RD Sharma Solutions for Class 12 Maths Chapter 30 Linear Programming Exercise 30.1, is provided here to help students come out with flying colours in their board examinations. Our expert faculty team has solved the questions in a step-by-step format, which helps students understand the concepts clearly. Start practising offline by downloading the PDF of RD Sharma Solutions for Class 12 now and these can be used for future reference, as well.
RD Sharma Solutions for Class 12 Maths Chapter 30 Exercise 1
Access RD Sharma Solutions for Class 12 Maths Chapter 30 Exercise 1
EXERCISE 30.1
Q1.
Solution:
Let us consider the data in the tabular form:
| Gadget | Foundry | Machine-shop | Profit |
| A | 10 | 5 | Rs 30 |
| B | 6 | 4 | Rs 20 |
| Firm’s capacity per week | 1000 | 600 |
Now, let x and y be the required weekly production of gadgets A and B.
Given:
Profit on each gadget A is = Rs 30
So, profit on x gadget of type A = 30x
Profit on each gadget B is = Rs 20
So, profit on x gadget of type B = 20y
Let total profit be ‘Z’
So, Z = 30x + 20y
Case I:
Given:
Production of one gadget A requires 10hours per week for foundry.
Production of one gadget B requires 6hours per week for foundry.
So, x units of gadget A requires 10x hours per week and
y unit of gadget B requires 6y hours per week.
Maximum capacity of foundry per week is 1000 hours, so => 10x + 6y ≤ 1000
Case II:
Given:
Production of one gadget A requires 5hours per week of machine-shop.
Production of one unit of gadget B requires 4hours per week of machine-shop.
So, x units of gadget A requires 5x hours per week and
y unit of gadget B requires 4y hours per week.
Maximum capacity of machine-shop per week is 600 hours, so => 5x + 4y ≤ 600
Hence, the required mathematical formulation of linear programming is:
Maximize Z = 30x + 20y
Subject to
10x + 6y ≤ 1000
5x + 4y ≤ 600
Where, x, y ≥ 0
Q2.
Solution:
Let us consider the data in the tabular form:
| Product | Machine hours | Labour hours | Profit |
| A | 1 | 1 | Rs 60 |
| B | – | 1 | Rs 80 |
| Total capacity
Minimum supply of product B is 200 units. |
400 for A | 500 |
Now, let x and y be the required production of products A and B.
Given:
Profit on one unit of product A is = Rs 60
So, profit on x unit of product A = 60x
Profit on one unit of product B is = Rs 80
So, profit on x unit of product B = 80y
Let total profit be ‘Z’
So, Z = 60x + 80y
First constraint:
Given, Minimum supply of product B is 200
So, y ≥ 200
Second constraint:
Given:
Production of one unit of product A requires 1hour per week of machine hours.
So, x units of product A requires 1x hour per week and
Total machine hours available for product A is 400hours,
So, x ≤ 400
Third constraint:
Given:
Production of one unit of product A requires 1hour per week of labour hours.
Production of one unit of product B requires 1hour per week of labour hours.
So, x units of product A requires 1x hour per week and
y units of product B requires 1x hour per week.
Total labour hours available is 500hours,
So, x + y ≤ 500
Hence, the required mathematical formulation of linear programming is:
Minimize Z = 60x + 80y
Subject to
x ≤ 400
x + y ≤ 500
Where, x, y ≥ 0
Q3.
Solution:
Let us consider the data in the tabular form:
| Product | Machine (M1) | Machine (M2) | Profit |
| A | 4 | 2 | 3 |
| B | 3 | 2 | 2 |
| C | 5 | 4 | 4 |
| Capacity maximum | 2000 | 2500 |
Now, let x, y and z units be the required production of products A,B and C.
Given:
Profit on one unit of product A is = Rs 3
So, profit on x unit of product A = 3x
Profit on one unit of product B is = Rs 20
So, profit on x unit of product B = 2y
Profit on one unit of product C is = Rs 4
So, profit on x unit of product C = 4z
Let total profit be ‘U’
So, U = 3x + 2y + 4z
First Constraint:
Given:
One unit of product A requires 4minutes on machine, M1
One unit of product B requires 3minutes on machine, M1
One unit of product C requires 5minutes on machine, M1
So,
x unit of product A requires 4x minutes on machine, M1
y unit of product B requires 3y minutes on machine, M1
z unit of product C requires 5z minutes on machine, M1
Total minutes on M1 = 2000 minutes
i.e., 4x + 3y + 5z ≤ 2000
Second constraint:
Given:
One unit of product A requires 2minutes on machine, M2
One unit of product B requires 2minutes on machine, M2
One unit of product C requires 4minutes on machine, M2
So,
x unit of product A requires 2x minutes on machine, M2
y unit of product B requires 2y minutes on machine, M2
z unit of product C requires 4z minutes on machine, M2
Total minutes on M2 = 2500 minutes
i.e., 2x + 2y + 4z ≤ 2500
Other constraints:
Given:
Firm must manufacture 100A’s, 200B’s and 50C’s but not more than 150A’s.
100 ≤ x ≤ 150
y ≥ 200
z ≥ 50
Hence, the required mathematical formulation of linear programming is:
Maximize U = 3x + 2y + 4z
Subject to
4x + 3y + 5z ≤ 2000
2x + 2y + 4z ≤ 2500
100 ≤ x ≤ 150
y ≥ 200
z ≥ 50
Where, x, y, z ≥ 0
Q4.
Solution:
Let us consider the data in the tabular form:
| Product | M1 | M2 | Profit |
| A | 1 | 2 | 2 |
| B | 1 | 1 | 3 |
| Capacity | 6 hours 40minutes = 400minutes | 10 hours = 600 minutes |
Now, let x and y be the required production of products A and B.
Given:
Profit on one unit of product A is = Rs 2
So, profit on x unit of product A = 2x
Profit on one unit of product B is = Rs 3
So, profit on x unit of product B = 3y
Let total profit be ‘Z’
So, Z = 2x + 3y
First Constraint:
Given:
One unit of product A requires 1minutes on machine, M1
One unit of product B requires 1minutes on machine, M1
So,
x unit of product A requires 1x minutes on machine, M1
y unit of product B requires 1y minutes on machine, M1
Total minutes on M1 = 2000 minutes
i.e., x + y ≤ 400
Second constraint:
Given:
One unit of product A requires 2minutes on machine, M2
One unit of product B requires 1minutes on machine, M2
So,
x unit of product A requires 2x minutes on machine, M2
y unit of product B requires 1y minutes on machine, M2
Total minutes on M2 = 2500 minutes
i.e., 2x + y ≤ 600
Hence, the required mathematical formulation of linear programming is:
Maximize Z = 2x + 3y
Subject to x + y ≤ 400
2x + y ≤ 600
Where, x, y ≥ 0
Q5.
Solution:
Let us consider the data in the tabular form:
| Plant | A | B | C | Cost |
| I | 50 | 100 | 100 | 2500 |
| II | 60 | 60 | 200 | 3500 |
| Monthly demand | 2500 | 3000 | 7000 |
Now, let x and y be the required days of plant I and II to minimize cost.
Given:
Plant I costs per day = Rs 2500
So, cost to run plant x requires per month = Rs 2500x
Plant II costs per day = Rs 3500
So, cost to run plant y requires per month = Rs 3500y
Let total cost per month be ‘Z’
So, Z = 2500x + 3500y
First Constraint:
Given:
Production of type A from plant I requires = 50
Production of type A from plant II requires = 60
So,
x unit of production of type A from plant I requires = 50x
y unit of production of type A from plant II requires = 60y
Total demand of type A per month = 2500
i.e., 50x + 60y ≥ 2500
Second Constraint:
Given:
Production of type B from plant I requires = 100
Production of type B from plant II requires = 60
So,
x unit of production of type B from plant I requires = 100x
y unit of production of type B from plant II requires = 60y
Total demand of type B per month = 3000
i.e., 100x + 60y ≥ 3000
Third Constraint:
Given:
Production of type C from plant I requires = 100
Production of type C from plant II requires = 200
So,
x unit of production of type C from plant I requires = 100x
y unit of production of type C from plant II requires = 200y
Total demand of type C per month = 7000
i.e., 100x + 200y ≥ 7000
Hence, the required mathematical formulation of linear programming is:
Minimize Z = 2500x + 3500y
Subject to
50x + 60y ≥ 2500
100x + 60y ≥ 3000
100x + 200y ≥ 7000
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