NCERT Solutions Class 12 Maths Chapter-12 (Linear Programming)Exercise 12.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-12 (Linear Programming)Exercise 12.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Class 12 Mathematics

Chapter-12 (Linear Programming)

Questions and answers given in practice

Chapter-12 (Linear Programming)

Exercise 12.1

Q1. Solve the Linear Programming Problems graphically:

Answer. The feasible region determined by the constraints, $x+y\le 4,x\ge 0,y\ge 0$ is as follows.  The corner points of the feasible region are 0 (0, 0), A (4, 0), and B (0, 4). The values of z at these points are as follows.  Therefore, the maximum value of Z is 16 at the point B (0 , 4).

Q2. Solve the Linear Programming Problems graphically:

Answer. The feasible region determined by the system of constraints, $x+2y\le 8,3x+2y\le 12,x\ge$ 0, and $y\ge 0,$ as follows.  The corner points of the feasible region are O (O, O), A (4, O), B (2, 3), and C (O, 4). The values of z at these corner points are as follows.  Therefore, the minimum value of Z is -12 at the point (4, 0).

Q3. Solve the Linear Programming Problems graphically:

Q4. Solve the Linear Programming Problems graphically:

Answer.   It can be seen that the feasible region is unbounded.  As the feasible region is unbounded, therefore, 7 may or may not be the minimum value For this, we draw the graph Of the inequality, 3x + 5y < 7, and check Whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 3x + 5y < 7 Therefore, The minimum value

Q5. Solve the Linear Programming Problems graphically:

Q6. Solve the Linear Programming Problems graphically:

Answer. The feasible region determined by the constraints, $2x+y\ge 3,x+2y\ge 6,x\ge 0$,$\ge$ is as follows.  The comer points of the feasible region are A (6,0) and B (0,3). The values of Z at these corner points are as follows.  It can be seen that 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

Q7. Show that the minimum of Z occurs at more than two points.

Answer.   The corner points of the feasible region are A (60, O), B (120, O), C (60, 30), and C) (40, 20) The values of Z at these corner points are as follows.  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).

Q8. Show that the minimum of Z occurs at more than two points.

Answer.   The corner points of the feasible region are A(O, 50), B(20, 40), C(50, 100), and D(0, 200). The values of Z at these corner points are as follows.  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).

Q9. Show that the minimum of Z occurs at more than two points.

Answer.   It can be seen that the feasible region is unbounded. The values of Z at comer points A (6, 0), B (4, 1), and C (3, 2) are as follows.  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.

Q10. Show that the minimum of Z occurs at more than two points. Z=x+y, subject to xy1,x+y0,x,y0

Answer. $x-y\le -1,-x+y\le 0,x,y\ge 0,$ is as follows