# Exam Question for Class 12 Mathematics Chapter 12 Linear Programming

Please refer to below Exam Question for Class 12 Mathematics Chapter 12 Linear Programming. These questions and answers have been prepared by expert Class 12 Mathematics teachers based on the latest NCERT Book for Class 12 Mathematics and examination guidelines issued by CBSE, NCERT, and KVS. We have provided Class 12 Mathematics exam questions for all chapters in your textbooks. You will be able to easily learn problems and solutions which are expected to come in the upcoming class tests and exams for standard 10th.

## Chapter 12 Linear Programming Class 12 Mathematics Exam Question

All questions and answers provided below for Exam Question Class 12 Mathematics Chapter 12 Linear Programming are very important and should be revised daily.

**Exam Question Class 12 Mathematics Chapter 12 Linear Programming**

**Question**. Find graphically, the maximum value of z = 2x + 5y,**subject to constraints given below:****2x + 4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4; x ≥ 0, y ≥ 0**** Answer.** Let l

_{1}: 2x + 4y = 8, l

_{2}: 3x + y = 6, l

_{3}: x + y = 4;

x = 0, y = 0

**Question**. Maximise z = 8x + 9y subject to the constraints**given below : 2x + 3y ≤ 6, 3x – 2y ≤ 6, y ≤ 1; x, y ≥ 0**** Answer.** Let l1 : 2x + 3y = 6, l2 : 3x – 2y = 6, l3 : y = 1;

x = 0, y = 0

**Question**. A manufacturer produces two products A and**B. Both the products are processed on two** **different machines. The available capacity of** **first machine is 12 hours and that of second machine is 9 hours per day. Each unit of** **product A requires 3 hours on both machines** **and each unit of product B requires 2 hours on** **first machine and 1 hour on second machine.** **Each unit of product A is sold at ₹7 profit and** **that of B at a profit of ₹4. Find the production** **level per day for maximum profit graphically.**** Answer.** The given information can be represented in the tabular form as below:

Let the manufacturer produces x units of product A and y units of product B per day.

∴ 3x + 2y ≤ 12 and 3x + y ≤ 9

Let Z denote the total profit.

∴ Z = 7x + 4y

Clearly x ≥ 0 and y ≥ 0.

Above LPP can be stated mathematically as:

Maximise Z = 7x + 4y

subject to the constraints

3x + 2y ≤ 12, 3x + y ≤ 9 and x, y ≥ 0

To solve graphically, we convert the inequations into equations to obtain the following lines:

3x + 2y = 12, 3x + y = 9, x = 0, y = 0

The line 3x + 2y = 12 meets the coordinate axes at A(4, 0) and B(0, 6). Similarly 3x + y = 9 meets the coordinate axes at C(3, 0) and D(0, 9)

The point of intersection of lines 3x + 2y = 12 and 3x + y = 9 is E(2, 3).

Coordinates of the corner points of the feasible region OCEB are O(0, 0), C(3, 0), E(2, 3), B(0, 6) Values of the objective function at corner points of the feasible region are

∴ Z is maximum at x = 2, y = 3 So, for maximum profit the manufacturer should manufacture 2 units of product A and 3 units of product B.

**Question**. A company manufactures three kinds of calculators : A, B and C in its two factories I and II. The company has got an order for manufacturing atleast 6400 calculators of kinds A, 4000 of kind B and 4800 of kind C. The daily output of factory I is of 50 calculators of kind A, 50 calculators of kind B and 30 calculators of kind C. The Daily output of factory II is of 40 calculators of kind A, 20 of kind B and 40 of kind C. The cost per day to run factory I is ₹ 12,000 and of factory II is ₹ 15,000. How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this problem as an LPP and solve graphically. ** Answer.** Let x and y be the number of days, factory I and factory II have to be in operation to produce the order, respectively.

The inequations thus formed based on the given problem are as follows :

50x + 40y ≥ 6400 ⇒ 5x + 4y ≥ 640,

50x + 20y ≥ 4000 ⇒ 5x + 2y ≥ 400,

30x + 40y ≥ 4800 ⇒ 3x + 4y ≥ 480; x, y ≥ 0

Let Z be the total cost of production.

∴ Z = 12000x + 15000y

So, the given LPP can be mathematically stated as

Minimise Z = 12000x + 15000y

Subject to constraints

5x + 4y ≥ 640, 5x + 2y ≥ 400, 3x + 4y ≥ 480; x, y ≥ 0

To solve the LPP graphically, we convert inequations into equations to obtain the following lines :

5x + 4y = 640, 5x +2y = 400, 3x + 4y = 480, x = 0 and y = 0

The line 5x + 4y = 640 meets the coordinate axes at A(128, 0) and B(0, 160)

The line 5x + 2y = 400 meets the coordinate axes at C(80, 0) and D(0, 200)

The line 3x + 4y = 480 meets the coordinate axes at E(160, 0) and F(0, 120)

The point of intersection of lines 5x + 4y = 640 and 5x + 2y = 400 is Q(32, 120).

The point of intersection of lines 5x + 4y = 640 and 3x + 4y = 480 is P(80, 60).

The shaded region DQPE represents the feasible region of the given LPP.

The corner points of the feasible region are D(0, 200), Q(32, 120), P(80, 60) and E(160, 0)

The values of the objective function at these points are given as follows:

From the table, we find that 1860000 is the minimum value of Z at P(80, 60). Since the region is unbounded we have to check that the inequality 12000x + 15000y < 1860000 in open half plane has any point in common or not. Since it has no point in common.

So, minimum value of Z is at P(80, 60) and the minimum value of Z is 18,60,000.

**Question**. In order to supplement daily diet, a person wishes to take X and Y tablets. The contents (in milligrams per tablet) of iron, calcium and vitamins in X and Y are given as below:

**The person needs to supplement at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamins. The price of each tablet of X and Y is Rs. 2 and Rs. 1 respectively. How many tablets of each type should the person take in order to satisfy the above requirement at the minimum cost? Make an LPP and solve graphically. **** Answer.** Let the person takes x tablets of type X and y tablets of type Y.

According to the given conditions, we have

6x + 2y ≥ 18 ⇒ 3x + y ≥ 9

3x + 3y ≥ 21 ⇒ x + y ≥ 7

2x + 4y ≥ 16 ⇒ x + 2y ≥ 8

Let z be the total cost of tablets.

∴ z = 2x + y

Hence, the given LPP is

Minimise Z = 2x + y

subject to the constraints

3x + y ≥ 9, x + y ≥ 7, x + 2y ≥ 8 and x, y ≥ 0

To solve graphically, we convert the inequations into equations.

3x + y = 9, x + y = 7, x + 2y = 8, x = 0, y = 0

The line 3x + y = 9 meets the coordinate axes at A(0, 9) and B(3, 0). Similarly, x + y = 7 meets the coordinate axes at C(0, 7) and D(7, 0). Also, line

x + 2y = 8 meets the coordinate axes at E(0, 4), F(8, 0)

The point of intersection of the lines 3x + y = 9 and x + y = 7 is G(1, 6).

Also, the point of intersection of the lines x + y = 7 and x + 2y = 8 is H(6, 1).

The shaded region AGHF represents the feasible region of the given LPP. The corner points of the feasible region are A(0, 9), G(1, 6), H(6, 1) and F(8, 0).

The values of the objective function at these points are given in the following table :

From the table, we find that 8 is the minimum value of Z at G(1, 6). Since the region is unbounded we have to check that the inequality 2x + y < 8 in open half plane has any point in common or not.

Since, it has no point in common. So, Z is minimum at G (1, 6) and the minimum value of Z is 8.

Hence, the person should take 1 tablet of type X and 6 tablets of type Y in order to meet the requirements at the minimum cost.

**Question**. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost ₹ 25,000 and ₹ 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit, if he does not want to invest more than ₹ 70 lakhs and his profit on the desktop model is ₹ 4500 and on the portable model is ₹ 5,000. Form it as an LPP and solve it graphically.**Answer.** Let the merchant stocks x desktop model computers and y portable model computers.

Hence, the mathematical formulation of the problem is

Maximize Z = 4500x + 5000y

subject to the constraints

x + y ≤ 250, 25000x + 40000y ≤ 7000000,

5x + 8y ≤ 1400; x, y ≥ 0

To solve LPP graphically we convert inequations into equations

The feasible region OAED is shown in the graph.

Here we observe that the feasible region is bounded.

The corner points are O(0, 0), A(250, 0), E(200, 50) and D(0, 175).

We find that maximum value of Z is 1150000 at E(200, 50). Hence the merchant should stock 200 units of desktop model and 50 units of portable model to realise maximum profit and maximum profit is ₹ 1150000.

**Question**. A library has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weigh 1 kg and 1(1/2) kg each respectively. The shelf is 96 cm long and at most can support a weight of 21 kg. How should the shelf be filled with the books of two types in order to include the greatest number of books ? Form it as an LPP and solve it graphically.**Answer.** Let x be the number of books of thickness 6 cm (type I) and y be the number of books of thickness 4 cm (type II) and x ≥ 0, y ≥ 0

∴The required LPP is

Maximise Z = x + y

Subject to constraints

6x + 4y ≤ 96 ⇒ 3x + 2y ≤ 48,

The shaded region OCEB is the feasible region. The corner points of feasible region are O(0, 0), C (0, 14), E (12, 6) and B (16, 0)

Clearly, Z is maximum at E(12, 6) i.e., 18.

Hence, 12 books of the type I and 6 books of type II can be arranged in the shelf.

** Question. One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it an LPP and solve it graphically.Answer.** Let x be the number of cakes of I kind and y be the number of cakes of II kind.

The required LPP is

Maximise Z = x + y subject to constraints

200x + 100y ≤ 5,000 ⇒ 2x + y ≤ 50

25x + 50y ≤ 1,000 ⇒ x + 2y ≤ 40

x ≥ 0, y ≥ 0

To solve LPP graphically, we convert inequations into equations.

2x + y = 50 …(i), x + 2y = 40 …(ii)

Lines (i) and (ii) intersect at P(20, 10).

Shaded region is the feasible region i.e. OAPD. The corner points of the feasible region are O(0, 0),

A(25, 0), P(20, 10), D(0, 20).

Clearly, the number of cakes is maximum at P(20, 10) i.e., when 20 cakes of I kind and 10 cakes of II kind are made.

**Question**. A manufacturer produces nuts and bolts. It take 2 hours work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 2 hours on machine B to produce a package of bolts. He earns a profit of ₹24 per package on nuts and ₹18 per package on bolts. How many package of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 10 hours a day. Make an LPP from above and solve it graphically? ** Answer.** Let x and y be the number of packages of nuts and bolts manufactured respectively by the manufacturer. Then by the given data, the required LPP is

Maximize Z = 24x + 18y

Subject to constraints

2x + 3y ≤ 10, 3x + 2y ≤ 10, x ≥ 0, y ≥ 0.

We have the lines

l1 : 2x + 3y = 10 and l2 : 3x + 2y = 10

These lines intersect at P(2, 2)

∴ Profit, Z = `84 will be maximum, when 2 packages nuts and 2 packages of bolts are manufactured.

**Question**. If a young man rides his motorcycle at 25 km per hour, he had to spend ₹2 per km on petrol with very little pollution in the air. If he rides it at a faster speed of 40 km per hour, the petrol cost increases to ₹5 per km and rate of pollution also increases. He has ₹100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this problem as an LPP. Solve it graphically to find the distance to be covered with different speeds. What value is indicated in this question?**Answer.**

**Question**. A cooperative society of farmers has 50 hectares of land to grow two crops A and B. e profits from crops A and B per hectare are estimated as ₹10,500 and ₹9,000 respectively. To control weeds a liquid herbicide has to be used for crops A and B at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. Keeping in mind that the protection of fish and other wildlife is more important than earning profit, how much land should be allocated to each crop so as to maximise the total profit? Form an LPP from the above information and solve it graphically. Do you agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment?**Answer.** Let x hectare of land to be allocated to crop A and y hectare to B.

Thus, the LPP can be formulated as

Maximise Z = 10,500x + 9,000y

subject to the constraints

x + y ≤ 50, 20x + 10y ≤ 800 ⇒ 2x + y ≤ 80, x ≥ 0, y ≥ 0

To solve LPP graphically, we convert inequations into equations.

l1 : x + y = 50, l2 : 2x + y = 80

x = 0, y = 0

Both the lines intersect at B(30, 20)

OABC is the feasible region which is bounded.

The corner points are O(0, 0), A(40, 0), B(30, 20), C(0, 50).

Hence society will allocate 30 hectares of land to crop A and 20 hectares of land to crop B to maximise the total profit.

Yes, protection of wildlife is necessary to preserve balance in environment because it will be a loss of biodiversity. The wild animals would get extinct ultimately if we would not provide protection to them. Wild life animals are being killed for their valuable ivory, skin, fur etc. to make products such as leather, meat etc.

**Question**. A decorative item dealer deals in two items A and B. He has ₹ 15,000 to invest and a space to store at the most 80 pieces. Item A costs him ₹ 300 and item B costs him ₹ 150. He can sell items A and B at respective profits of ₹ 50 and ₹ 28. Assuming he can sell all he buys, formulate the linear programming problem in order to maximise his profit and solve it graphically.**Answer.** Let the number of items of the type A and B be x and y respectively. Then the LPP is

Maximise Z = 50x + 28y

Subject to the constraints,

x + y ≤ 80,

300x + 150y ≤ 15000 ⇒ 2x + y ≤ 100,

x ≥ 0, y ≥ 0

To solve LPP graphically, we convert inequations into equations.

l_{1} : x + y = 80, l_{2} : 2x + y = 100

Their point of intersection is P(20, 60). The feasible region OBPC is shown shaded in the figure with corner points O(0, 0), B(50, 0), P(20, 60) and C(0, 80)

Thus, the maximum profit of ` 2680 is at P (20, 60)

i.e., when 20 items of type A and 60 items of type B are purchased and sold.