Math, asked by shrutishri221, 1 day ago

9. A vertex of a feasible region by the linear constraints 2x + y greater than equal to 30, x + 2y is greater than equal to 24 and x, y is small than equal to 0 is (a) (0,10) (b) (12,6) (d) (14,0) (c) (0,2)​

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Answered by sreenivas011
1

Answer:

A vertex of a feasible region by the linear constraints 2x + y greater than equal to 30, x + 2y is greater than equal to 24 and x, y is small than equal to 0 is (a) (0,10) (b) (12,6) (d) (14,0) (c) (0,2)​

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Answered by qwsuccess
0

The correct option is (b).

Given:

2x + y ≥ 30

x + 2y ≥ 24

x ≥ 0, y ≥ 0

To find: a possible vertex of the feasible region

Solution:

Required point is a vertex

⇒ Required point is a part of feasible region and satisfies all inequalities.

We have four options - (0,10), (12,6), (14,0), (0,2)​.

Taking inequality 1, we get

2(0) + 10 = 10 < 30 ⇒ (0, 10) is rejected

2(12) + 6 = 30 ≥ 30 ⇒ (12, 6) is selected

2(14) + 0 = 28 < 30 ⇒ (14, 0) is rejected

2(0) + 2 = 2 < 30 ⇒ (0, 2) is rejected

Now, we shall see if the point (12, 6) satisfies the remaining inequalities

12 + 2(6) = 24 ≥ 24 ⇒ satisfied

12 ≥ 0 ⇒ satisfied

6 ≥ 0 ⇒ satisfied

⇒ (12, 6) is the only point that fulfils all inequalities

⇒ It is the only point that can be a vertex

∴ The correct option is (b)

SPJ2

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