Class 12 NCERT Solutions- Mathematics Part ii – Chapter 12– LINEAR PROGRAMMING Exercise 12.1

Question 1. Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

Solution:

Maximise Z = 3x + 4y subject to the constraints:

• x + y ≤ 4
• x ≥ 0
• y ≥ 0

Constraint y ≥ 0 & x ≥ 0 ⇒ Feasible region in the 1^st quadrant.

Table of values for line x + y = 4 corresponding to constraint :-

x	0	4
y	4	0

So let us draw the line joining the pts. (4, 0) & (0, 4).

Now let us test the origin (x = 0, y = 0) in the constraint x + y ≤ 4. This gives us 0 ≤ 4 which is true. Therefore, region for constraint is on the origin side of the line.

The shaded region of the figure is the feasible region determined by the system of constraints. The feasible region OAB is obtained.

The coordinates of the corner points O, A & C are (0, 0) , (4, 0) & (0, 4) respectively.

Now we evaluate Z at each corner point.

Corner points	Z = 3x + 4y
O(0, 0)	0
A(4, 0)	12
B(0, 4)	16 (Maximum)

Therefore, the maximum value of Z is 16 at the point B (0, 4).

Question 2. Minimise Z= -3x + 4y subject to the constraints : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Solution:

Maximise Z=3x + 4y subject to the constraints:

• x + 2y ≤ 8
• 3x + 2y ≤ 12
• x ≥ 0
• y ≥ 0

Constraint y ≥ 0 & x ≥ 0 ⇒ Feasible region in the 1st quadrant.

Table of values for line x + 2y = 8 corresponding to constraint :-

x	0	8
y	4	0

So let us draw the line joining the pts. (4, 0) & (8, 0).

Now let us test for origin (0, 0) in constraint which gives 0 ≤ 8 which is true.

Therefore, Region for constraint is on the origin side of the line.

Table of values for line 3x + 2y = 12 of constraint :

x	0	4
y	6	0

Let us draw the line, joining the points (0, 6) and (4, 0).

Now let us test for origin in constraint, which gives 0 ≤ 12 and which is true.

Therefore, the region for constraint (3) is also on the origin side of the line. The shaded region in the figure is the visible region determined by the system of constraints from (2) to (4). The feasible region, OABC is bounded.

The coordinates of the corner points O, A & C are (0, 0) , (4, 0) & (0, 4) respectively.

Now let us find corner point B, intersection of lines :-

x + 2y = 8 & 3x + 2y = 12

Subracting 2x = 4 ⇒ x = 4/2 ⇒ x =2.

Putting x = 2 in the above equation = 2+2y = 8 ⇒ 2y = 6 ⇒ y = 3.

Therefore, Corner point B is (2, 3).

Now we evaluate Z at each corner point.

Corner points	Z = -3x+4y
O(0, 0)	0
A(4, 0)	-12 (Minimum)
B(2, 3)	6
C(0,4)	16

Therefore, the minimum value of Z is −12 at the point (4, 0).

Question 3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Solution:

Maximise Z = 5x + 3y subject to the constraints :-

• 3x + 5y ≤ 15
• 5x + 2y ≤ 10
• x ≥ 0
• y ≥ 0

Constraint y ≥ 0 & x ≥ 0 ⇒ Feasible region in the 1st quadrant.

Table of values for line 3x + 5y = 15 corresponding to constraint :-

x	0	5
y	3	0

So let us draw the line joining the pts. (0, 3) & (5, 0).

Now let us test for origin (0, 0) in constraint which gives 0 ≤ 15 which is true.

Therefore, Region for constraint contains the origin.

Table of values for line 5x + 2y = 10 of constraint :

x	0	2
y	5	0

Let us draw the line, joining the points (0, 5) and (2, 0).

Now let us test for origin in constraint, which gives 0 ≤ 10 and which is true.

Therefore, the region for constraint (3) is also contains the origin. The shaded region in the figure is the visible region determined by the system of constraints from (2) to (4). The feasible region, OABC is bounded.

The coordinates of the corner points O,A & C are (0, 0) , (2, 0) & (0, 3) respectively.

Now let us find corner point B, intersection of lines :-

3x + 5y = 15 & 5x + 2y = 10

[1st equation *2 – 2nd equation] gives -19x = -20 ⇒ x =20/19.

Putting x = 20/19 in the first equation ⇒ 60/19 + 5y = 15

⇒ 5y = 15 – 60/19 = 225/19

⇒ y = 45/19.

Therefore, Corner point B is (20/19, 45/19).

Now we evaluate Z at each corner point.

Corner Points	Z = 5x+3y
O(0, 0)	0
A(2, 0)	10
B(20/19, 45/19)	235/19 (Maximum)
C(0, 3)	9

Therefore, the maximum value of Z is 235/19 at the point (20/19, 45/19).

Question 4. Minimise Z= 3x + 5y subject to the constraints : x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0.

Solution:

Minimise Z = 3x + 5y subject to the constraints :-

• x + 3y ≥ 3
• x + y ≥ 2
• x ≥ 0
• y ≥ 0

The feasible region determined by the the above constraints is given by :-

It can seen that the feasible region is unbounded.

The corner points of the feasible region are : A(3, 0) , B(3/2, 1/2) and C(0, 2). The values of Z at these corner points are as follows :-

Corner Points	Z = 3x + 3y
A(3, 0)	9
B(3/2, 1/2)	7 (Smallest)
C(0, 2)	10

As the visible region is unbounded, therefore 7 may or may not be the minimum value of Z.

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, minimum value of Z is 7 at B(3/2, 1/2).

Question 5. Maximise Z= 3x + 2y subject to the constraints : x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, y ≥ 0.

Solution:

Maximise Z = 3x + 5y subject to the constraints :-

• x + 2y ≤ 10
• 3x + y ≤ 15
• x ≥ 0
• y ≥ 0

Constraint y ≥ 0 & x ≥ 0 ⇒ Feasible region in the 1st quadrant. The feasible region determined by the the above constraints is given by :-

The corner points of the feasible region are A(5, 0) , B(4, 3) and C(0, 5). The values of Z at these corner points are as follows.

Corner Points	Z = 3x + 2y
A(5, 0)	15
B(4, 3)	18 (Maximum)
C(0, 5)	10

Therefore, the maximum value of Z is 18 at the point (4, 3).

Question 6. Minimise Z = x + 2y subject to the constraints : 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0.

Solution:

Minimise Z = x + 2y subject to the constraints :-

• 2x + y ≥ 3
• x + 2y ≥ 6
• x ≥ 0
• y ≥ 0

The feasible region determined by the the above constraints is given by :-

The corner points of the feasible region are A(6, 0) & B(0, 3). The values of Z at these corner points are as follows.

Corner Points	Z = x + 2y
A(6, 0)	6
B(0, 3)	6

It can be seen that the value of Z at points A and B is same. If we take any other points, such as (2, 2), on the line x + 2y = 6, then Z is equal to 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.

Question 7. Maximise and Minimise Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x ≥ 0, y ≥ 0.

Solution:

Maximise and Minimise Z = 5x + 10y subject to constraint as follow :-

• x + 2y ≤ 120
• x + y ≥ 60
• x ≥ 0
• y ≥ 0

The feasible region determined by the the above constraints is given by :-

The corner points of the feasible region are A(60, 0) , B(120, 0) , C(60, 30) & D(40, 20).

The values of Z at these corner points are as follows.

Corner Points	Z = 5x+10y
A(60, 0)	300 (Minimum)
B(120, 0)	600 (Maximum)
C(60, 30)	600 (Maximum)
D(40, 20)	400

The minimum value of Z is 300 at (60, 0) & the maximum value of Z is 600 at all the points on the line segment joining B(120, 0) & C(60, 30).

Question 8. Maximise and Minimise Z = 5x + 2y subject to x + 2y ≥ 120, 2x – y ≤ 0, 2x + y ≤ 200, y ≥ 0.

Solution:

Maximise and Minimise Z = 5x + 10y subject to constraint as follow :-

• x + 2y ≥ 120
• 2x – y ≤ 0
• 2x + y ≤ 200
• y ≥ 0

The feasible region determined by the the above constraints is given by :-

The corner points of the feasible region are A(0, 50) , B(20, 40) , C(50, 100) & D(0, 200).

The values of Z at these corner points are as follows.

Corner point	Z = x + 2y
A(0, 50)	100 (Minimum)
B(20, 40)	100 (Minimum)
C(50, 100)	250
D(0, 200)	400 (Maximum)

The minimum value of Z is 400 at (0, 200) & the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) & B(20, 40).

Question 9. maximise Z=-x + 2y, subject to the constraints x ≥ 3, x + 2y ≥ 6, y ≥ 0.

Solution:

Maximise Z = 3x + 5y subject to the constraints :-

• x ≥ 3
• x + y ≥ 5x
• x + 2y ≥ 6 y
• y ≥ 0

The feasible region determined by the the above constraints is given by :-

It can seen that the feasible region is unbounded.

The values of Z at corner points A(6, 0) , B(4, 1) and C(3, 2) are as follows :

Corner Points	Z = -x + 2y
A(6, 0)	-6
B(4, 1)	-2
C(3, 2)	1

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.

Question 10. Maximise Z = x + y, subject to x – y ≤ -1, -x + y ≤ 0, x ≥0, y ≥ 0.

Solution:

The region is determined by the constraints :-

• x – y ≤ -1
• -x + y ≤ 0
• x ≥ 0
• y ≥ 0

The feasible region determined by the the above constraints is given by :-

There is no feasible region and thus, Z has no Maximum value.

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