Question 1

The first step in formulating a linear programming problem is

Accepted Answer

Identify the decision variables — [{"id": "d19c2a0f-e0ca-edc3-3c6a-ff182e44d756", "type": "html", "value": " The first step in formulating a linear programming problem is Identify the decision variables Ans=Option 4 "}]

Question 2

A feasible solution of LPP

Accepted Answer

Must satisfy all the constraints simultaneously — [{"id": "ae4f41d8-e2c1-17d4-8044-2ff21f5c52d1", "type": "html", "value": " A feasible solution of LPP must satisfy all the constraints simultaneously Ans:Option 1 "}]

Question 3

The value of objective function is maximum under linear constraints

Accepted Answer

At the centre of feasible region — [{"id": "60e9882b-1716-6fb4-449b-68508d26ce72", "type": "html", "value": " The value of objective function is maximum under linear constraints at the centre of feasible region. Ans-Option 1 "}]

Question 4

Which of the following statement is correct?

Accepted Answer

If a LPP admits two optimal solutions it has an infinite number of optimal solutions — [{"id": "9b46f4e0-4e13-bda7-8d94-e0049dd44afc", "type": "html", "value": " If a LPP admits two optimal solutions it has an infinite number of optimal solution Ans=Option 3 "}]

Question 5

The optimal value of the objective function is attained at the points

Accepted Answer

Given by corner points of the feasible region — [{"id": "dc567ab4-354f-6d46-0f73-12233343c2af", "type": "html", "value": " The optimal value of the objective function is attained at the points given by corner points of the feasible region. Ans= Option 3 "}]

Question 6

Maximise  $Z = 3x + 4y$ subject to the constraints:  $x + y 64; 4, x 64; 0, y 64; 0$ .

Accepted Answer

To maximise Z = 3x + 4y subject to x + y 64; 4, x 64; 0, y 64; 0, follow these steps:

1. Identify the feasible region defined by the constraints:
   - x + y 64; 4
   - x 64; 0
   - y 64; 0

2. Plot the line x + y = 4 on the xy-plane. The feasible region is the triangle bounded by x=0, y=0, and x + y = 4.

3. Find the corner points of the feasible region:
   - (0,0)
   - (4,0)
   - (0,4)

4. Evaluate Z at each corner point:
   - At (0,0): Z = 3*0 + 4*0 = 0
   - At (4,0): Z = 3*4 + 4*0 = 12
   - At (0,4): Z = 3*0 + 4*4 = 16

5. The maximum value of Z is 16 at (0,4).

Hence, the maximum value of Z is 16 when x=0 and y=4. — The problem is a linear programming problem with two variables and linear constraints. The feasible region is a triangle in the first quadrant bounded by x + y 64; 4, x 64; 0, and y 64; 0. The maximum or minimum of the objective function occurs at the vertices of the feasible region. Evaluating Z at each vertex gives the maximum value at (0,4).

Question 7

Minimise $Z = -3x + 4y$ subject to $x + 2y 64; 8$, $3x + 2y 64; 12$, $x 64; 0$, $y 64; 0$.

Accepted Answer

To minimise Z = -3x + 4y subject to the constraints:

x + 2y 64; 8
3x + 2y 64; 12
x 64; 0
 y 64; 0

Steps:
1. Plot the constraints to find the feasible region.

2. Find the intersection points (corner points) of the feasible region:
- Intersection of x + 2y = 8 and 3x + 2y = 12:
  Subtracting equations:
  (3x + 2y) - (x + 2y) = 12 - 8
  2x = 4 => x = 2
  Substitute x=2 into x + 2y = 8:
  2 + 2y = 8 => 2y = 6 => y = 3
  So, point A = (2,3)

- Intersection with axes:
  For x + 2y = 8:
  x=0 => y=4 (point B)
  y=0 => x=8 (point C)

For 3x + 2y = 12:
  x=0 => y=6 (point D)
  y=0 => x=4 (point E)

3. The feasible region is bounded by points where all constraints are satisfied and x,y 64; 0.

4. Evaluate Z at corner points of feasible region:
- At (0,4): Z = -3*0 + 4*4 = 16
- At (2,3): Z = -3*2 + 4*3 = -6 + 12 = 6
- At (4,0): Z = -3*4 + 4*0 = -12

5. Minimum value of Z is -12 at (4,0).

Hence, the minimum value of Z is -12 at x=4, y=0. — The feasible region is determined by the intersection of the constraints and the non-negativity conditions. The minimum or maximum of the objective function occurs at the vertices of the feasible region. Evaluating Z at these points shows the minimum at (4,0).

Question 8

Maximise $Z = 5x + 3y$ subject to $3x + 5y 64; 15$, $5x + 2y 64; 10$, $x 64; 0$, $y 64; 0$.

Accepted Answer

To maximise Z = 5x + 3y subject to:
3x + 5y 64; 15
5x + 2y 64; 10
x 64; 0
 y 64; 0

Steps:
1. Find corner points of feasible region by solving the constraints:

- Intersection of 3x + 5y = 15 and 5x + 2y = 10:
  Multiply first by 2: 6x + 10y = 30
  Multiply second by 5: 25x + 10y = 50
  Subtract: (25x - 6x) + (10y - 10y) = 50 - 30
  19x = 20 => x = 20/19
  Substitute x into 3x + 5y = 15:
  3*(20/19) + 5y = 15
  60/19 + 5y = 15
  5y = 15 - 60/19 = (285/19) - (60/19) = 225/19
  y = 225/(19*5) = 225/95 = 45/19

- Intercepts:
  For 3x + 5y = 15:
  x=0 => y=3
  y=0 => x=5

For 5x + 2y = 10:
  x=0 => y=5
  y=0 => x=2

2. Feasible region is bounded by these constraints and x,y 64; 0.

3. Evaluate Z at corner points:
- (0,0): Z=0
- (0,3): Z=5*0 + 3*3=9
- (5,0): Z=5*5 + 3*0=25
- (2,0): Z=5*2 + 3*0=10
- (20/19, 45/19): Z=5*(20/19) + 3*(45/19) = (100/19) + (135/19) = 235/19  12.37

4. Maximum Z is 25 at (5,0).

Hence, maximum value of Z is 25 at x=5, y=0. — The feasible region is determined by the intersection of the constraints and non-negativity conditions. The objective function is evaluated at each vertex of the feasible region to find the maximum value.

Question 9

Minimise $Z = 3x + 5y$ such that $x + 3y 2; 3$, $x + y 2; 2$, $x, y 64; 0$.

Accepted Answer

To minimise Z = 3x + 5y subject to:
x + 3y 2; 3
x + y 2; 2
x, y 64; 0

Steps:
1. Since constraints are 2; (greater than or equal to), the feasible region lies above or on the lines.

2. Find intersection points of the constraints:
- Intersection of x + 3y = 3 and x + y = 2:
  Subtract second from first:
  (x + 3y) - (x + y) = 3 - 2
  2y = 1 => y = 0.5
  Substitute y=0.5 into x + y = 2:
  x + 0.5 = 2 => x = 1.5
  So, point A = (1.5, 0.5)

3. Check intercepts:
- For x + 3y = 3:
  x=0 => y=1
  y=0 => x=3
- For x + y = 2:
  x=0 => y=2
  y=0 => x=2

4. Feasible region is the intersection of half-planes x + 3y 2; 3 and x + y 2; 2 with x,y 64; 0.

5. Evaluate Z at corner points:
- At (1.5, 0.5): Z = 3*1.5 + 5*0.5 = 4.5 + 2.5 = 7
- At (0,2): Z = 3*0 + 5*2 = 10
- At (3,0): Z = 3*3 + 5*0 = 9

6. Minimum Z is 7 at (1.5, 0.5).

Hence, minimum value of Z is 7 at x=1.5, y=0.5. — The feasible region is the intersection of the half-planes defined by the inequalities and the non-negativity constraints. The minimum of the objective function occurs at one of the vertices of the feasible region. Evaluating Z at these vertices shows the minimum at (1.5, 0.5).

Question 10

Maximise $Z = 3x + 2y$ subject to $x + 2y 64; 10$, $3x + y 64; 15$, $x, y 64; 0$.

Accepted Answer

To maximise Z = 3x + 2y subject to:
x + 2y 64; 10
3x + y 64; 15
x, y 64; 0

Steps:
1. Find corner points by solving constraints:

- Intersection of x + 2y = 10 and 3x + y = 15:
  From first: x = 10 - 2y
  Substitute into second:
  3(10 - 2y) + y = 15
  30 - 6y + y = 15
  30 - 5y = 15
  5y = 15
  y = 3
  Then x = 10 - 2*3 = 10 - 6 = 4
  So point A = (4,3)

- Intercepts:
  For x + 2y = 10:
  x=0 => y=5
  y=0 => x=10

For 3x + y = 15:
  x=0 => y=15
  y=0 => x=5

2. Feasible region is bounded by these constraints and x,y 64; 0.

3. Evaluate Z at corner points:
- (0,0): Z=0
- (0,5): Z=3*0 + 2*5=10
- (4,3): Z=3*4 + 2*3=12 + 6=18
- (5,0): Z=3*5 + 2*0=15

4. Maximum Z is 18 at (4,3).

Hence, maximum value of Z is 18 at x=4, y=3. — The feasible region is the intersection of the half-planes defined by the inequalities and the non-negativity constraints. The maximum of the objective function occurs at one of the vertices of the feasible region. Evaluating Z at these vertices shows the maximum at (4,3).

Question 11

Minimise $Z = x + 2y$ subject to $2x + y 2; 3$, $x + 2y 2; 6$, $x, y 64; 0$. Show that the minimum of $Z$ occurs at more than two points.

Accepted Answer

Given:
Minimise Z = x + 2y
Subject to:
2x + y 2; 3
x + 2y 2; 6
x, y 64; 0

Steps:
1. Since constraints are 2; (greater than or equal to), feasible region lies above or on the lines.

2. Find intersection points:
- Intersection of 2x + y = 3 and x + 2y = 6:
  Multiply second by 2: 2x + 4y = 12
  Subtract first: (2x + 4y) - (2x + y) = 12 - 3
  3y = 9 => y = 3
  Substitute y=3 into 2x + y = 3:
  2x + 3 = 3 => 2x = 0 => x = 0
  So point A = (0,3)

3. Check intercepts:
- For 2x + y = 3:
  x=0 => y=3
  y=0 => x=1.5
- For x + 2y = 6:
  x=0 => y=3
  y=0 => x=6

4. The feasible region is the intersection of half-planes above these lines and x,y 64; 0.

5. Evaluate Z = x + 2y on the line segment joining (0,3) and (0,3) (actually a single point) and check along the boundary.

6. Note that the objective function Z = x + 2y can be rewritten as Z = x + 2y.

7. On the line 2x + y = 3, express y = 3 - 2x.

8. Substitute into Z:
Z = x + 2(3 - 2x) = x + 6 - 4x = 6 - 3x

9. For x 64; 0, Z 64; 6, minimum at x = 1.5, y = 0 (since y = 3 - 2x)

10. On the line x + 2y = 6, y = (6 - x)/2

11. Substitute into Z:
Z = x + 2 * (6 - x)/2 = x + 6 - x = 6

12. So on the line x + 2y = 6, Z = 6 constant.

13. The feasible region includes points where Z = 6 on the line segment between (0,3) and (6,0).

14. Hence, minimum value of Z is 6 and occurs at all points on the line segment x + 2y = 6 within the feasible region.

Therefore, the minimum of Z occurs at more than two points (all points on the line segment of the feasible region where Z=6). — The objective function Z = x + 2y is constant along the line x + 2y = 6, which is part of the boundary of the feasible region. Since the feasible region includes this entire line segment, the minimum value of Z occurs at infinitely many points along this segment, not just at one or two points.

Question 12

Minimise and Maximise $Z = 5x + 10y$ subject to $x + 2y 64; 120$, $x + y 2; 60$, $x - 2y 2; 0$, $x, y 64; 0$.

Accepted Answer

Given:
Minimise and Maximise Z = 5x + 10y
Subject to:
x + 2y 64; 120
x + y 2; 60
x - 2y 2; 0
x, y 64; 0

Steps:
1. Identify feasible region bounded by constraints.

2. Find intersection points of constraints:
- Intersection of x + 2y = 120 and x + y = 60:
  From x + y = 60 => y = 60 - x
  Substitute into x + 2y = 120:
  x + 2(60 - x) = 120
  x + 120 - 2x = 120
  -x + 120 = 120
  -x = 0 => x = 0
  Then y = 60 - 0 = 60
  So point A = (0,60)

- Intersection of x + y = 60 and x - 2y = 0:
  From x - 2y = 0 => x = 2y
  Substitute into x + y = 60:
  2y + y = 60 => 3y = 60 => y = 20
  Then x = 2*20 = 40
  So point B = (40,20)

- Intersection of x + 2y = 120 and x - 2y = 0:
  From x - 2y = 0 => x = 2y
  Substitute into x + 2y = 120:
  2y + 2y = 120 => 4y = 120 => y = 30
  Then x = 2*30 = 60
  So point C = (60,30)

3. Check non-negativity constraints.

4. Evaluate Z at corner points:
- At (0,60): Z = 5*0 + 10*60 = 600
- At (40,20): Z = 5*40 + 10*20 = 200 + 200 = 400
- At (60,30): Z = 5*60 + 10*30 = 300 + 300 = 600

5. Also check at (0,0): Z=0

6. Minimum Z is 0 at (0,0)

7. Maximum Z is 600 at (0,60) and (60,30)

Hence, minimum value of Z is 0 at (0,0) and maximum value of Z is 600 at (0,60) and (60,30). — The feasible region is bounded by the given constraints and non-negativity conditions. The objective function is evaluated at all vertices of the feasible region to find minimum and maximum values. The maximum occurs at two points, minimum at the origin.

Question 13

Minimise and Maximise $Z = x + 2y$ subject to $x + 2y 2; 100$, $2x - y 64; 0$, $2x + y 64; 200$; $x, y 64; 0$.

Accepted Answer

Given:
Minimise and Maximise Z = x + 2y
Subject to:
x + 2y 2; 100
2x - y 64; 0
2x + y 64; 200
x, y 64; 0

Steps:
1. Express constraints:
- x + 2y 2; 100 => feasible region is above or on the line x + 2y = 100
- 2x - y 64; 0 => y 64; 2x
- 2x + y 64; 200
- x, y 64; 0

2. Find intersection points:
- Intersection of x + 2y = 100 and 2x - y = 0:
  From 2x - y = 0 => y = 2x
  Substitute into x + 2y = 100:
  x + 2(2x) = 100
  x + 4x = 100
  5x = 100 => x = 20
  y = 2*20 = 40
  Point A = (20,40)

- Intersection of 2x - y = 0 and 2x + y = 200:
  From 2x - y = 0 => y = 2x
  Substitute into 2x + y = 200:
  2x + 2x = 200
  4x = 200 => x = 50
  y = 2*50 = 100
  Point B = (50,100)

- Intersection of x + 2y = 100 and 2x + y = 200:
  Multiply first by 1: x + 2y = 100
  Multiply second by 2: 4x + 2y = 400
  Subtract first from second:
  (4x + 2y) - (x + 2y) = 400 - 100
  3x = 300 => x = 100
  Substitute into x + 2y = 100:
  100 + 2y = 100 => 2y = 0 => y = 0
  Point C = (100,0)

3. Check non-negativity constraints.

4. Evaluate Z = x + 2y at points:
- (20,40): Z = 20 + 2*40 = 20 + 80 = 100
- (50,100): Z = 50 + 2*100 = 50 + 200 = 250
- (100,0): Z = 100 + 0 = 100
- (0,0): Z=0

5. Feasible region is above x + 2y = 100, so (0,0) is not feasible.

6. Minimum Z is 100 at (20,40) and (100,0)

7. Maximum Z is 250 at (50,100)

Hence, minimum value of Z is 100 at points (20,40) and (100,0), maximum value is 250 at (50,100). — The feasible region is defined by the intersection of the given inequalities and non-negativity constraints. The objective function is evaluated at the vertices of the feasible region to find minimum and maximum values.

Question 14

Maximise $Z = -x + 2y$, subject to the constraints: $x 64; 3$, $x + y 2; 5$, $x + 2y 2; 6$, $y 64; 0$.

Accepted Answer

Given:
Maximise Z = -x + 2y
Subject to:
x 64; 3
x + y 2; 5
x + 2y 2; 6
y 64; 0

Steps:
1. Identify feasible region bounded by constraints.

2. Find intersection points:
- Intersection of x + y = 5 and x + 2y = 6:
  Subtract first from second:
  (x + 2y) - (x + y) = 6 - 5
  y = 1
  Substitute y=1 into x + y = 5:
  x + 1 = 5 => x = 4
  Point A = (4,1)

- Check x 64; 3 constraint:
  x=4 violates x 64; 3, so point A is not feasible.

3. Check intersection of x = 3 and x + y = 5:
  x=3 => y = 2
  Point B = (3,2)

4. Check intersection of x = 3 and x + 2y = 6:
  x=3 => 3 + 2y = 6 => 2y = 3 => y = 1.5
  Point C = (3,1.5)

5. Check intersection of x + y = 5 and y=0:
  y=0 => x=5
  But x 64; 3, so not feasible.

6. Check intersection of x + 2y = 6 and y=0:
  y=0 => x=6
  x 64; 3, so not feasible.

7. Feasible region vertices are:
- (3,1.5)
- (3,2)
- (0,3) from x 64; 3 and x + y 2; 5
- (0,0)

8. Evaluate Z at feasible vertices:
- (3,1.5): Z = -3 + 2*1.5 = -3 + 3 = 0
- (3,2): Z = -3 + 4 = 1
- (0,3): Z = 0 + 6 = 6
- (0,0): Z = 0

9. Maximum Z is 6 at (0,3).

Hence, maximum value of Z is 6 at x=0, y=3. — The feasible region is bounded by the constraints and non-negativity conditions. The maximum value of the objective function occurs at one of the vertices of the feasible region. Evaluating Z at these vertices shows the maximum at (0,3).

Question 15

Maximise $Z = x + y$, subject to $x - y 64; -1$, $-x + y 64; 0$, $x, y 64; 0$.

Accepted Answer

Given:
Maximise Z = x + y
Subject to:
x - y 64; -1
-x + y 64; 0
x, y 64; 0

Steps:
1. Rewrite constraints:
- x - y 64; -1 => y 2; x + 1
- -x + y 64; 0 => y 64; x
- x, y 64; 0

2. From inequalities:
 y 2; x + 1 and y 64; x

3. For y to satisfy both, x + 1 64; y 64; x
 This implies x + 1 64; x
 Which is impossible unless 1 64; 0, which is false.

4. Hence, no feasible region exists satisfying all constraints with x,y 64; 0.

5. Therefore, the problem has no feasible solution and Z cannot be maximised.

Hence, no solution exists for the given constraints. — The constraints are contradictory when combined with the non-negativity conditions, leading to an empty feasible region. Therefore, the linear programming problem has no feasible solution.

Linear Programming

Linear Programming — Study Notes

12.1 Introduction

12.2 Linear Programming Problem and its Mathematical Formulation

12.2.2 Graphical method of solving linear programming problems

Practice Questions — Linear Programming

All 7 Chapters in Mathematics Part-II