# Important Questions Class 12 Maths Chapter 12

### Important Questions For CBSE Class 12 Mathematics Chapter 12 – Linear Programming

Important Questions Class 12 Mathematics Chapter 12 provided by Extramarks are prepared by subject experts to give students an idea of different types of questions that can be asked from this chapter in the board exams.

### CBSE Class 12 Mathematics Chapter-12 Important Questions

Study Important Questions For Class 12 Mathematics Chapter 12 – Linear Programming

A few of the Important Questions for Class 12 Mathematics Chapter 12 are given below. Students can click on the link given to view the entire set of questions.

Q1. Solve the Linear Programming Problem graphically.

Maximise and minimise: z = 3x + 9y.

Subject to constraints x + 3y ≤ 60

x + y ≥ 10

x ≤ y

x ≥ 0; y ≥ 0.

Solution: Maximise z = 3x + 9y; Minimise z = 3x + 9y.

Subject to x + 3y ≤ 60

x + y ≥ 10

x ≤ y

x ≥ 0; y ≥ 0.

x + 3y ≤ 60

 x 0 60 y 20 0

x + y ≥ 10

 x 0 10 y 10 0

x ≤ y

 x 0 20 y 0 20
 Corner Points Values of Z (0, 10) 90 (5, 5) 60 (15, 15) 180 (0, 20) 180

Therefore, Z = 60 is maximum at point (5, 5).

Again, Z = 60 is maximum at other two points, (15, 15) and (0, 20).

Z= 180 is maximum at all the points that join (15, 15) and (0, 20).

Q2. Find graphically the minimum value of z = – 50x + 20y, subject to the constraints

2x−y⩾−5

3x+y⩾3

2x−3y⩽12

x⩾0, y⩾0

Solution: Minimum value of z = 50x + 20y.

Subject to the constraints,

2x−y⩾−5

3x+y⩾3

2x−3y⩽12

x⩾0, y⩾0

2x – – y ≥ – 5

 x 0 1 y 5 7

3x – – 3y ≥ 3

 x 0 0 y 1 3

2x – -3y ≤ 12

 x 6 9 y 0 2

 Corner Points Value of Z (0, 5) 100 (0, 3) 60 (1, 0) -50 (6, 0) -300

As the feasible region is unbounded – 300 may or may not be the minimum value of Z. Therefore, to find the minimum value of Z we have to graph inequality

-50x + 20y < -300

 X 6 8 y 0 5

As there are some common points between the feasible regions, -300 cannot be the minimum value. Therefore, it is evident that Z does not have any minimum value.

Q3. A trader sells two types of tea, A and B. Type A tea is made up of 10% Assam tea and 6% Darjeeling tea, while type B is made up of 5% Assam tea and 10% Darjeeling tea. The trader finds that he needs 14 kg of Assam tea and 14 kg of Darjeeling tea for mixing tea leaves to increase his profit. If A costs Rs. 6 per kg and B costs Rs. 5 per kg, find how much of each type of tea should be mixed so that the requirements are met at the minimum cost. Calculate the minimum cost.

Solution: Suppose, requirement of type A tea = x kg; and requirement of type B tea = y kg.

 Mixed Tea Assam Tea Darjeeling Tea Quantity Cost A 10 % 6 % X kg Rs. 6 per kg B 5 % 10 % Y kg Rs. 5 per kg Total 14 kg 14 kg

According to the question,

 Assam Tea Darjeeling Tea Assam Tea present in A = 10 % Assam Tea present in B = 5 % Minimum amount of Assam Tea required for the new mixture of tea = 14 kg Therefore, it can be said 10 % of x + 5 % of y ≥ 14 2x + y ≥ 280 Darjeeling Tea present in A = 6 % Darjeeling Tea present in B = 10 % Minimum amount of Assam Tea required for the new mixture of tea = 14 kg Therefore, it can be said 6 % of x + 10 % of y ≥ 14 3x + 5y ≥ 700

Also, x ≥ 0, y ≥ 0.

As the minimum cost is required, the minimum value of Z should be found. Therefore, the function used here is to minimise z.

Cost of tea A = Rs. 6 per kg

Cost of tea B = Rs. 5 per kg

Minimise z = 6x + 5y

Combining all the constraints Min z = 6x + 5y

Subject to the constraints

2x + y ≥ 280

3x + 5y ≥ 700

x ≥ 0, y ≥ 0

2x + y ≥ 280

 x 140 0 y 10 280

3x + 5y ≥ 700

 X 0 700/3 y 140 0

 Corner Points Value of z (0, 180) 1400 (100, 800 1000 (700/3, 0) 1400

As the area that is feasible is unbounded, 1000 may or may or may not be the minimum value of z. To find the minimum value, we have to graph the inequality.

 X 500/3 0 y 0 200

As there is no common point between the feasible region and 6x + 5y < 1000 minimum cost will be required, if type A is used 100 kg; and type B is used 80 kg.

The minimum cost will be Rs. 1000.

Q4. Jimmy is a young biker. He finds that if he rides his bike at 25 km/hr., he has to spend Rs. 2 per km on petrol. If he increases his speed to 40 km/hr. his petrol cost also increases to Rs 5 per km. Jimmy decides to spend Rs 100 on petrol cost but he also wants to travel the maximum distance within one hour. Express this problem as L.L.P. and solve it graphically.

Solution: Suppose, Jimmy drives x km at a speed of 25 km/hr. and y km at 40 km/hr. speed.

x ≥ 0, y ≥ 0.

When the speed is 25 km/hr. The cost of petrol is Rs. 2 per km.

Therefore, when the distance is x km, the petrol cost is Rs. 2x.

It is given that, when the speed is 40 km/hr. The petrol cost is Rs. 5 per km.

Therefore, for travelling by km, the required petrol cost is Rs. 5y.

Jimmy has only 100 rupees to spend on petrol.

Therefore, 2x + 5y ≤ 100 ………….(i)

Time taken by Jimmy to travel x km = x25 hr. (when speed is 25 km/hr.)

Time taken by him to travel y km = y40 hr. (when speed is 40 km/hr.)

Jimmy wants to cover the total distance within 1 hour.

Therefore, the total time available = 1 hour.

Therefore,

x25+ y40 ≤ 1

Or, 40x + 25y ≤ 1000 ……………………… (ii)

Let, the total distance covered = z

Therefore, z = x+ y ……………………… (iii)

As Jimmy wants to travel the maximum distance the maximum value of z is required.

The mathematical expression of the given L.L.P. is Max Z = x + y which is subject to the constraints

2x + 5y ≤ 100 ………………… (i)

40x + 25y ≤ 1000 …………………… (ii)

x, y ≥0 ……………………… (iv)

The converted equations are as follows:

2x + 5y = 100

40x + 25y = 1000

x = 0; y = 0

The region represented by 2x + 5y ≤ 100:

The line 2x + 5y = 100 cuts x axis and y axis at A (5, 0) and B (0, 20) respectively. By joining these points, we get the line 2x + 5y = 100. It is clear that (0, 0) satisfies the 2x + 5y = 100. Therefore, the region containing the origin presents the solution set of the equation 2x + 5y ≤ 100.

The region represented by 40x + 25y ≤ 1000:

The line 40x + 25y = 1000 meets the x axis and y axis at C (25, 0) and D (0, 40) respectively. If these points are joined we get the line, 2x + y = 12. The equation 40x + 25y = 1000 is satisfied by (0, 0). So, the region containing the origin is the solution set for the inequation 40x + 25y ≤ 1000.

The region represented by x ≥ 0, y ≥ 0:

These equations are satisfied by every point in the first quadrant. Hence, the first quadrant is the region represented by equations x ≥ 0, and y ≥ 0.

Therefore, the feasible region determined by the given constraints are as follows:

 Corner Points Value of z (0, 0) 20 (0,20) 20 (50/3, 40/3) 30 (25, 0) 25

The value of z is maximum at the point having the position (50/3, 40/3).

Thus, the maximum distance that can be covered by Jimmy is 30 km. given that he drives 503 km. at the speed of 25 km/hr. and 403 km. at the speed of 40 km/hr.

Q 4. A company manufactures two types of steel pipes, A and B. Three machines are required to manufacture them, and the time taken by each is given in the chart below.

 Type of Steel Pipes Time Required By Each Pipe (In Minute) Machine 1 Machine 2 Machine 3 A 12 18 6 B 6 0 9

Each machine is available for a maximum 6 hours per day. If the profit on each pipe of Type A and Type B is 75 paise and 50 paise respectively, find out how many pipes must be manufactured per day to maximise the profit.

Solution:

 Type of Steel Pipes Time Required By Each Pipe (In Minute) Machine 1 Machine 2 Machine 3 A 12 18 6 B 6 0 9

Let, the number of required steel pipes of type A = x;

And the number of required steel pipes of type B = y.

It is to be maximised, Z = 1.50x + 1y.

Z = 1.50x + y

Subject to the constraints,

12x+6y⩽360

18x+0.y⩽360

6x+9y⩽360

x⩾0

y⩾0

The given problem can be arranged in the following manner

Max Z=1.50x + y

Subject to

2x + y ≤ 60

x ≤ 20

And 2x + 3y ≤ 120

x⩾0

y⩾0

Consider the equation 2x+y=60

When the value of x = 0, then the value of y = 60

When the value of y becomes 0, the value of x = 30.

The table of solution as follows:

 x 0 30 y 60 0

Now, consider the equation 2x + 3y = 120

When x = 0, y = 40

When y = 0, x = 60

The table of solution for this equation is given below:

 x 0 60 y 40 0

The corner points of the feasible region from the graph are:

O (0, 0)

A (20, 0)

B (20, 20)

C (15, 30)

D (0, 40)

At the point O (0, 0),

Z = 1.5 (0) + (0) = 0;

At the point A (20, 0),

Z = 1.5 (20) + (0) = 30;

At the point B (20, 20),

Z = 1.5 (20) + (20) =50;

At the point C (15, 30),

Z = 1.5 (15) + (30) = 52.50

At the point D (0, 40),

Z = 1.5 (0) + (40) = 40.

The maximum profit is at point C (15, 30) and the amount is Rs. 52.50.

Therefore, 15 steel pipes of Type A and 30 steel pipes of Type b should be manufactured to maximise the profit of the company.

Q1.

Solve the following linear programming problem graphically:
Maximise Z = 60x+15y
Subject to constraints
x + y ? 50
3x + y ? 90
x, y ? 0

Opt.

We have,
Maximise Z = 60x + 15y
Subject to constraints,
x + y ? 50 ? (i)
3x + y ? 90 ? (ii)
x ? 0, y ? 0   ? (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

From the graph, it is clear that the shaded portion is the optimum solution. Possible points
Ans.

We have,
Maximise Z = 60x + 15y
Subject to constraints,
x + y 50 (i)
3x + y 90 (ii)
x 0, y 0 (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

Q2.

A man has Rs. 1,500 for purchase of rice and wheat. A bag of rice and a bag of wheat cost Rs. 180 and Rs. 120 respectively. He has a storage capacity of 10 bags only. He earns a profit of Rs. 11 and Rs. 9 respectively per bag of rice and wheat. Formulate it as a linear programming problem and solve it graphically for maximum profit.

Opt.

Let the man purchases x bags of rice and y bags of wheat. So, the objective function is to minimise Z = 11x + 9y
Subject to constraints,
x + y ? 10 ? (i)
180x + 120y ? 1500
i.e., 3x + 2y ? 25 ? (ii)
x ? 0, y ? 0? (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

Ans.

Let the man purchases x bags of rice and y bags of wheat. So, the objective function is to minimise Z = 11x + 9y
Subject to constraints,
x + y 10 (i)
180x + 120y 1500
i.e., 3x + 2y 25 (ii)
x 0, y 0 (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

The feasible region of L.P.P. is the shaded region.
Maximum profit can be at the points A(25/2,0), P(5,5) and B(0,10).
The value of the objective function at these points is given:

Z is maximum at x = 5 and y = 5.
Hence, for the maximum profit, he will purchase 5 bags of rice and 5 bags of wheat.

Q3.

A manufacturer manufactures two types of steel trunks. He has two machines ? A and B. For completing the first type of trunk, it requires 3 hours on machine A and 1 hour on machine B. Machine A can work for 18 hours and B for 8 hours only per day. There is a profit of Rs.30 on the first type of trunk and Rs. 48 on the second type of trunk. How many trunks of each type should be manufactured every day to earn maximum profit? Solve graphically.

Opt.

Let the number of A type steel trunk be x per day and number of B type steel trunk be y per day. According to the given conditions, L.P.P. is
x ? 0, y ? 0   ? (1)
3x + 3y ? 48 ? x + y ? 6 ? (ii)
x + 2y ? 8 ? (iii)
and to maximise Z = 30x + 48y
Plotting these inequations on the graph, we get

We get a shaded portion as the optimum solution. Maximum pr
Ans.

Let the number of A type steel trunk be x per day and number of B type steel trunk be y per day. According to the given conditions, L.P.P. is
x 0, y 0 (1)
3x + 3y 48 x + y 6 (ii)
x + 2y 8 (iii)
and to maximise Z = 30x + 48y
Plotting these inequations on the graph, we get

We get a shaded portion as the optimum solution. Maximum profit can be at the points A(6, 0), B(4, 2) and C(0, 4).

We notice maximum profit is at B (4, 2), i.e., x = 4, y = 2.
Hence, 4 trunks of type I and 2 trunks of type II must be manufactured each day to get a maximum profit of Rs. 216.