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  • Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 28 maths

Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 28 maths

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Answer: the minimum value of Z is 150

Hint: plot the point on the graph

Given: minimize z=6 x+3 y

Solution: the given constraints are 4 x+y \geq 80, x+y \geq 115,3 x+2 y \geq 150, x, y \geq 0

Converting the given equations into equations we get 4 x+y=80, x+y=115,3 x+2 y=150, x, y=0

The lines are drawn on the graph and the shaded region A,B,C represents the feasible region of the given LPP

It can be observed that the feasible region is bounded thus the coordinates of the corner points of the feasible region A(2, 22)\; \; B(12, 30) and C(40, 15). The value of function Z at the corner points are given in the following table;

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Values of the objective } Z=6 x+2 y \\ \hline A(2,22), & z=6(2)+3(22)=228 \\ B(15,20) & z=6(15)+3(20)=150 \\ C(40,15) & z=6(40)+3(15)=285 \\ \hline \end{array}

From the table, Z is minimum at x=15 and  y=20  and the minimum value of Z is 150

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