Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 7 Linear Programming Ex 7.2 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.2
I) Find the feasible solution of the following inequations graphically.
Question 1.
3x + 2y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 2y = 18 and 2x + y = 10 respectively.
The feasible solution is OCPBO which is shaded in the graph.
Question 2
Maharashtra Board Solution
2x + 3y ≤ 6, x + y ≥ 2, x ≥ 0, y ≥ 0
Solution & Step-by-Step Answer:
First we draw the lines AB and CB whose equations are 2x + 3y = 6 and x + y = 2 respectively. The feasible solution is ∆ABC which is shaded in the graph.
Question 3
Maharashtra Board Solution
3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0
Solution & Step-by-Step Answer:
First we draw the lines AB, CD and EF whose equations are 3x + 4 y = 12, 4x + 7y = 28 and y = 1 respectively. The feasible solution is PQDBP. which is shaded in the graph.
Question 4
Maharashtra Board Solution
x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution & Step-by-Step Answer:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively. The feasible solution is OCPQBO. which is shaded in the graph.
Question 5
Maharashtra Board Solution
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5, 2x + y ≥ 4
Solution & Step-by-Step Answer:
First we draw the lines AB, CD, EF and GH whose equations are x + y = 5, 2x + y = 4, x = 3 and y = 3 respectively. The feasible solution is CEPQRC. which is shaded in the graph.
Question 6
Maharashtra Board Solution
x – 2y ≤ 2, x + y ≥ 3, -2x + y ≤ 4, x ≥ 0, y ≥ 0
Solution & Step-by-Step Answer:
First we draw the lines AB, CD and EF whose equations are x – 2y = 2, x + y = 3 and -2x + y = 4 respectively. The feasible solution is shaded in the graph.
Question 7
Maharashtra Board Solution
A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find the feasible solution graphically.
Solution & Step-by-Step Answer:
Let the company produces x units of article A and y units of article B. The given data can be tabulated as: Inequations are : x + 2y ≤ 4 and 3x + 2y ≤ 6 x and y are number of items, x ≥ 0, y ≥ 0 First we draw the lines AB and CD whose equations are x + 2y = 4 and 3x + 2y = 6 respectively. The feasible solution is OCPBO. which is shaded in the graph.
Question 8
Maharashtra Board Solution
A furniture dealer deals in tables and chairs. He has Rs.1,50,000 to invest and a space to store at most 60 pieces. A table costs him Rs.1500 and a chair Rs.750. Construct the inequations and find the feasible
Solution & Step-by-Step Answer:
Question is modified A furniture dealer deals in tables and chairs. He has ₹ 15,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 150 and a chair ₹ 750. Construct the inequations and find the feasible solution. Solution: Let x be the number of tables and y be the number of chairs. Then x ≥ 0, y ≥ 0. The dealer has a space to store at most 60 pieces. ∴ x + y ≤ 60 Since, the cost of each table is ₹ 150 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 150x + 750y. Since the dealer has ₹ 15,000 to invest, 150x + 750y ≤ 15,000 Hence the system of inequations are x + y ≤ 60, 150x + 750y ≤ 15000, x ≥ 0, y ≥ 0. First we draw the lines AB and CD whose equations are x + y = 60 and 150x + 750y = 15,000 respectively. The feasible solution is OAPDO. which is shaded in the graph.
