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Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 27

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Answer: the maximum value of Z is 200

Hint: plot the point on the graph.

Given: maximize z=20 x+10 y

Solution: the given constraints are x+2 y \leq 28,3 x+y \leq 24, x \geq 2, x, y \geq 0

Converting the inequalities into equations, we obtain the following equations;

                x+2 y=28,3 x+y=24, x=0, y=0

These equations represents straight lines in Xy palne

The line x+2 y=28   must coordinate axes at A(18, 0) and B(0, 14). Join these to obtain the line x+2 y=28 .

The line 3 x+y=24  meets the coordinates axes at A_{2}(0,0), B_{2}(0,24)  join these points to obtain the line 3 x+y=24  . The line u=2  is parallel to y-axis passes through the point A_{2}(2,0)

Also, u=0  is the y-axis and y=o  is the x-axis.the feasible region of the LPP is shaded

The point of intersection of lines x+2 y=28,3 x+y=24   is D(-1,12)

The point of intersection of line u=2  and  x+2 y=28  R(2,0)

The coordinates of the corner points of the feasible region we have A_{2}(0,0), B_{2}(0,24)  the value of the objective function of these points are given in the following table;

\begin{array}{|l|l|} \hline \text { Points } & \text { Values of the objective z } \\ \hline A_{2}(2,0), & z=20(3)+10(0)=40 \\ A_{2}(8,0) . & z=20(8)+10(0)=160 \\ Q(4,12) & z=20(4)+10(12)=200 \\ R(2,3) & z=20(2)+10(13)=170 \\ \hline \end{array}

Clearly , the maximum value of Z is 200

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