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  • 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

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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

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