, subject to the constraints: \[\mathbf{x\ge 3}\], \[\mathbf{x+y\ge 5}\], \[\mathbf{x+2y\ge 6}\], \[\mathbf{y\ge 0}\].
Ans: The given constraints are, \[x\ge 3\], \[x+y\ge 5\], \[x+2y\ge 6\], and \[y\ge 0\] and the feasible region which is in accordance with the given constraints is
Feasible Region having points A (6, 0), B (4, 1), and C (3, 2) at the corners
We can see that the feasible region is not bounded.
The points at the corners in the feasible region are A (6, 0), B (4, 1), and C (3, 2) . Z assumes the following values on these points.
Since the feasible region is unbounded, we cannot be sure that 1 is the maximum value of Z. To confirm this, we need to sketch the graph of the inequality, \[-x+2y>1\], and see if the resulting plane has any point in common with the feasible region.
From the graph that we sketched, we can see that there are common points between feasible regions and the sketched inequality. Thus, Z = 1 is not the maximum value. Z has no maximum value.
10. Maximize Z=x+y , subject to\[\mathbf{x-y\le -1}\], \[\mathbf{-x+y\le 0}\], \[\mathbf{x,y\ge 0}\].
Ans: The given constraints are, \[x-y\le -1\], \[-x+y\le 0\], \[x,y\ge 0\]and the feasible region which is in accordance with the given constraints is
Graph having No feasible region
We can see from the graph that there is no feasible region, hence Z has no maximum value.
Overview of Deleted Syllabus for CBSE Class 12 Maths Linear Programming
Class 12 Maths Chapter 12 : Exercises Breakdown
Conclusion
Chapter 12 Linear Programming of Class 12 Maths, focuses on the formulation and graphical solution of linear programming problems (LPP). It emphasizes the practical applications of optimizing a linear objective function subject to a set of linear inequalities or constraints. This chapter is crucial for understanding how mathematical techniques can be applied to real-world problems involving optimization. From previous year's question papers, around 4 questions are typically asked from this chapter. Understanding and practising these solutions will help students score well on their exams.
Other Study Material for CBSE Class 12 Maths Chapter 12
NCERT Solutions for Class 12 Maths | Chapter-wise List
Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.
Related Links for NCERT Class 12 Maths in Hindi
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FAQs on NCERT Solutions For Class 12 Maths Chapter 12 Linear Programming - 2025-26
1. How do you correctly formulate a Linear Programming Problem (LPP) as per the NCERT Class 12 Maths Chapter 12 methodology?
To formulate an LPP correctly as per the NCERT solutions, you must follow these steps:
- Identify Decision Variables: Determine the quantities to be found (e.g., x and y).
- Define the Objective Function: Write a linear equation (e.g., Z = ax + by) that you need to maximise or minimise.
- State the Constraints: Formulate a system of linear inequalities based on the conditions and limitations described in the problem.
- Include Non-Negativity Constraints: Always add the constraints x ≥ 0 and y ≥ 0, as the variables usually represent quantities that cannot be negative.
2. What are the key steps to solve an LPP graphically, as demonstrated in the NCERT solutions?
The NCERT solutions for Class 12 Maths Chapter 12 demonstrate a clear graphical method with the following steps:
- Plot the Constraints: Treat each inequality as an equation to draw a straight line. Shade the region that satisfies the inequality.
- Identify the Feasible Region: Determine the common area on the graph that satisfies all the constraints simultaneously. This is the feasible region.
- Find the Corner Points: Identify the coordinates of all vertices (corner points) of the feasible region.
- Evaluate the Objective Function: Substitute the coordinates of each corner point into the objective function (Z) to find its value.
- Determine the Optimal Solution: The point that gives the maximum or minimum value, as required by the problem, is the optimal solution.
3. How do the NCERT Solutions for Chapter 12 apply the Corner Point Method theorem to find the optimal value?
The NCERT solutions apply the Corner Point Method theorem by establishing that for any LPP, if an optimal solution exists, it must occur at a vertex (or corner point) of the feasible region. The solutions systematically identify all corner points of the shaded feasible region and then calculate the value of the objective function Z at each of these points. For a bounded region, the largest of these values is the maximum and the smallest is the minimum.
4. How can you correctly identify the feasible region when solving problems from NCERT Class 12 Maths Chapter 12?
To correctly identify the feasible region, first plot the line for each constraint. To determine which side of the line to shade, pick a test point, usually the origin (0, 0), if the line does not pass through it. Substitute this point into the inequality. If the inequality holds true, shade the side containing the origin. If it is false, shade the opposite side. The feasible region is the common overlapping area that satisfies all the constraints, including x ≥ 0 and y ≥ 0 (the first quadrant).
5. What does it signify if an NCERT solution states that an LPP has 'no feasible region'?
If the NCERT solution concludes that there is no feasible region, it means that there is no point (x, y) that can satisfy all the given constraints at the same time. The shaded regions for the inequalities do not overlap. Consequently, the problem has no solution, and the objective function cannot be maximised or minimised under the given conditions.
6. When solving an LPP from the NCERT textbook, how do you correctly handle an unbounded feasible region?
For an unbounded feasible region, you first find the optimal value at the corner points. However, this may not be the final answer. To confirm:
- For a maximisation problem with a supposed maximum value M, you must graph the inequality ax + by > M. If this new region has any points in common with the feasible region, there is no maximum value.
- For a minimisation problem with a supposed minimum value m, you must graph ax + by < m. If this region has no points in common with the feasible region, then m is the true minimum value.
7. Why is it sufficient to check only the corner points of a bounded feasible region to find the optimal solution in an LPP?
This is based on the Fundamental Theorem of Linear Programming. The theorem states that if a linear objective function has an optimal value (maximum or minimum) in a bounded, convex feasible region, that value must occur at one of its vertices (corner points). The linear nature of the objective function ensures it doesn't curve or dip between these points; its extreme values are always found at the boundaries, specifically at the corners where constraint lines intersect.
8. What is the key difference in the solving process for a 'maximization' versus a 'minimization' problem in the NCERT solutions?
While the initial steps of plotting constraints and finding the feasible region are identical, the final step differs significantly:
- For a maximization problem, you must identify the corner point that yields the highest value for the objective function Z.
- For a minimization problem, you must identify the corner point that results in the lowest value for Z.
This distinction is also crucial when testing for optimal solutions in unbounded regions.
9. According to the CBSE syllabus for 2025-26, how many exercises are in NCERT Class 12 Maths Chapter 12, Linear Programming?
As per the updated syllabus for the 2025-26 academic session, the NCERT Class 12 Maths Chapter 12 on Linear Programming primarily contains one main exercise (Exercise 12.1) and a miscellaneous exercise. The solutions are focused on the core skills of problem formulation and solving two-variable problems using the graphical method.
10. What does it mean if the NCERT solution for an LPP shows that the optimal value occurs at more than one corner point?
If the optimal value (maximum or minimum) of the objective function Z is the same at two adjacent corner points, it indicates that the LPP has infinitely many optimal solutions. Every point on the line segment connecting these two vertices will also be an optimal solution, providing the same maximum or minimum value for Z.
