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# NCERT Solutions class 12 Maths Exercise 12.1

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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

### 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

### 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 .

### 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

### 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 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

### 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.

### 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

### 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

### 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

### 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

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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### 2 thoughts on “NCERT Solutions class 12 Maths Exercise 12.1”

1. Awesome explanation of questions

2. please clear 9 number graph otherwise all good..