RD Sharma Solutions for Class 12 Maths Chapter 30 Linear Programming Exercise 30.2, is provided here. For students who aim to secure an excellent score, solving RD Sharma Solutions Class 12 is a must. These solutions will also help you in obtaining knowledge and strong command over the concept. Students can refer and download the PDF which is readily available from the links given below.
RD Sharma Solutions for Class 12 Maths Chapter 30 Exercise 2
Access RD Sharma Solutions for Class 12 Maths Chapter 30 Exercise 2
EXERCISE 30.2
Q1.
Solution:
Firstly, let us convert the given inequations into equations, we get
3x + 5y = 15,
5x + 2y = 10,
x = 0,
y = 0
Region represented by 3x + 5y ≤ 15:
The line meets the coordinate axes at A2(5,0) and B2(0,3). By joining these points we obtain the line 3x + 5y = 15.
So, (0,0) satisfies the inequation 3x + 5y ≤ 15. Hence, the region containing the origin represents the solution set of the inequation 3x + 5y ≤ 15.
Region represented by 5x + 2y ≤ 10:
The line meets the coordinate axes at A1(2,0) and B1(0,5). By joining these points we obtain the line 5x + 2y = 10.
So, (0,0) satisfies the inequation 5x + 2y ≤ 10. Hence, the region containing the origin represents the solution set of the inequation 5x + 2y ≤ 10.
Region represented by x ≥ 0, y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, y ≥ 0.
The coordinates of the corner points of shaded region are O(0,0), A(2,0), P(20/19, 45/19), B2(0,3)
Q2.
Solution:
Firstly let us convert the given Inequations into equations, we get
2x + 3y = 13,
3x + y = 5,
x = 0,
y = 0
Region represented by 2x + 3y ≤ 13:
The line meets the coordinate axes at A1(13/2,0) and B1(0,13/3). By joining these points we obtain the line 2x + 3y = 13.
So, (0,0) satisfies the inequation 2x + 3y ≤ 13. Hence, the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 13.
Region represented by 3x + y ≤ 5:
The line meets the coordinate axes at A2(5/3,0) and B2(0,5). By joining these points we obtain the line 3x + y = 5.
So, (0,0) satisfies the inequation 3x + y ≤ 5. Hence, the region containing the origin represents the solution set of the inequation 3x + y ≤ 5.
Region represented by x ≥ 0, y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, y ≥ 0.
The coordinates of the corner points of shaded region are O(0,0), A(5/3,0), P(2/7, 29/7), B2(0,13/3)
Q3.
Solution:
Firstly let us convert the given Inequations into equations, we get
4x + y = 20,
2x + 3y = 30,
x = 0,
y = 0
Region represented by 4x + y ≥ 20:
The line meets the coordinate axes at A1(5,0) and B1(0,20). By joining these points we obtain the line 4x + y = 20.
So, (0,0) satisfies the inequation 4x + y ≥ 20. Hence, the region containing the origin represents the solution set of the inequation 4x + y ≥ 20.
Region represented by 2x + 3y ≥ 30:
The line meets the coordinate axes at A2(15,0) and B2(0,10). By joining these points we obtain the line 2x + 3y = 30.
So, (0,0) satisfies the inequation 2x + 3y ≥ 30. Hence, the region containing the origin represents the solution set of the inequation 2x + 3y ≥ 30.
Region represented by x ≥ 0, y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, y ≥ 0.
The coordinates of the corner points of shaded region are A2(15,0), P(3,8), B1(0,20)
It is clear that, x = 3 and y = 8 is optimal. Hence, Minimum value of Z is 134 at points (3,8).
Q4.
Solution:
Let us plot these points, we get lines AB and CD.
We know that the feasible area is the unbounded area D-E-12
| Corner point | Value of Z = 50x + 30y |
| 10, 2 | 560 |
| 11.3, 17 | 1076.66 |
The maximum value of Z = 50x + 30y which occurs at x = 34/3, y = 17
Since we have an unbounded region as the feasible area plot 50x + 30y > 1076.66
The region D-F-B has common points with region D-E-12 the problem has no optimal maximum value.
Q5.
Solution:
3x + 4y ≤ 24; when x = 0, y = 6 and when y = 0, x = 8, line AB.
8x + 6y ≤ 48; when x = 0, y = 8 and when y = 0, x = 6, line CD.
Let us plot these points, x ≤ 5 we get line EF and y ≤ 6 we get line AG.
We know that the feasible area is the unbounded area O, O-C-H-G-E
| Corner point | Value of Z = 4x + 3y |
| 0,0 | 0 |
| 0,6 | 18 |
| 3.4, 3.4 | 24 |
| 5, 1 | 23 |
| 5, 0 | 20 |
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