NCERT Solutions class 12 Maths Exercise 12.1

NCERT Solutions class 12 Maths Exercise 12.1 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

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NCERT Solutions class 12 Maths Linear Programming

Solve the following Linear Programming Problems graphically:

1.  Maximize Z =  subject to the constraints: .

Ans.  As  therefore we shall shade the other inequalities in the first quadrant only.

Now 

Let 

 

Thus the line has 4 and 4 as intercepts along the axes. Now, (0, 0) satisfies the inequation, i.e.,   Therefore, shaded region OAB is the feasible solution.

Its corners are O (0, 0), A (4, 0), B (0, 4)

At O (0, 0)    Z = 0

At A (4, 0)    Z = 3 x 4 = 12

At B (0, 4)    Z = 4 x 4 = 16

Hence, max Z = 16 at

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2.  Minimize Z =  subject to  

Ans.  Consider 

Let

 

 

Since, (0, 0) satisfies the inequaitons

Therefore, its solution contains (0, 0)

Again 

Let 

 

Again, (0, 0) satisfies

Therefore its solution contains (0, 0).

The feasible region is the solution set which is double shaded and is OABCO.

At  O (0, 0)  Z = 0

At  A (4, 0)  Z = –3 x 4 = –12

At  B (2, 3)  Z = –3 x 2 + 4 x 3 = 6

At  C (0, 4)  Z = 4 x 4 = 16

Hence, minimum Z = –12 at

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NCERT Solutions class 12 Maths Exercise 12.1

3.  Maximize Z =  subject to  

Ans.  We first draw the graph of equation

 

For     

And for    

	0	5
	3	0

 

	2	0
	0	5

Similarly, for equation , the points are (2, 0) and (0, 5).

As (0, 0) satisfies both the inequations and also  then the feasible require contains the half-plane containing (0, 0).

Therefore, the feasible portion is OABC which is shown as shaded in the graph.

Co-ordinates of point B can be obtained by solving  and  and it is B

Thus, co-ordinates of O, A, B and C are (0, 0), (2, 0),  and (0, 3).

Z =  (if )

Z = 5 x 2 + 3 x 0 = 10 (if )

Z = 5 x  + 3 x  =

(if )

Z = 5 x 0 + 3 x 3 = 9 (if )

Hence, Z =  is maximum when .

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NCERT Solutions class 12 Maths Exercise 12.1

4.  Minimize Z =  such that  

Ans.  For plotting the graphs of  and , we have the following tables:

	1	0
	1	2

 

	0	3
	1	0

The feasible portion represented by the inequalities

 and  is ABC which is shaded

 

in the figure. The coordinates of point B are

Which can be obtained by solving  and .

At A (0, 2)

Z = 3 x 0 + 5 x 2 = 10

At B

Z =

At C (3, 0)

Z = 3 x 3 + 5 x 0 = 9

Hence, Z is minimum is 7 when  and

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NCERT Solutions class 12 Maths Exercise 12.1

5.  Maximize Z =  subject to  

Ans.  Consider 

Let   

  

Since, (0, 0) satisfies the inequation, therefore the half plane containing (0, 0) is the required plane.

Again 

Let 

 

It also satisfies by (0, 0) and its required half plane contains (0, 0).

Now double shaded region in the first quadrant contains the solution.

Now OABC represents the feasible region.

Z =

At  O (0, 0)    Z = 3 x 0 + 2 x 0 = 0

At  A (5, 0)    Z = 3 x 5 + 2 x 0 = 15

At  B (4, 3)    Z = 3 x 4 + 2 x 3 = 18

At  C (0, 5)    Z = 3 x 0 + 2 x 5 = 10

Hence, Z is maximum i.e., 18 at

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NCERT Solutions class 12 Maths Exercise 12.1

6.  Minimize Z =  subject to

Ans.  Consider 

Let       

	0	3	–1
	3	–1	5

(0, 0) is not contained in the required half plane as (0, 0) does not satisfy the inequation .

Again 

Let 

 

Here also (0, 0) does not contain the required half plane. The double shaded region XABY is the solution set. Its corners are A (6, 0) and B (0, 3).

At A (6, 0)  Z = 6 + 0 = 6

At B (0, 3)  Z = 0 + 2 x 3 = 6

Therefore, at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z = 6 which is also minimum.

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NCERT Solutions class 12 Maths Exercise 12.1

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

7.  Minimize and Maximize Z =  subject to

Ans.  Consider 

Let   

 

The half plane containing(0, 0) is the required half plane as (0, 0) makes , true.

	0	30	60
	60	30	0

Again  

Let   

Also the half plane containing (0, 0) does not make  true.

Therefore, the required half plane does not contain (0, 0).

Again   

Let      

Let test point be (30, 0).

	0	30	60
	0	15	30

      30 – 2 x 0  0    It is true.

Therefore, the half plane contains (30, 0).

The region CFEKC represents the feasible region.

At  C (60, 0)    Z = 5 x 60 = 300

At  F (120, 0)    Z = 5 x 120 = 600

At  E (60, 30)    Z = 5 x 60 + 10 x 30 = 600

At  K (40, 20)    Z = 5 x 40 + 10 x 20 = 400

Hence, minimum Z = 300 at  and maximum Z = 600 at  or

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NCERT Solutions class 12 Maths Exercise 12.1

8.  Minimize and Maximize Z = subject to  

Ans.  Consider 

Let          

 represents which does not include (0, 0) as it does not made it true.

	0	25	50	100
	0	50	100	200

Again consider 

Let         

Let the test point be (10, 0).

 2 x 10 – 0  0 which is false.

Therefore, the required half does not contain (10, 0).

Again consider

Let 

 

Now (0, 0) satisfies

Therefore, the required half place contains (0, 0).

Now triple shaded region is ABCDA which is the required feasible region.

At  A (0, 50)

Z =  = 0 + 2 x 50 = 100

At  B (20, 40)  Z = 20 + 2 x 40 = 100

At  C (50, 100)  Z = 50 + 2 x 100 = 250

At  D (0, 200)  Z = 0 + 2 x 200 = 400

Hence maximum Z = 400 at  and minimum Z = 100 at  or

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NCERT Solutions class 12 Maths Exercise 12.1

9.  Maximize Z =  subject to the constraints:  

Ans.  Consider

Let  which is a line parallel to axis at a positive distance of 3 from it.

Since , therefore the required half-plane does not contain (0, 0).

Now consider  5

Let 

 

Now (0, 0) does not satisfy  5, therefore the required half plane does not contain (0, 0).

Again consider

Let 

 

Here also (0, 0) does not satisfy , therefore the required half plane does not contain (0, 0).

The corners of the feasible region are A (6, 0), B (4, 1) and C (3, 2).

At  A (6, 0)  Z = –6  + 2 x 0 = –6

At  B (4, 1)  Z = –4 + 2 x 1 = –2

At  C (3, 2)  Z = –3 + 2 x 2 = 1

Hence, maximum Z = 1 at

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NCERT Solutions class 12 Maths Exercise 12.1

10.  Maximize Z =  subject to  

Ans.  Consider 

Let 

 

	A	B	C	D
		0	0	3
	0	1	3	4

If (0, 0) is the test point then    which is false and thus the required plane does not include (0, 0).

Again 

Let 

 

	O	E	F
	0	1	2
	0	1	2

For (1, 0) – 1  0 which is true, therefore the required half-plane include (1, 0).

It is clear that the two required half planes do not intersect at all, i.e., they do not have a common region.

Hence there is no maximum Z.

NCERT Solutions class 12 Maths Exercise 12.1

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