Table of Contents
Ans. (c) Zero
Explanation:
As not all the constraints are satisfied. That is why, no solution exist or is infeasible.
Ans. (c) Concave region
Explanation:
Optimal solution, feasible solution and objective function are the terms used in linear programming problem.
Explanation:
The given data can be put in tabular form as under
Profit | ₹5 | ₹8 | |
---|---|---|---|
Machine/Products | Chair | Table | Time available |
A | 3 hrs. | 6 hrs. | 15 hrs. |
B | 5 hrs. | 2 hrs. | 24 hrs. |
Let x and y number of chairs and tables be produced respectively.
Then, total profit to be maximized
Max z = 5x + 8y
Given, A chair requires 3 hrs. on machine A and A table requires 6 hrs. on machine A. Machine A cannot work for more than 15 hrs.
∴ 3x + 6y ≤ 15
Similarly, A chair requires 5 hrs. and a table requires 2 hrs. on machine B. Machine B can not work for more than 24 hrs.
∴ 5x + 2y ≤ 24
Since the number of chairs and tables can not be negative
∴ x ≥ 0, 1 ≥ 0
Hence, the given LPP is
Max Z = 5x + 8y
Subject to the constraints
3x + 6y ≤ 15
5x + 2y ≤ 24
x ≥ 0 y ≥ 0
Explanation:
Minimise Z = 200 x + 500 y … (1)
subject to the constraints:
x + 2y ≥ 10 … (2)
3x + 4y ≤ 24 … (3)
x ≥ 0, y ≥ 0 … (4)
Let us draw the graph of x + 2y = 10 and 3x + 4y = 24 as below.
The shaded region in the above figure is the feasible region ABC determined by the
system of constraints (2) to (4), which is bounded. The coordinates of corner point A, B and C are (0,5), (4,3) and (0,6) respectively.
Calculation of Z = 200x + 500y at these points.
Corner point | Value of Z |
(0, 5) | 2500 |
(4, 3) | 2300← Minimum |
(0, 6) | 3000 |
Hence, the minimum value of Z is 2300 is at the point (4, 3).
Explanation:
Let x and y denote, respectively, the number of black and white sets and coloured sets made each week.
Thus x ≥ 0, y ≥ 0
The company can make at most 300 sets a week, therefore, x + y ≤ 300.
Weekly cost (in Rs) of manufacturing the set is 1800x + 2700y and the company can spend up to Rs. 648000.
Therefore, 1800x + 2700y ≤ 648000
or
2x + 3y ≤ 720
The total profit on x black and white sets and y coloured sets is Rs (510x + 675y).
Let the objective function be Z = 510x + 675y.
Therefore, the mathematical formulation of the problem is as follows.
Maximise Z = 510x + 675y subject to the constraints :
x + y ≤ 300
2x + 3y ≤ 720
x ≥ 0, y ≥ 0
The graph of x + y = 30 and 2x + 3y = 720 is given below.
Corner point | Value of Z |
A(300, 0) | 153000 |
B(180, 120) | 172800 = Maximum |
C(0, 240) | 162000 |
Hence, the maximum profit will occur when 180 black & white sets and 120 coloured sets are produced.
Chapter No. | Chapter Name |
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Chapter 1 | Relations and Functions |
Chapter 2 | Inverse Trigonometric Functions |
Chapter 3 | Matrices |
Chapter 4 | Determinants |
Chapter 5 | Continuity and Differentiability |
Chapter 6 | Applications of Derivatives |
Chapter 7 | Integrals |
Chapter 8 | Applications of the Integrals |
Chapter 9 | Differential Equations |
Chapter 10 | Vectors |
Chapter 11 | Three - dimensional Geometry |
Chapter 12 | Linear Programming |
Chapter 13 | Probability |
Chapter Wise Important Questions for CBSE Board Class 12 Maths |
---|
Relations and Functions |
Inverse Trigonometric Functions |
Matrices |
Determinants |
Continuity and Differentiability |
Applications of Derivatives |
Integrals |
Applications of the Integrals |
Differential Equations |
Vectors |
Three - dimensional Geometry |
Linear Programming |
Probability |
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