Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 16 maths

Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 16 maths

Answers (1)

best_answer

Answer: The optimal value of 2 is 0

Hint: Plot the points on the graph

Given: z=3 x_{1}+5 x_{2}

Solution: First, we will convert the given equation into equations, we obtain the following equations:

The following equations:\lambda 1+3 \lambda_{2}=3, \lambda_{1}+\lambda_{2}=2, \lambda_{1}=0 \text { and } \lambda_{2}=0

Region represented by\lambda_{1}+3 \lambda_{2} \geq 3 . The line \lambda_{1}+3 \lambda_{2}=3 meets the coordinate axes at A (3,0) and B(0,1) respectively. By joining these points we obtain the line \lambda_{1}+3 \lambda_{2}=3

Clearly (0,0) does not satisfy the equation \lambda_{1}+3 \lambda_{2} \geq 3,so the region in the plane which does not contain the origin represents the solution set of the equation \lambda_{1}+3 \lambda_{2} \geq 3

Region represented by \lambda_{1}+3 \lambda_{2} \geq 2 The line \lambda_{1}+\lambda_{2}=2 meets the coordinate axis at c (2,0) and D (0,2) respectively. By joining these points we obtain the line λ1+ λ2=2 clearly (0,0) does not satisfy the equation

\lambda_{1}+\lambda_{2} \geq 2 .So the region containing origin represents the solution set of the equation\lambda_{1}+\lambda_{2} \geq 2.Region represented by \lambda_{1} \geq 0 \text { and } \lambda_{2} \geq 0, since every point in the first quadrant satisfies these in equations ,so

the first quadrant is the region represented by the equation  \lambda_{1} \geq 0 \text { and } \lambda_{2} \geq 0

The feasible region determined by the system of the constraints, \lambda 1+3 \lambda_{2} \geq 3, \lambda_{1}+\lambda_{2}=2, \lambda_{1} \geq 0  and \lambda_{2}=0 are as follows:

The correct points of the feasible region \mathrm{A}(0,0) \; \; \mathrm{B}(0,1) \; \; \mathrm{C}\left(\frac{3}{2}, \frac{1}{2}\right) \text { and } \mathrm{c}(2,0)The values of 2 at these common points are as follows:

    Common points     \mathrm{Z}=3 \mathrm{x}_{1}+5 \mathrm{x}_{2}
    0(0,0)     3(0)+5(0)=0
    B(0,1)     3(0)+5(1)=5
    \mathrm{E}\left(\frac{3}{2}, \frac{1}{2}\right)     \frac{3}{2}+5\left(\frac{1}{2}\right)=7
    \mathrm{C}(2,0)     3(2)+5(2)=6

 

Therefore the minimum value of 2 is  0 at the point (9,0),Hence , \lambda_{1}=0 \text { and } \lambda_{2}=0 is the optional value of 2 is 0

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question