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

Last updated at August 12, 2026 by

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

Let F = 4x + 6y be the objective function. Maximum of F – Minimum of F - NCERT Exemplar - MCQs

part 2 - Question 10 (Case Based MCQ) - NCERT Exemplar - MCQs - Serial order wise - Chapter 12 Class 12 Linear Programming
part 3 - Question 10 (Case Based MCQ) - NCERT Exemplar - MCQs - Serial order wise - Chapter 12 Class 12 Linear Programming

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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)

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