1. Chapter 12 Class 12 Linear Programming (Term 1)
  2. Serial order wise
  3. NCERT Exemplar - MCQs

Question 4 (Case Based MCQ) - NCERT Exemplar - MCQs - Chapter 12 Class 12 Linear Programming (Term 1)

Last updated at Nov. 19, 2021 by

The feasible solution for a LPP is shown in Fig. 12.12. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at

(A) (0, 0)                                (B) (0, 8)   

(C) (5, 0)                               (D) (4, 10)

 

This question is similar to Ex 12.2, 11 - Chapter 12 Class 12 - Linear Programming

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Slide13.JPG

  1. Chapter 12 Class 12 Linear Programming (Term 1)
  2. Serial order wise

    3. NCERT Exemplar - MCQs

    Linear equations given - Bounded →

Transcript

Question 4 The feasible solution for a LPP is shown in Fig. 12.12. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at (A) (0, 0) (B) (0, 8) (C) (5, 0) (D) (4, 10) Checking value of Z = 3x – 4y at corner points Thus, Min Z = −32 at (0, 8) So, the correct answer is (B)

Chapter 12 Class 12 Linear Programming (Term 1)
Serial order wise

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