Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

Mathematics Part-II

Book: Mathematics Part-II

Chapter: 12. Linear Programming

Subject: Maths - Class 12^th

Q. No. 1 of Exercise 12.1

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1

Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

It is given in the question that,

Z = 3x + 4y

We have to subject on the following equation:

x ≥ 0, y ≥ 0, x + y ≤ 4

(x, y) = (0, 4), (4, 0)

We can clearly see that Z is maximum at (0, 4). Hence, maximum value of Z will be 16

1

Chapter Exercises

Exercise 12.1

Exercise 12.2

Miscellaneous Exercise

1

Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

2

Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

3

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

4

Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

5

Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

6

Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. Show that the minimum of Z occurs at more than two points.

7

Minimize and Maximize Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.

8

Minimize and Maximize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

9

Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

10

Maximize Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.