For an objective function 𝑍 = 𝑎𝑥 + 𝑏𝑦, where 𝑎, 𝑏 > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is:

(a) 𝑏 − 3𝑎 = 0  (b) 𝑎 = 3𝑏

(c) 𝑎 + 2𝑏 = 0  (d) 2𝑎 − 𝑏 = 0

 

This question is inspired from Ex 12.2, 11 (MCQ) - Chapter 12 Class 12 - Linear Programming

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Question 41 For an objective function 𝑍 = 𝑎𝑥 + 𝑏𝑦, where 𝑎, 𝑏 > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is: (a) 𝑏 − 3𝑎 = 0 (b) 𝑎 = 3𝑏 (c) 𝑎 + 2𝑏 = 0 (d) 2𝑎 − 𝑏 = 0 Since maximum Z occurs at both the points (30, 30) and (0, 40) Value of Z at both these points will be same Therefore 30a + 30b = 40b 30a = 40b − 30b 30a = 10b Dividing by 10 both sides 3a = b 0 = b − 3a b − 3a = 0 So, the correct answer is (A)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo