Linear Programming

Given:

Z=11x+7y

It is subject to constraints

2x+y≤6, x≤2, x≥0, y≥0

Now let us convert the given inequalities into equation.

We obtain the following equation

2x+y≤6

⇒ 2x+y=6

x≤2

⇒ x=2

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 2x+y≤6:

The line 2x+y=6 meets the coordinate axes (3,0) and (0,6) respectively. We will join these points to obtain the line 2x+y=6. It is clear that (0,0) satisfies the inequation 2x+y≤6. So the region containing the origin represents the solution set of the inequation 2x+y≤6

The region represented by x≤2:

The line is parallel to y-axis and meets the x-axis at x=2.It is clear (0,0) satisfies the inequation x≤2. So the region containing the origin represents the solution set of the inequation x≤2

Region represented by x≥0 and y≥0 is first quadrant, Since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

The shaded region OBDE is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), B (0, 6), D (2, 2) and E (2, 0).

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 42 at the point (0, 6).

Given:

Z=3x + 4y

It is subject to constraints

x + y ≤ 1, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + y ≤ 1

⇒ x + y=1

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x+y≤1:

The line x + y=1 meets the coordinate axes (0,1) and (1,0) respectively. We will join these points to obtain the line x + y=1. It is clear that (0,0) satisfies the inequation x+y≤1. So the region containing the origin represents the solution set of the inequation x+y≤1

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

The shaded region OBC shows the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), B (0, 1) and C (1, 0).

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 4 at the point (0, 1).

Given:

Z=11x + 7y

It is subject to constraints

x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x ≤ 3

⇒ x=3

y ≤ 2

⇒ y=2

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x≤3:

The line is parallel to y-axis and meets the x-axis at x=3. It is clear (0,0) satisfies the inequation x≤2. So the region containing the origin represents the solution set of the inequation x≤3.

The region represented by y≤2:

The line is parallel to x-axis and meets the y-axis at y=2. It is clear (0,0) satisfies the inequation y≤2. So the region containing the origin represents the solution set of the inequation y≤2.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), B (0, 2), C (3, 2) and D (3, 0)

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 47 at the point (3, 2).

Given:

Z=13x – 15y

It is subject to constraints

x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + y ≤ 7

⇒ x + y=7

2x – 3y + 6 ≥ 0

⇒ 2x-3y+6=0

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x+y≤7:

The line x + y=7 meets the coordinate axes (7,0) and (0,7) respectively. We will join these points to obtain the line x + y=7. It is clear that (0,0) satisfies the inequation x+y≤7. So the region containing the origin represents the solution set of the inequation x+y≤7

The region represented by 2x – 3y + 6 ≥ 0:

The line 2x-3y+6=0 meets the coordinate axes (-3,0) and (0,2) respectively. We will join these points to obtain the line 2x-3y+6=0. It is clear that (0,0) satisfies the inequation 2x – 3y + 6 ≥ 0. So, the region containing the origin represents the solution set of the inequation 2x – 3y + 6 ≥ 0

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

The shaded region OBCD is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), B (0, 2), C (3, 4) and D (7,0)

Now we will substitute these values in Z at each of these corner points, we get

Hence, the minimum value of Z is -30 at the point (0, 2).

Given:

Z=3x + 4y

From the given figure it is subject to constraints

x + 2y ≤ 76, 2x +y ≤ 104, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + 2y ≤ 76

⇒ x+2y=76

2x +y ≤ 104

⇒ 2x +y = 104

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x + 2y ≤ 76:

The line x + 2y=76 meets the coordinate axes (76,0) and (0,38) respectively. We will join these points to obtain the line x + 2y=76. It is clear that (0,0) satisfies the inequation x + 2y ≤ 76. So the region containing the origin represents the solution set of the inequation x+2y≤76

The region represented by 2x +y ≤ 104:

The line 2x +y=104 meets the coordinate axes (52,0) and (0,104) respectively. We will join these points to obtain the line 2x +y=104. It is clear that (0,0) satisfies the inequation 2x +y ≤ 104. So the region containing the origin represents the solution set of the inequation 2x +y ≤ 104

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

The graph of these equations is given.

The shaded region ODBA is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), D (0, 38), B (44,16) and A (52,0)

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 196 at the point (44,16).

Given:

Z=5x+7y

The shaded region OABD in the given figure is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), A (7,0), B (3,4) and D (0,2)

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 43 at the point (3,4).

Given:

Z=11x + 7y

From the given figure it is subject to constraints

x + y ≤ 5, x +3y ≥ 9, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + y ≤ 5

⇒ x + y = 5

x +3y ≥ 9

⇒ x +3y = 9

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x + y ≤ 5:

The line x+y=5 meets the coordinate axes (5,0) and (0,5) respectively. We will join these points to obtain the line x+y=5. It is clear that (0,0) satisfies the inequation x + y ≤ 5. So the region containing the origin represents the solution set of the inequation x + y ≤ 5

The region represented by x +3y ≥ 9:

The line x+3y=9 meets the coordinate axes (9,0) and (0,3) respectively. We will join these points to obtain the line x+3y=9. It is clear that (0,0) satisfies the inequation x +3y ≥ 9. So the region that doesn’t contain the origin represents the solution set of the inequation x +3y ≥ 9.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.

Corner Points are B (0, 3), E (0,5) and C (3,2)

Now we will substitute these values in Z at each of these corner points, we get

Hence, the minimum value of Z is 21 at the point (0, 3).

Given:

Z=11x + 7y

From the given figure it is subject to constraints

x + y ≤ 5, x +3y ≥ 9, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + y ≤ 5

⇒ x + y = 5

x +3y ≥ 9

⇒ x +3y = 9

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x + y ≤ 5:

The line x+y=5 meets the coordinate axes (5,0) and (0,5) respectively. We will join these points to obtain the line x+y=5. It is clear that (0,0) satisfies the inequation x + y ≤ 5. So the region containing the origin represents the solution set of the inequation x + y ≤ 5

The region represented by x +3y ≥ 9:

The line x+3y=9 meets the coordinate axes (9,0) and (0,3) respectively. We will join these points to obtain the line x+3y=9. It is clear that (0,0) satisfies the inequation x +3y ≥ 9. So the region that doesn’t contain the origin represents the solution set of the inequation x +3y ≥ 9.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.

Corner Points are B (0, 3), E (0,5) and C (3,2)

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 47 at the point (3, 2).

Given:

Z=4x + y

From the given figure it is subject to constraints

x + 2y ≥ 4, x +y ≥ 3, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + 2y ≥ 4

⇒ x + 2y = 4

x +y ≥ 3

⇒ x +y = 3

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x + 2y ≥ 4:

The line x + 2y=4 meets the coordinate axes (4,0) and (0,2) respectively. We will join these points to obtain the line x+2y=4. It is clear that (0,0) does not satisfy the inequation x + 2y ≥ 4. So the region that doesn’t contain the origin represents the solution set of the inequation x + 2y ≥ 4

The region represented by x +y ≥ 3:

The line x +y=3 meets the coordinate axes (3,0) and (0,3) respectively. We will join these points to obtain the line x +y=3. It is clear that (0,0) does not satisfy the inequation x +y ≥ 3. So the region that doesn’t contain the origin represents the solution set of the inequation x +y ≥ 3.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region ABC is the feasible region is unbounded, and minimum value will occur at a corner point of the feasible region.

Corner Points are A (0, 3), B (2, 1) and C (4, 0)

Now we will substitute these values in Z at each of these corner points, we get

Note that here we see that, the region is unbounded, therefore 3 may not be the minimum value of Z.

To decide this issue, we graph the inequality 4x + y < 3 and check whether the resulting open half plane has no point in common with feasible region otherwise, Z has no minimum value.
From the shown graph above, it is clear that there is no point in common with feasible region and hence Z has minimum value 3 at (0, 3).

Hence, the minimum value of Z is 3 at the point (0, 3).

Given:

Z=x + 2y

From the given figure, the shaded region PRQS is the feasible region is bounded, so maximum and minimum value will occur at a corner point of the feasible region.

Corner Points are

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 9 at the point .

And the minimum value of Z is at the point .

Let the manufacturer produces x units of type A circuits and y units of type B circuits. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=50x+60y

Now, we have to maximize the profit, i.e., maximize Z=50x+60y

The constraints so obtained, i.e., subject to the constraints,

20x+10y≤ 200 [this is resistor constraint]

Now will divide throughout by 10, we get

⇒ 2x+y≤ 20…………..(i)

And 10x+20y≤ 120 [this is transistor constraint]

Now will divide throughout by 10, we get

⇒ x+2y≤ 12…………..(ii)

And 10x+30y≤ 150 [this is capacitor constraint]

Now will divide throughout by 10, we get

⇒ x+3y≤ 15…………..(iii)

And x≥0, y≥0 [non-negative constraint]

So, maximize profit, Z=50x+60y,

subject to 2x+y≤ 20,
x+2y≤ 12

x+3y≤ 15

x≥0, y≥0

Let the firm has x number of large vans and y number of small vans. We make the following table from the given data:

Thus according to the table, the cost becomes, Z=400x+200y

Now, we have to minimize the cost, i.e., minimize Z=400x+200y

The constraints so obtained, i.e., subject to the constraints,

200x+80y≥ 1200

Now will divide throughout by 40, we get

⇒ 5x+2y≥ 30…………..(i)

And 400x+200y≤3000

Now will divide throughout by 200, we get

⇒ 2x+y≤ 15…………..(ii)

Also given the number of large vans cannot exceed the number of small vans

⇒ x≤ y……………..(iii)

And x≥0, y≥0 [non-negative constraint]

So, minimize cost we have to minimize Z=400x+200y subject to

5x+2y≥ 30

2x+y≤ 15

x≤ y

x≥0, y≥0

Let the company manufactures x boxes of type A screws and y boxes of type B screws. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=100x+170y

Now, we have to maximize the profit, i.e., maximize Z=100x+170y

The constraints so obtained, i.e., subject to the constraints,

2x+8y≤ 3600 [time constraints for threading machine]

Now will divide throughout by 2, we get

⇒ x+4y≤ 1800…………..(i)

And 3x+2y≤3600 [time constraints for slotting machine]

⇒ 3x+2y≤3600…………..(ii)

And x≥0, y≥0 [non-negative constraint]

So, to maximize profit we have to maximize Z=100x+170y subject to

x+4y≤ 1800

3x+2y≤3600

x≥0, y≥0

Let the company manufactures x number of type A sweaters and y number of type B. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=200x+120y

Now, we have to maximize the profit, i.e., maximize Z=200x+120y

The constraints so obtained, i.e., subject to the constraints,

The company spends at most Rs 72000 a day.
∴ 360x + 120y ≤ 72000

Divide throughout by 120, we get
=> 3x+y≤ 600 …(i)
Also, company can make at most 300 sweaters.
∴ x+y≤ 300 …(ii)
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100

i.e., y-x≤ 100

⇒ y≤ 100+x……….(iii)
And x≥0, y≥0 [non-negative constraint]

So, to maximize profit we have to maximize Z=200x+120y, subject to

3x+y≤ 600

x+y≤ 300

y≤ 100+x

x≥0, y≥0

Let the man rides his motorcycle for a distance of x km at a speed of 50km/hr then he has to spend Rs. 2/km on petrol.

let the man rides his motorcycle for a distance of y km at a speed of 80 km/hr then he has to spend Rs. 3/km on petrol.

He has at most Rs 120 to spend on petrol for total distance covered so the constraint becomes,

2x+3y≤120…………(i)

Now also given he has at most one hour’s time for total distance to be covered, so the constraint becomes

{as distance=speed×time}

Now taking the LCM as 400, we get

⇒ 8x+5y≤400……………(ii)

And x≥0, y≥0 [non-negative constraint]

He want to find out the maximum distance travelled, here total distance, Z =x+y

Now, we have to maximize the distance, i.e., maximize Z=x+y

So, to maximize distance we have to maximize, Z=x+y, subject to

2x+3y≤120

8x+5y≤400

x≥0, y≥0

Referring to the exercise 11, we get following data

Let the manufacturer produces x units of type A circuits and y units of type B circuits. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=50x+60y

Now, we have to maximize the profit, i.e., maximize Z=50x+60y

The constraints so obtained, i.e., subject to the constraints,

20x+10y≤ 200 [this is resistor constraint]

Now will divide throughout by 10, we get

⇒ 2x+y≤ 20…………..(i)

And 10x+20y≤ 120 [this is transistor constraint]

Now will divide throughout by 10, we get

⇒ x+2y≤ 12…………..(ii)

And 10x+30y≤ 150 [this is capacitor constraint]

Now will divide throughout by 10, we get

⇒ x+3y≤ 15…………..(iii)

And x≥0, y≥0 [non-negative constraint]

So, maximize profit, Z=50x+60y, subject to 2

x+y≤ 20

x+2y≤ 12

x+3y≤ 15

x≥0, y≥0

Now let us convert the given inequalities into equation.

We obtain the following equation

2x+y≤ 20

⇒ 2x+y= 20

x+2y≤ 12

⇒ x+2y= 12

x+3y≤ 15

⇒ x+3y= 15

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 2x+y≤ 20:

The line 2x+y=20 meets the coordinate axes (10,0) and (0,20) respectively. We will join these points to obtain the line 2x+y=20. It is clear that (0,0) satisfies the inequation 2x+y≤ 20. So the region that contain the origin represents the solution set of the inequation 2x+y≤ 20

The region represented by x+2y≤ 12:

The line x+2y=12 meets the coordinate axes (12,0) and (0,6) respectively. We will join these points to obtain the line x+2y=12. It is clear that (0,0) satisfies the inequation x+2y≤ 12. So the region that doesn’t contain the origin represents the solution set of the inequation x+2y≤ 12.

The region represented by x+3y≤ 15:

The line x+3y=15 meets the coordinate axes (15,0) and (0,5) respectively. We will join these points to obtain the line x+3y=15. It is clear that (0,0) satisfies the inequation x+3y≤ 15. So the region that contain the origin represents the solution set of the inequation x+3y≤ 15

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region OABCD is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.

Corner Points are O(0,0), A (0, 5), B (6, 3), C (9.3, 1.3) and D(10,0)

Now we will substitute these values in Z at each of these corner points, we get

So from the above table the maximum value of Z is at point (9.3, 1.3), but as the manufacturer is required to produce two type of circuits, so the parts of resistors, transistors and capacitors cannot be decimals. So we will consider the next maximum number.

Hence, the maximum value of Z is 480 at the point (6, 3) i.e., the manufacturer should produce 6 circuits of type A and 3 circuits of type B so as to maximize his profit.

Referring to the exercise 12, we get following data

Let the firm has x number of large vans and y number of small vans. We make the following table from the given data:

Thus according to the table, the cost becomes, Z=400x+200y

Now, we have to minimize the cost, i.e., minimize Z=400x+200y

The constraints so obtained, i.e., subject to the constraints,

200x+80y≥ 1200

Now will divide throughout by 40, we get

⇒ 5x+2y≥ 30…………..(i)

And 400x+200y≤3000

Now will divide throughout by 200, we get

⇒ 2x+y≤ 15…………..(ii)

Also given the number of large vans cannot exceed the number of small vans

⇒ x≤ y……………..(iii)

And x≥0, y≥0 [non-negative constraint]

So, minimize cost we have to minimize, Z=400x+200y, subject to

5x+2y≥ 30

2x+y≤ 15

x≤ y

x≥0, y≥0

Now let us convert the given inequalities into equation.

We obtain the following equation

5x+2y≥ 30

⇒ 5x+2y=30

2x+y≤ 15

⇒ 2x+y=15

x≤ y

⇒ x=y

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 5x+2y≥ 30:

The line 5x+2y=30 meets the coordinate axes (6,0) and (0,15) respectively. We will join these points to obtain the line 5x+2y=30. It is clear that (0,0) does not satisfy the inequation 5x+2y≥ 30. So the region that does not contain the origin represents the solution set of the inequation 5x+2y≥ 30

The region represented by 2x+y≤ 15:

The line 2x+y=15 meets the coordinate axes (7.5,0) and (0,15) respectively. We will join these points to obtain the line 2x+y=15. It is clear that (0,0) satisfies the inequation 2x+y≤ 15. So the region that contain the origin represents the solution set of the inequation 2x+y≤ 15

The region represented by x≤y:

The line x=y is a line that passes through the origin and doesn’t touch any coordinate axes at any other point except (0,0). We will join these points to obtain the line x=y. It is clear that (0,0) satisfies the inequation x≤y. So the region that contain the origin represents the solution set of the inequation x≤y

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region ABC represents the feasible region is bounded, and minimum value will occur at a corner point of the feasible region.

Corner Points are , B (0, 15) and C(5, 5)

Now we will substitute these values in Z at each of these corner points, we get

So from the above table the minimum value of Z is at point ,

Hence, the minimum cost of the firm is Rs. 2571.43.

Referring exercise 13, we get the following data:

Let the company manufactures x boxes of type A screws and y boxes of type B screws. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=100x+170y

Now, we have to maximize the profit, i.e., maximize Z=100x+170y

The constraints so obtained, i.e., subject to the constraints,

2x+8y≤ 3600 [time constraints for threading machine]

Now will divide throughout by 2, we get

⇒ x+4y≤ 1800…………..(i)

And 3x+2y≤3600 [time constraints for slotting machine]

⇒ 3x+2y≤3600…………..(ii)

And x≥0, y≥0 [non-negative constraint]

So, to maximize profit we have to maximize, Z=100x+170y, subject to

x+4y≤ 1800

3x+2y≤3600

x≥0, y≥0

Now let us convert the given inequalities into equation.

We obtain the following equation

x+4y≤ 1800

⇒ x+4y=1800

3x+2y≤3600

⇒ 3x+2y=3600

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x+4y≤ 1800:

The line x+4y=1800 meets the coordinate axes (1800,0) and (0,450) respectively. We will join these points to obtain the line x+4y=1800. It is clear that (0,0) satisfies the inequation x+4y≤ 1800. So the region that contain the origin represents the solution set of the inequation x+4y≤ 1800

The region represented by 3x+2y≤3600:

The line 3x+2y=3600 meets the coordinate axes (1200,0) and (0,1800) respectively. We will join these points to obtain the line 3x+2y≤3600 . It is clear that (0,0) satisfies the inequation 3x+2y≤3600. So the region that contain the origin represents the solution set of the inequation 3x+2y≤3600

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region OBCD is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.

Corner Points are O(0, 0), B(0, 450), C(1080, 180) and D(1200, 0)

Now we will substitute these values in Z at each of these corner points, we get

So from the above table the maximum value of Z is at point (1080,180).

Hence, the maximum profit to the manufacturer is Rs. 1,38,600.

Referring to exercise 15, we get the following data:

Let the company manufactures x number of type A sweaters and y number of type B. We make the following table from the given data:

Thus according to the table, the profit becomes, Z=200x+120y

Now, we have to maximize the profit, i.e., maximize Z=200x+120y

The constraints so obtained, i.e., subject to the constraints,

The company spends at most Rs 72000 a day.
∴ 360x + 120y ≤ 72000

Divide throughout by 120, we get
=> 3x+y≤ 600 …(i)
Also, company can make at most 300 sweaters.
∴ x+y≤ 300 …(ii)
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100

i.e., y-x≤ 100

y≤ 100+x………. (iii)
And x≥0, y≥0 [non-negative constraint]

So, to maximize profit we have to maximize, Z=200x+120y, subject to

3x+y≤ 600

x+y≤ 300

y≤ 100+x

x≥0, y≥0

Now let us convert the given inequalities into equation.

We obtain the following equation

3x+y≤ 600

⇒ 3x+y=600

x+y≤ 300

⇒ x+y=300

y≤ 100+x

⇒ y=100+x

⇒ x-y=-100

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 3x+y≤ 600:

The line 3x+y=600 meets the coordinate axes (200,0) and (0,600) respectively. We will join these points to obtain the line 3x+y=600. It is clear that (0,0) satisfies the inequation 3x+y≤ 600. So the region that contain the origin represents the solution set of the inequation 3x+y≤ 600

The region represented by x+y≤ 300:

The line x+y=300 meets the coordinate axes (300,0) and (0,300) respectively. We will join these points to obtain the line x+y=300. It is clear that (0,0) satisfies the inequation x+y≤300. So the region that contain the origin represents the solution set of the inequation x+y=300

The region represented by y≤ 100+x:

The line y= 100+x meets the coordinate axes (-100,0) and (0,100) respectively. We will join these points to obtain the line y= 100+x. It is clear that (0,0) satisfies the inequation y≤ 100+x. So the region that contain the origin represents the solution set of the inequation y≤ 100+x

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region OBCDE is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.

Corner Points are O(0, 0), B(0, 100), C(100, 200), D(150, 150) and E(200,0)

Now we will substitute these values in Z at each of these corner points, we get

So from the above table the maximum value of Z is at point (150,150).

Hence, the maximum profit to the manufacturer is Rs. 48,000, for making 150 sweaters each of type A and type B.

Referring to the exercise 15, we get the following data:

Let the man rides his motorcycle for a distance of x km at a speed of 50km/hr then he has to spend Rs. 2/km on petrol.

let the man rides his motorcycle for a distance of y km at a speed of 80 km/hr then he has to spend Rs. 3/km on petrol.

He has at most Rs 120 to spend on petrol for total distance covered so the constraint becomes,

2x+3y≤120…………(i)

Now also given he has at most one hour’s time for total distance to be covered, so the constraint becomes

{as distance=speed×time}

Now taking the LCM as 400, we get

⇒ 8x+5y≤400……………(ii)

And x≥0, y≥0 [non-negative constraint]

He want to find out the maximum distance travelled, here total distance, Z =x+y

Now, we have to maximize the distance, i.e., maximize Z=x+y

So, to maximize distance we have to maximize, Z=x+y, subject to

2x+3y≤120

8x+5y≤400

x≥0, y≥0

Now let us convert the given inequalities into equation.

We obtain the following equation

2x+3y≤120⇒ 2x+3y=120

8x+5y≤400 ⇒ 8x+5y=400

x ≥ 0 ⇒ x=0

y ≥ 0 ⇒ y=0

The region represented by 2x+3y≤120:

The line 2x+3y=120 meets the coordinate axes (60,0) and (0,40) respectively. We will join these points to obtain the line 2x+3y=120. It is clear that (0,0) satisfies the inequation 2x+3y≤120. So the region that contain the origin represents the solution set of the inequation 2x+3y≤120

The region represented by 8x+5y≤400:

The line 8x+5y=400 meets the coordinate axes (50,0) and (0,80) respectively. We will join these points to obtain the line 8x+5y=400. It is clear that (0,0) satisfies the inequation 8x+5y≤400. So the region that contain the origin represents the solution set of the inequation 8x+5y≤400

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of these equations is given.

The shaded region OBCD is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.

Corner Points are O(0, 0), B(0, 40), and D(50, 0)

Now we will substitute these values in Z at each of these corner points, we get

So from the above table the maximum value of Z is at point .

Hence, the maximum distance the man can travel is or 54.3km.

Given-

Z = x + y

It is subject to constraints

x + 4y ≤ 8

2x + 3y ≤ 12

3x + y ≤ 9

x ≥ 0, y ≥ 0

We need to maximize Z, subject to the above constraints.

Now let us convert the given inequalities into equation.

We obtain the following equation

x + 4y ≤ 8

⇒ x + 4y = 8

2x + 3y ≤ 12

⇒ 2x + 3y = 12

3x + y ≤ 9

⇒ 3x + y = 9

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x + 4y ≤ 8:

The line x + 4y = 8 meets the coordinate axes (8,0) and (0,2) respectively. We will join these points to obtain the line x + 4y = 8. It is clear that (0,0) satisfies the inequation x + 4y ≤ 8. So, the region containing the origin represents the solution set of the inequation x + y ≤ 8.

The region represented by 2x + 3y ≤ 12:

The line 2x + 3y = 12 meets the coordinate axes (6,0) and (0,4) respectively. We will join these points to obtain the line 2x + 3y = 12. It is clear that (0,0) satisfies the inequation 2x + 3y ≤ 12. So, the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 12

The region represented by 3x + y ≤ 9:

The line 3x + y = 9 meets the coordinate axes (3,0) and (0,9) respectively. We will join these points to obtain the line 3x + y = 9. It is clear that (0,0) satisfies the inequation 3x + y ≤ 9. So, the region containing the origin represents the solution set of the inequation 3x + y ≤ 9

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

Feasible region is ABCD

Value of Z at corner points A, B, C and D –

So, value of Z is maximum at B (2.54, 1.36), the maximum value is 3.90.

Let number of bikes per week of model X and Y be x and y respectively.

Given that model X takes 6 man-hours.

So, time taken by x bikes of model X = 6x hours.

Given that model Y takes 10 man-hours.

So, time taken by y bikes of model X = 10y hours.

Total man-hour available per week = 450

So, 6x + 10y ≤ 450

⇒ 3x + 5y ≤ 225

Handling and marketing cost of model X and Y is Rs 2000 and Rs 1000 per unit respectively.

So, total handling and marketing cost of x units of model X and y units of model y is 2000x + 1000y

Maximum amount available for handling and marketing per week is Rs 80000.

So, 2000x + 1000y ≤ 80000

⇒ 2x + y ≤ 80

Profits per unit for Models X and Y are Rs 1000 and Rs 500, Respectively.

Let total profit = Z

So, Z = 1000x + 500y

Also, as units will be positive numbers so x, y ≥ 0

So, we have,

Z = 1000x + 500y

With constraints,

3x + 5y ≤ 225

2x + y ≤ 80

x, y ≥ 0

We need to maximize Z, subject to the given constraints.

Now let us convert the given inequalities into equation.

We obtain the following equation

3x + 5y ≤ 225

⇒ 3x + 5y = 225

2x + y ≤ 80

⇒ 2x + y = 80

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 3x + 5y ≤ 225:

The line 3x + 5y = 225 meets the coordinate axes (75,0) and (0,45) respectively. We will join these points to obtain the line 3x + 5y = 225. It is clear that (0,0) satisfies the inequation 3x + 5y ≤ 225. So, the region containing the origin represents the solution set of the inequation 3x + 5y ≤ 225.

The region represented by 2x + y ≤ 80:

The line 2x + y = 80 meets the coordinate axes (40,0) and (0,80) respectively. We will join these points to obtain the line

2x + y = 80.

It is clear that (0,0) satisfies the inequation 2x + y ≤ 80. So, the region containing the origin represents the solution set of the inequation 2x + y ≤ 80

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

Feasible region is ABCD

Value of Z at corner points A, B, C and D –

So, value of Z is maximum on-line BC, the maximum value is 40000.So manufacturer must produce 25 number of models X and 30 number of model Y.

Let number of tablet X be x and number of tablet Y be y.

Iron content in X and Y tablet is 6 mg and 2 mg respectively.

So, total iron content from x and y tablets = 6x + 2y

Minimum of 18 mg of iron is required. So, we have

6x + 2y ≥ 18

⇒ 3x + y ≥ 9

Similarly, calcium content in X and Y tablet is 3 mg each respectively.

So, total calcium content from x and y tablets = 3x + 3y

Minimum of 21 mg of calcium is required. So, we have

6x + 2y ≥ 21

⇒ x + y ≥ 7

Also, vitamin content in X and Y tablet is 2 mg and 4 mg respectively.

So, total vitamin content from x and y tablets = 2x + 4y

Minimum of 16 mg of vitamin is required. So, we have

2x + 4y ≥ 16

⇒ x + 2y ≥ 8

Also, as number of tablets should be non-negative so, we have,

x, y ≥ 0

Cost of each tablet of X and Y is Rs 2 and Re 1 respectively.

Let total cost = Z

So, Z = 2x + y

Finally, we have,

Constraints,

3x + y ≥ 9

x + y ≥ 7

x + 2y ≥ 8

x, y ≥ 0

Z = 2x + y

We need to minimize Z, subject to the given constraints.

Now let us convert the given inequalities into equation.

We obtain the following equation

3x + y ≥ 9

⇒ 3x + y = 9

x + y ≥ 7

⇒ x + y = 7

x + 2y ≥ 8

⇒ x + 2y = 8

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 3x + y ≥ 9:

The line 3x + y = 9 meets the coordinate axes (3,0) and (0,9) respectively. We will join these points to obtain the line 3x + y = 9. It is clear that (0,0) does not satisfy the inequation 3x + y ≥ 9. So, the region not containing the origin represents the solution set of the inequation 3x + y ≥ 9.

The region represented by x + y ≥ 7:

The line x + y = 7 meets the coordinate axes (7,0) and (0,7) respectively. We will join these points to obtain the line x + y = 7. It is clear that (0,0) does not satisfy the inequation x + y ≥ 7. So, the region not containing the origin represents the solution set of the inequation x + y ≥ 7.

The region represented by x + 2y ≥ 8:

The line x + 2y = 8 meets the coordinate axes (8,0) and (0,4) respectively. We will join these points to obtain the line x + 2y = 8. It is clear that (0,0) does not satisfy the inequation x + 2y ≥ 8. So, the region not containing the origin represents the solution set of the inequation x + 2y ≥ 8.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

Feasible region is the region to the right of ABCD

Feasible region is unbounded.

Value of Z at corner points A, B, C and D –

Now, we check if 2x + y<8, to check if resulting open half has any point common with feasible region.

The region represented by 2x + y<8:

The line 2x + y=8meets the coordinate axes (4,0) and (0,8) respectively. We will join these points to obtain the line 2x + y=8. It is clear that (0,0) satisfies the inequation 2x + y<8. So, the region not containing the origin represents the solution set of the inequation 2x + y<8.

Clearly, 2x + y = 8 intersects feasible region only at B.

So, 2x + y<8 does not have any point inside feasible region.

So, value of Z is minimum at B (1, 6), the minimum value is 8.

So, number of tablets that should be taken of type X and Y is 1, 6 Respectively.

Let number of days for which factory I operates be x and number of days for which factory II operates be y.

Number of calculators made by factory I and II of model A are 50 and 40 respectively.

Minimum number of calculators of model A required = 6400

So, 50x + 40y ≥ 640

⇒ 5x + 4y ≥ 640

Number of calculators made by factory I and II of model B are 50 and 20 respectively.

Minimum number of calculators of model B required = 4000

So, 50x + 20y ≥ 4000

⇒ 5x + 2y ≥ 400

Number of calculators made by factory I and II of model C are 30 and 40 respectively.

Minimum number of calculators of model C requires = 4800

So, 30x + 40y ≥ 4800

⇒ 3x + 4y ≥ 480

Operating costs is Rs 12000 and Rs 15000 each day to operate factory I and II respectively.

Let Z be total operating cost so we have Z = 12000x + 15000y

Also, number of days are non-negative so, x, y ≥ 0

So, we have,

Constraints,

5x + 4y ≥ 640

5x + 2y ≥ 400

3x + 4y ≥ 480

x, y ≥ 0

Z = 12000x + 15000y

We need to minimize Z, subject to the given constraints.

Now let us convert the given inequalities into equation.

We obtain the following equation

5x + 4y ≥ 640

⇒ 5x + 4y = 640

5x + 2y ≥ 400

⇒ 5x + 2y = 400

3x + 4y ≥ 480

⇒ 3x + 4y = 480

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by 5x + 4y ≥ 640:

The line 5x + 4y = 640 meets the coordinate axes (128,0)

and (0,160) respectively. We will join these points to obtain the line 5x + 4y = 640. It is clear that (0,0) does not satisfy the inequation 5x + 4y ≥ 640. So, the region not containing the origin represents the solution set of the inequation 5x + 4y ≥ 640.

The region represented by 5x + 2y ≥ 400:

The line 5x + 2y = 400 meets the coordinate axes (80,0) and (0,200) respectively. We will join these points to obtain the line x + y = 7. It is clear that (0,0) does not satisfy the inequation 5x + 2y ≥ 400. So, the region not containing the origin represents the solution set of the inequation 5x + 2y ≥ 400.

The region represented by 3x + 4y ≥ 480:

The line 3x + 4y = 480 meets the coordinate axes (160,0) and (0,120) respectively. We will join these points to obtain the line 3x + 4y = 480. It is clear that (0,0) does not satisfy the inequation 3x + 4y ≥ 480. So, the region not containing the origin represents the solution set of the inequation 3x + 4y ≥ 480.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

Feasible region is the region to the right of ABCD

Feasible region is unbounded.

Value of Z at corner points A, B, C and D –

Now, we plot 12000x + 15000y<1860000, to check if resulting open half has any point common with feasible region.

The region represented by 12000x + 15000y<1860000:

The line 12000x + 15000y=1860000 meets the coordinate axes (155,0) and (0,124) respectively. We will join these points to obtain the line 12000x + 15000y=1860000. It is clear that (0,0) satisfies the inequation 12000x + 15000y<1860000. So, the region containing the origin represents the solution set of the inequation 12000x + 15000y<1860000.

Clearly, 12000x + 15000y<1860000 intersects feasible region only at C.

So, value of Z is minimum at C (80, 60), the minimum value is 1860000.

So, number of days factory I is required to operate is 80 and number of days factory II should operate is 60 to minimize the cost.

We have constraints,

x – 2y ≤ 0

– 3x + y ≤ 4

x – y ≤ 6

x, y ≥ 0

Z = 3x – 4y

We need to maximize and minimize Z, subject to the given constraints.

Now let us convert the given inequalities into equation.

We obtain the following equation

x – 2y ≤ 0

⇒ x - 2y = 0

– 3x + y ≤ 4

⇒ -3x + y = 4

x – y ≤ 6

⇒ x - y = 6

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x – 2y ≤ 0:

The line x - 2y = 0 meets the coordinate axes at origin and slope of the line is . We will construct a line passing through origin and whose slope is . As point (1,1) satisfies the inequality. So, the side of line which contains (1,1) is feasible. Hence, the solution set of the inequation x – 2y ≤ 0 is the side which contains (1,1).

The region represented by – 3x + y ≤ 4:

The line – 3x + y = 4 meets the coordinate axes and (0,4) respectively. We will join these points to obtain the line x + y = 7. It is clear that (0,0) satisfies the inequation – 3x + y ≤ 4. So, the region containing the origin represents the solution set of the inequation – 3x + y ≤ 4.

The region represented by x – y ≤ 6:

The line x – y = 6meets the coordinate axes (6,0) and (0,-6) respectively. We will join these points to obtain the line x – y = 6. It is clear that (0,0) satisfies the inequation x – y ≤ 6. So, the region containing the origin represents the solution set of the inequation x – y ≤ 6.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

The feasible region is region between line -3x + y = 4 and x – y = 6, above BC and to the right of y – axis as shown.

Feasible region is unbounded.

Corner points are A, B, C

So, maximum value of Z at corner points is 12 at C and minimum is -16 at A.

Value of Z at corner points A, B, C and D –

So, to check if the solution is correct, we plot 3x – 4y > 12 and 3x – 4y < -16 for maximum and minimum respectively.

The region represented by 3x – 4y > 12:

The line 3x – 4y = 12 meets the coordinate axes (4,0) and (0,-3) respectively. We will join these points to obtain the line 3x – 4y > 12. It is clear that (0,0) does not satisfy the inequation 3x – 4y > 12. So, the region not containing the origin represents the solution set of the inequation 3x – 4y > 12.

The region represented by 3x – 4y <-16:

The line 3x – 4y = -16 meets the coordinate axes and (0,4) respectively. We will join these points to obtain the line 3x – 4y <-16. It is clear that (0,0) does not satisfy the inequation 3x – 4y <-16. So, the region not containing the origin represents the solution set of the inequation 3x – 4y <-16.

We get,

Clearly, 3x – 4y = 12 has no point inside feasible region, but 3x -4y = -16 passes through the feasible region.

Therefore, Z has no minimum value it has only a maximum value which is 12.

A. The quantity in column A is greater

Z = 4x + 3y

Corner points- (0, 0), (0, 40), (20, 40), (60, 20), (60, 0)

As feasible region is bounded.

Value of Z at corner points-

At (0, 0), Z = 0

At (0, 40), Z = 120

At (20, 40), Z = 200

At (60, 20), Z = 360

At (60, 0), Z = 240

Clearly, maximum value of Z is 360, which is greater than 325.

Value of Z = 3x – 4y, at corner points are –

At (0, 0) = 0

At (0, 8) = -32

At (4, 10) = -28

At (6, 8) = -14

At (6, 5) = -2

At (5, 0) = 15

So, clearly minimum value is at (0, 8)

Value of Z = 3x – 4y, at corner points are –

At (0, 0) = 0

At (0, 8) = -32

At (4, 10) = -28

At (6, 8) = -14

At (6, 5) = -2

At (5, 0) = 15

So, clearly maximum value is at (5, 0)

Value of Z = 3x – 4y, at corner points are –

At (0, 0) = 0

At (0, 8) = -32

At (4, 10) = -28

At (6, 8) = -14

At (6, 5) = -2

At (5, 0) = 15

So, clearly maximum value is at (5, 0)

Maximum value = 15,

Minimum value = -32

So, maximum + minimum = 15 -32

= -17

F = 3x – 4y

Corner points are (0, 0), (0, 4), (12, 6)

Value of F at corner points –

At (0, 0), F = 0

At (0, 4), F = -16

At (12, 6), F = 12

Clearly, maximum value of F = 12.

F = 3x – 4y

Corner points are (0, 0), (0, 4), (12, 6)

Value of F at corner points –

At (0, 0), F = 0

At (0, 4), F = -16

At (12, 6), F = 12

Clearly, minimum value of F = – 16.

F = 4x + 6y

Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).

Value of F at corner points –

At (0, 2), F = 12

At (3, 0), F = 12

At (6, 0), F = 24

At (6, 8), F = 72

At (0, 5), F = 30

Feasible region –

As, feasible region to be bounded so it is a closed polygon.

So, minimum value of F = 12 are at (3, 0) and (0, 2).

Therefore, minimum value of F occurs at, any point on the line segment joining the points (0, 2) and (3, 0).

F = 4x + 6y

Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).

Value of F at corner points –

At (0, 2), F = 12

At (3, 0), F = 12

At (6, 0), F = 24

At (6, 8), F = 72

At (0, 5), F = 30

Feasible region –

Considering, feasible region to be bounded so it is a closed polygon.

Minimum value of F = 12

Maximum value of F = 72

So, Maximum of F – Minimum of F = 60.

Z = px + qy

Given, minimum occours at (3, 0) and (1, 1).

For minimum to occur at two points the value of Z at both points should be same.

So, value of Z at (3, 0) = value of Z at (1, 1)

⇒ 3p = p + q

⇒ 2p = q

So, option B is correct.

In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

In a LPP, the objective function is always linear.

If the feasible region for a LPP is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same maximum value

A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle.

A corner point of a feasible region is a point in the region which is the intersection of two boundary lines.

The feasible region for an LPP is always a convex polygon.

True.

If the feasible region is unbounded then we may or may not have a maximum or minimum of objective function, but if we have a maximum or a minimum value then it must be at one of the corner points only.

False.

Maximum value or minimum value can occur at more than one points. In such all the points lie on a line segment and is part of boundary of the feasible region.

False.

Minimum value of objective function can also be negative if the coefficient of x or y is negative. So, it is not necessary that minimum value of objective function will be zero.

False.

In a LPP, the maximum value of the objective function Z = ax + by may or may not be finite. It depends on the feasible region.

If feasible region is unbounded then we can also have infinite maximum value of objective function.