Tamilnadu State Board New Syllabus Samcheer Kalvi 11th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 10 Operations Research Ex 10.1

### Samacheer Kalvi 11th Business Maths Operations Research Ex 10.1 Text Book Back Questions and Answers

Question 1.

A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are ₹5 and ₹3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.

Solution:

(i) Variables: Let x_{1} and x_{2} denotes the number of pens in type A and type B.

(ii) Objective function:

Profit on x_{1} pens in type A is = 5x_{1}

Profit on x_{2} pens in type B is = 3x_{2}

Total profit = 5x_{1} + 3x_{2}

Let Z = 5x_{1} + 3x_{2}, which is the objective function.

Since the B total profit is to be maximized, we have to maximize Z = 5x_{1} + 3x_{2}

(iii) Constraints:

Raw materials required for each pen A is twice as that of pen B.

i.e., for pen A raw material required is 2x_{1} and for B is x_{2}.

Raw material is sufficient only for 1000 pens per day

∴ 2x_{1} + x_{2} ≤ 1000

Pen A requires 400 clips per day

∴ x_{1} ≤ 400

Pen B requires 700 clips per day

∴ x_{2} ≤ 700

(iv) Non-negative restriction:

Since the number of pens is non-negative, we have x_{1} > 0, x_{2} > 0.

Thus, the mathematical formulation of the LPP is

Maximize Z = 5x_{1} + 3x_{2}

Subj ect to the constrains

2x_{1} + x_{2} ≤ 1000, x_{1} ≤ 400, x_{2} ≤ 700, x_{1}, x_{2} ≥ 0

Question 2.

A company produces two types of products say type A and B. Profits on the two types of product are ₹ 30/- and ₹ 40/- per kg respectively. The data on resources required and availability of resources are given below.

Formulate this problem as a linear programming problem to maximize the profit.

Solution:

(i) Variables: Let x_{1} and x_{2} denote the two types products A and B respectively.

(ii) Objective function:

Profit on x_{1} units of type A product = 30x_{1}

Profit on x_{2} units of type B product = 40x_{2}

Total profit = 30x_{1} + 40x_{2}

Let Z = 30x_{1} + 40x_{2}, which is the objective function.

Since the profit is to be maximized, we have to maximize Z = 30x_{1} + 40x_{2}

(iii) Constraints:

60x_{1} + 120x_{2} ≤ 12,000

8x_{1} + 5x_{2} ≤ 600

3x_{1} + 4x_{2} ≤ 500

(iv) Non-negative constraints:

Since the number of products on type A and type B are non-negative, we have x_{1}, x_{2} ≥ 0

Thus, the mathematical formulation of the LPP is

Maximize Z = 30x_{1} + 40x_{2}

Subject to the constraints,

60x_{1} + 120x_{2} ≤ 12,000

8x_{1} + 5x_{2} ≤ 600

3x_{1} + 4x_{2} ≤ 500

x_{1}, x_{2} ≥ 0

Question 3.

A company manufactures two models of voltage stabilizers viz., ordinary and autocut. All components of the stabilizers are purchased from outside sources, assembly and testing is carried out at company’s own works. The assembly and testing time required for the two models are 0.8 hour each for ordinary and 1.20 hours each for auto-cut. Manufacturing capacity 720 hours at present is available per week. The market for the two models has been surveyed which suggests maximum weekly sale of 600 units of ordinary and 400 units of auto-cut. Profit per unit for ordinary and auto-cut models has been estimated at ₹ 100 and ₹ 150 respectively. Formulate the linear programming problem.

Solution:

(i) Variables : Let x_{1} and x_{2} denote the number of ordinary and auto-cut voltage stabilized.

(ii) Objective function:

Profit on x_{1} units of ordinary stabilizers = 100x_{1}

Profit on x_{2} units of auto-cut stabilized = 150x_{2}

Total profit = 100x_{1} + 150x_{2}

Let Z = 100x_{1} + 150x_{2}, which is the objective function.

Since the profit is to be maximized. We have to

Maximize, Z = 100x_{1} + 15x_{2}

(iii) Constraints: The assembling and testing time required for x_{1} units of ordinary stabilizers = 0.8x_{1} and for x_{2} units of auto-cut stabilizers = 1.2x_{2}

Since the manufacturing capacity is 720 hours per week. We get 0.8x_{1} + 1.2x_{2} ≤ 720

Maximum weekly sale of ordinary stabilizer is 600

i.e., x_{1} ≤ 600

Maximum weekly sales of auto-cut stabilizer is 400

i.e., x_{2} ≤ 400

(iv) Non-negative restrictions:

Since the number of both the types of stabilizer is non-negative, we get x_{1}, x_{2} ≥ 0.

Thus, the mathematical formulation of the LPP is,

Maximize Z = 100x_{2} + 150x_{2}

Subject to the constraints

0.8x_{1} + 1.2x_{2} ≤ 720, x_{1} ≤ 600, x_{2} ≤ 400, x_{1}, x_{2} ≥ 0

Question 4.

Solve the following linear programming problems by graphical method.

(i) Maximize Z = 6x_{1} + 8x_{2} subject to constraints 30x_{1} + 20x_{2} ≤ 300; 5x_{1} + 10x_{2} ≤ 110; and x_{1}, x_{2} ≥ 0.

(ii) Maximize Z = 22x_{1} + 18x_{2} subject to constraints 960x_{1} + 640x_{2} ≤ 15360; x_{1} + x_{2} ≤ 20 and x_{1}, x_{2} ≥ 0.

(iii) Minimize Z = 3x_{1} + 2x_{2} subject to the constraints 5x_{1} + x_{2} ≥ 10; x_{1} + x_{2} > 6; x_{1}+ 4x_{2} ≥ 12 and x_{1}, x_{2} ≥ 0.

(iv) Maximize Z = 40x_{1} + 50x_{2} subject to constraints 3x_{1} + x_{2} ≤ 9; x_{1} + 2x_{2} ≤ 8 and x_{1}, x_{2} ≥ 0.

(v) Maximize Z = 20x_{1} + 30x_{2} subject to constraints 3x_{1} + 3x_{2} ≤ 36; 5x_{1} + 2x_{2} ≤ 50; 2x_{1} + 6x_{2} ≤ 60 and x_{1}, x_{2} ≥ 0.

(vi) Minimize Z = 20x_{1} + 40x_{2} subject to the constraints 36x_{1} + 6x_{2} ≥ 108; 3x_{1} + 12x_{2} ≥ 36; 20x_{1} + 10x_{2} ≥ 100 and x_{1}, x_{2} ≥ 0.

Solution:

(i) Given that 30x_{1} + 20x_{2} ≤ 300

Let 30x_{1} + 20x_{2} = 300

Therefore

3x_{1} + 2x_{2} = 30

Also given that 5x_{1} + 10x_{2} ≤ 110

Let 5x_{1} + 10x_{2} = 110

x_{1} + 2x_{2} = 22

To get point of intersection, (i.e., the to get eo-ordinates of B)

3x_{1} + 2x_{2} = 30 …….(1)

x_{1} + 2x_{2} = 22 ……..(2)

(1) – (2) ⇒ 2x_{1} = 8

x_{1} = 4

x_{1} = 4 substitute in (1),

x_{1} + 2x_{2} = 22

4 + 2x_{2} = 22

2x_{2} = 18

x_{2} = 9

i.e., B is (4, 9)

The feasible region satisfying all the given conditions is OABC.

The co-ordinates of the points are O(0, 0), A(10, 0), B(4, 9), C(0, 11).

The maximum value of Z occurs at B.

∴ The optimal solution is x_{1} = 4, x_{2} = 9 and Z_{max} = 96

(ii) Given that 960x_{1} + 640x_{2} ≤ 15360

Let 960x_{1} + 640x_{2} = 15360

3x_{1} + 2x_{2} = 48

Also given that x_{1} + x_{2} ≤ 20

Let x_{1} + x_{2} = 20

To get point of intersection

3x_{1} + 2x_{2}= 48 …..(1)

x_{1} + x_{2} = 20 ……(2)

(2) × -2 ⇒ -2x_{1} – 2x_{2} = -40 …..(3)

(1) + (3) ⇒ x_{1} = 8

x_{1} = 8 substitute in (2),

8 + x_{2} = 20

x_{2} = 12

The feasible region satisfying all the given conditions is OABC.

The co-ordinates of the comer points are O(0, 0), A(16, 0), B(8,12) and C(0, 16).

The maximum value of Z occurs at B(8, 12).

∴ The optimal solution is x_{1} = 8, x_{2} = 12 and Z_{max} = 392

(iii) Given that 5x_{1} + x_{2} ≥ 10

Let 5x_{1} + x_{2} = 10

Also given that x_{1} + x_{2} ≥ 6

Let x_{1} + x_{2} = 6

Also given that x_{1} + 4x_{2} ≥ 12

Let x_{1} + 4x_{2} = 12

To get C

5x_{1} + x_{2} = 10 ……..(1)

x_{1} + x_{2} = 6 ………(2)

(1) – (2) ⇒ 4x_{1} = 4

⇒ x_{1} = 1

x = 1 substitute in (2)

⇒ x_{1} + x_{2} = 6

⇒ 1 + x_{2} = 6

⇒ x_{2} = 5

∴ C is (1, 5)

To get B

x_{1} + x_{2} = 6

x_{1} + 4x_{2} = 12

(1) – (2) ⇒ -3x_{2} = -6

x_{2} = 2

x_{2} = 2 substitute in (1), x_{1} = 4

∴ B is (4, 2)

The feasible region satisfying all the conditions is ABCD.

The co-ordinates of the comer points are A(12, 0), B(4, 2), C(1, 5) and D(0, 10).

The minimum value of Z occours at C(1, 5).

∴ The optimal solution is x_{1} = 1, x_{2} = 5 and Z_{min} = 13

(iv) Given that 3x_{1} + x_{2} ≤ 9

Let 3x_{1} + x_{2} = 9

Also given that x_{1} + 2x_{2} ≤ 8

Let x_{1} + 2x_{2} = 8

3x_{1} + x_{2} = 9 ………(1)

x_{1} + 2x_{2} = 8 ……..(2)

(1) × 2 ⇒ 6x_{1} + 2x_{2} = 18 ……..(3)

(2) + (3) ⇒ -5x_{1} = -10

x_{1} = 2

x_{1} = 2 substitute in (1)

3(2) + x_{2} = 9

x_{2} = 3

The feasible region satisfying all the conditions is OABC.

The co-ordinates of the corner points are O(0, 0), A(3, 0), B(2, 3), C(0, 4)

The maximum value of Z occurs at (2, 3).

∴ The optimal solution is x_{1} = 2, x_{2} = 3 and Z_{max} = 230

(v) Given that 3x_{1} + 3x_{2} ≤ 36

Let 3x_{1} + 3x_{2} = 36

Also given that 5x_{1} + 2x_{2} ≤ 50

Let 5x_{1} + 2x_{2} = 50

x_{1} + x_{2} = 12 ……(1)

5x_{1} + 2x_{2} = 50 ……….(2)

(1) × 2 ⇒ 2x_{1} + 2x_{2} = 24 ………(3)

(2) – (3) ⇒ 3x_{1} = 26

x_{1} = \(\frac{26}{3}\) = 8.66

put x_{1} = \(\frac{26}{3}\) substitute in (1)

x_{1} + x_{2} = 12

x_{2} = 12 – x_{1}

x_{2} = 26 – \(\frac{26}{3}\) = \(\frac{10}{3}\) = 3.33

Also given that 2x_{1} + 6x_{2} ≤ 60

Let 2x_{1} + 6x_{2} = 60

x_{1} + 3x_{2} = 30

x_{1} + x_{2} = 12 …….(1)

x_{1} + 3x_{2} = 30 …….(2)

(1) – (2) ⇒ -2x_{2} = -18

x_{2} = 9

x_{2} = 9 substitute in (1) ⇒ x_{1} = 3

The feasible region satisfying all the given conditions is OABCD.

The co-ordinates of the comer points are O(0, 0), A(10, 0), B(\(\frac{26}{3}, \frac{10}{3}\)), and C = (3, 9) and D (0, 10)

The maximum value of Z occurs at C(3, 9)

∴ The optimal solution is x_{1} = 3, x_{2} = 9 and Z_{max} = 330

(vi) Given that 36x_{1} + 6x_{2} ≥ 108

Let 36x_{1} + 6x_{2} = 108

6x_{1} + x_{2} = 18

Also given that 3x_{1} + 12x_{2} ≥ 36

Let 3x_{1} + 12x_{2} = 36

x_{1} + 4x_{2} = 12

Also given that 20x_{1} + 10x_{2} ≥ 100

Let 20x_{1} + 10x_{2} = 100

2x_{1} + x_{2} = 10

The feasible region satisfying all the conditions is ABCD.

The co-ordinates of the comer points are A(12, 0), B(4, 2), C(2, 6) and D(0, 18).

The minimum value of Z occurs at B(4, 2)

∴ The optimal solution is x_{1} = 4, x_{2} = 2 and Z_{min} = 160