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Question 10 (Case Based MCQ) - NCERT Exemplar - MCQs - Chapter 12 Class 12 Linear Programming

Last updated at Dec. 16, 2024 by Teachoo

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).

Let F = 4x + 6y be the objective function.

 Maximum of F – Minimum of F =

(A)60                          (B) 48

(C) 42                         (D) 18

This question is similar to Question 38 Choice 2 - Sample Paper 2021 Class 12 - Linear Programming

Transcript

Question 10 Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. Maximum of F – Minimum of F = 60 (B) 48 (C) 42 (D) 18 We need to find Max and Min F = 4x + 6y Corner Points are (0, 2), (3, 0), (6, 0), (6, 8), (0, 5) Checking value of F at corner points Thus, Max Z = 72 Min Z = 12 Therefore, Max value of Z − Min value of Z = 72 − 12 = 60 So, the correct answer is (D)
  1. Chapter 12 Class 12 Linear Programming
  2. Serial order wise

    3. NCERT Exemplar - MCQs

    Forming equations and solving →
Forming equations and solving →
Chapter 12 Class 12 Linear Programming
Serial order wise

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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