Plus Two Maths Chapter Wise Previous Questions Chapter 12 Linear Programming are part of Plus Two Maths Chapter Wise Previous Year Questions and Answers. Here we have given Plus Two Maths Chapter Wise Previous Chapter 12 Linear Programming.

## Kerala Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 12 Linear Programming

Question 1.

A manufacturer produces nuts and bolts. It takes 1 hour of work on Machine A and 3 hours on Machine B to produce a package of nuts. It take 3 hours on Machine A and 1 hour on Machine B to produce a package of bolts. He earns a profit of ? 17.50 per package on nuts and ? 7.00 per package on bolts. Formulate the above L.P.P, if the machines operates for at most 12 hours a day. **[March-2018]
**Answer:

i. Let x = no. of packets of nuts

y = no. of packets of bolts

ii. LPP is maximise, z = 17.5x + 7y subject to x + 3y ≤ 12, 3x + y ≤ 12 x ≥ 0, y ≥ 0

(4, 0),(3,3), (0, 4), (0,0) are the comers among these profit is maximum at (3, 3). Maximum profit = 17.5 x 3+7 x 3=73.5

Question 2.

Solve the L.P.P. given below graphically:

Minimise Z=-3x+4y

Subject to x+2y ≤8

3x+2y ≤12

x≥0.y≥0 **[March-2018]**

Answer:

Minimize z= -3x+4y

subject to x+2y≤ 8

3x+2y≤12

x≥0.y≥0

x+2y≤8 3x+2y≤2

x 0 8 x 0 4

y 4 0 y 6 0

Hence, z=-12 is minimum at (4,0)

Question 3.

Consider the linear programming problem: Maximize Z = 50x + 40y Subject to the constraints

x + 2y ≥ 10

3x + 4y ≤ 24

x ≥ 0, y ≥ 0

a. Find the feasible region.

b. Find the corner points of the feasible region.

c. Find the maximum value of Z.

Maximize Z = 50x + 40y Subject to the constraints

x + 2y ≥ 10

3x + 4y ≤ 24

x ≥ o ≥ y ≥ o** [March-2017] **

Answer:

z = 50 x + 40 y …………….(1)

x + 2y≥10 …………..(2)

3x + 4y≤ 24 …….(3)

x≥0, y≥0 …………(4)

Answer:

a. The shaded region in the graph determined by the system of constraints (2) to (4).

x + 2y = 10 3x + 4y = 24

Here OABC is the feasible region.

b. The corner points are A(0, 5), B(4, 3), C(0, 6)

c. Comer points z = 50 x + 40 y

A(0,5) 200

B(4,3) 320

C(0,6) 240

∴ is maximum at the point A(8, 0). maximum value of z = 320

Question 4.

Consider the following L.P.P. Maximize Z = 3x+2y

Subject to the constraints

x+2y≤10

3 x 4-y≤15

x,y≥0

a. Draw its feasible region.

b. Find the corner points of the feasible region

c. Find the maximum value of Z.** [March-2016]
**Answer:

Question 5.

Consider the linear inequalities 2x + 3y ≤ 6, 2x + y ≤ 4, x ≥ 0, y ≥ 0.

a. Mark the feasible region.

b. Maximise the function z = 4x + 5y subject to the given constraints. **[March-2015]**

Answer:

b. Z = 4×+ 5y

at (2,0), Z=4× 2+5×0=8

at (0,2), Z = 4×0 + 5×2 = 10

at (3/2,1), Z = 4×3/2 + 5×1 =11

Z is maximum at (3/2, 1)

Question 6.

In a factory, there are two machines A and B producing toys. They respectively produce 60 and 80 units in one hour. A can run a maximum of 10 hours and B a maximum of 7 hours a day. The cost of their running per hour respectively amounts to 2,000 and 2,500 rupees. The total duration of ; working these machines cannot exceed 12 hours a day. If the total cost cannot exceed rupees 25,000 per day and the total daily production is a least 800 units, then formulate the problem mathematically. **[March-2015]**

Answer:

Let machine A works x hours a day,

machine B works y hours a day.

Total production per day = 60x + 80y x≤ 10, y≤7

Total cost of working = 2000x + 2500y x + y≤ 12 ; 2000x + 2500y ≤ 25000

60x + 80y ≥ 800

Minimize, Z = x + y, subject to the above linear constraints.

Question 7.

Consider the linear programming problem:

Maximise z = 5x + 3y, subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y≥ 0.

a. Draw its feasible region.

b. Find the corner points of the feasible region.

c. Find the corner at which z attains its maximum. **[March-2013]
**Answer:

Question 8.

Consider the LPP, Maximize z = x + y, subject to 2x + y – 3 ≥ 0, x – 2y + 1 ≤ 0, y≤3, x ≥ 0,y ≥ 0.

i. Draw the feasible region.

ii. Find the co-ordinates of the corner points of the feasible region.

iii. Solve the LPP. **[June-2012]**

Answer:

ii. Comer points are (0, 5), (4, 3) & (0,6) Among these minimum value occurs at (4,3).

iii. Minimum value of

z = 200 x 4 + 500 x 3= 2300

Question 9.

Consider the LPP, Maximize z = x + y, subject to 2x + y – 3 ≥ 0, x – 2y + 1≤0, y≤3, x≥0,y≥0.

i. Draw the feasible region.

ii. Find the co-ordinates of the corner points of the feasible region.

iii. Find the corner at which z attains maximum.

Answer:

ii. Comer points are (0, 3), (1 .1), (5, 3)

iii. Z attains maximum at (5,5)

Maximum value of z = 5 + 3 = 8

Question 10.

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a packets of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a packet of bolts. He earns a profit of Rs 17.50 per packet on nuts and Rs 7.00 per packet on bolts. How many packets of each should be produced each day so as to maximise his profit? If he operates his machine for at the most 12 hours a day.

i. By suitably defining the variables, write the objective function of the problem.

ii. Formulate the problem as Linear programming problem.

iii. Solve the problem by graphical method and find the number of packets of nuts and bolts to be manufactured. **[June-2011]
**Answer:

i. Let x = no. of packets of nuts

y = no. of packets of bolts

ii. LPP is maximise, z = 17.5x + 7y subject to x + 3y ≤ 12, 3x + y ≤ 12 x ≥ 0, y ≥ 0

(4, 0),(3,3), (0, 4), (0,0) are the comers among these profit is maximum at (3, 3). Maximum profit = 17.5 x 3+7 x 3=73.5

Question 11.

The graph of LPP is given below. The shaded region is the feasible region. The objective function is Max. z = px +qy

a. What are the co-ordinates of the corners of feasible region?

b. Write the constraints.

c. If the Max. Z occurs at A and B, what is the relation between p and q?

d. If q = 1, write the objective function.

e. Find the Max. Z. **[June-2012]
**Answer:

a. (0, 0), (5, 0), (0, 5) and (3, 4)

b. x, y ≥ 0, x + 3y ≤ 15 and 2x + y ≤ 10

c. Max Z occurs at (5, 0) and (3, 4).

z = 5p + Oq and z = 3p + 4q, but both the values are the same.

i.e., 5p = 3p + 4q ; 2p = 4q; p = 2q

d. z = 2x + y [if q = 1, p = 2]

e. Z = 2 × 5 + 0 = 10

Question 12.

A company produces two types of cricket balls A and B. The production time of one ball of type B is double the type A. The company has the time to produce a maximum of 2000 balls per day. The supply of raw materials is sufficient for the production of 1500 balls (both A and B) per day. The company wants to make maximum profit by making profit of Rs 3 from ball of type A and Rs 5 from ball of type B. Then

a. Write the objective function.

b. Write the constraints.

c. How many balls should be produced in each type per day to get maximum profit? **[June – 2010]
**Answer:

a. No. of balls of type A = x

No. of balls of type B = y

b. Profit = 3x + 5y; x+2y ≤ 2000

x + 2y≤ 2000 ; x + y≤ 1500

x ≥ 0 ; y ≥ 0

x + 2y ≤ 2000; x + y≤ 1500

Max. profit = 3 x 1000 + 5 × 500 = 5500

Question 13.

A furniture dealer sells only tables and chairs. He has Rs 12000 to invest and a space to store 90 pieces. A table costs him Rs 400 and a chair Rs 100. He can sell a table at a profit of Rs 75 and chair at a profit of Rs 25. Assume that he can sell all the items. The dealer wants to get maximum profit.

a. By defining suitable variables,write the objective function. Write the constraints.

b. Maximise the objective function graphically. **[March- 2010]
**Answer:

a. No. of tables = x, No. of chairs = y

Objective function Z = 75x + 25y

b. x + y≤ 90; 400x + 100y ≤ 12000

i.e. 4x + y ≤ 120; x≥ 0; y≥ 0

c. Comer points are (0,90), (30,0), (10,80)

at (0,90), z = 75 × 0 + 25 × 90 = 2250

at (30,0), z = 75 ×30 + 25× 0 = 2250

at (10,80), z = 75 × 10 + 25×80 = 2750

Z is maximum at (10,80).

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