Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 18

Answers (1)

Answer: the objective value Z is \frac{20}{3}

Hint: Plot the points on the graph

Given: z=-x_{1}+2 x_{2}

Solution: first, we will convert the given inequalities , we obtain the following equations;

-x_{1}+3 x_{2}=10, x_{1}+x_{2}=6, x_{1}+x_{2}=2, u_{1}=0, u_{2}=0

Region represented by  -x_{1}+3 x_{2}=10 : The line -x_{1}+3 x_{2}=10 meets the coordinates axes at  A(-10,0) \text { and } B\left(0, \frac{10}{3}\right)    respectively. By joining the points we obtain the line -x_{1}+3 x_{2}=10   clearly

(0, 0) satisfies the equation -x_{1}+3 x_{2}<10  . So the region in the plane contains the origin represents the solution set of the equation -x_{1}+3 x_{2} \leq 10

Region represented by x_{1}+x_{2} \leq 6

Region represented by x_{1} \geq 0 \text { and } x_{2} \geq 0 ; since every point in the list and gradient satisfies these equations, so the first quadrant is the region represented by the equations x_{1} \geq 0 \text { and } x_{2} \geq 0

The feasible region determined by the system of construction. -x_{1}+3 x_{2}=10, x_{1}+x_{2}=6, x_{1}+x_{2}=2, u_{1}=0, u_{2}=0

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

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