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 

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 

Ans.  We first draw the graph of equation  

 

For     

And for    

	0	5
	3	0

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

	2	0
	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 .

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

	0	3
	1	0

 

	1	0
	1	2

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 

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 

Ans.  Consider   

Let       

	0	1	–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.

Ans.  Consider   

Let    

 

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

	0	30	60
	60	45	30

Again   x+y≥60x+y≥60

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 

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 

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 

Ans.  Consider   

Let  

 

	A	B	C	D
		0	2	3
	0	1	2	4

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

Again  

Let  −x+y=0−x+y=0

  

	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.