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  • Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 20 maths

Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 20 maths

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Answer: the feasible region of the given LPP does not exist

Hint: plot the points on the graph

Given: z=3 x_{1}+4 x_{2}

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

\lambda_{1}-\lambda_{2}=-1, \bar{a} \lambda_{1}+\lambda_{2}=0, \lambda_{1}=0, \lambda_{1}=0

Region represented by the \lambda_{1}-\lambda_{2} \leq-1 : the line \lambda_{1}-\lambda_{2}=-1  meets the coordinates axes A(-1, 0) and B(0, 1) respectively. By joining these points we obtain the line \lambda_{1}-\lambda_{2}=-1

Clearly (0, 0) does not satisfy the equation \lambda_{1}-\lambda_{2} \leq-1 . So the region is the plane which does not contain the origin represents the solution set of the equation \lambda_{1}-\lambda_{2} \leq-1 .

Region represented by \lambda_{1}+\lambda_{2} \leq 0, \text { or } u_{1} \geq u_{2} . The line -\lambda_{1}+\lambda_{2}=0 \text { or } u_{1}=u_{2}  is the line passing through (0, 0). The region to the right of the line u_{1}=u_{2}  will satisfy the given equation .\lambda_{1}+\lambda_{2} \leq 0 . if

we take a point (1, 3) to the left of the line u_{1}=u_{2} . Here 1 \leq 3  which is not satisfy the equation u_{1} \geq u_{2}   . therefore , the region to the right of the line \lambda_{1}=\lambda_{2}   will satisfy the given inequalition \lambda_{1}+\lambda_{2} \leq 0

Region represented by  \lambda_{1} \geq 0 \text { and } \lambda_{2} \geq 0: since every point is the first quadrant is the region represented by the equations \lambda_{1} \geq 0 \text { and } \lambda_{2} \geq 0

The feasible region determined by the system of constraints \lambda_{1}-\lambda_{2}=-1, \bar{a} \lambda_{1}+\lambda_{2}=0, \lambda_{1}=0, \lambda_{1}=0

We observe that the feasible region of the given LPP does not exist

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