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  • Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 3

Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 3

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Answer:

Thus the optimal value of z is 134.

Hint:

Plot the points on the graph.

Given:

2=18x+10y

Solution:

First, we will convert the given in equations into equations, we obtain the following equations

4x+y=20, 2x+3y=0, x=0 and  y=0

Region presented by 4x+y\geq 20 . The line 4x+y=20  meets the coordinate axes at A(5,0)  and B(0,20)  respectively. By joining these points are obtain the line 4x+y=20. Clearly, (0,0)  does not satisfies the equation 4x+y\geq 204x+y\geq 20 . So, the region in xy plane which does not contain the origin represents the solution set of the equation 4x+y\geq 20.

Region presented by 2x+3y\geq 30 . The line 2x+3y=30 meets the coordinate axes at C(15,0) andD(0,10)  respectively. By joining these points are obtain the line 2x+3y=0.Clearly, (0,0 ) does not satisfies the equation 2x+3y\geq 30 . So, the region in xy plane which does not contain the origin represents the solution set of the equation 2x+3y\geq 30.

Region presented by x\geq 0 and y\geq 0. Since the every point in the first quadrant satisfies these equations. Som the first quadrant is the region represented by the equations x\geq 0 and y\geq 0

The feasible region determined by the system of constraints,
4x+y\geq 20, \; 2x+3y\geq 30, \; x\geq 0  and  y\geq 0 are as follows

The corner points of the feasible region are

B(0,20), C(15,0), E(3,8) \text { and } C(15,0)

The value of z at these corner points are as follows

\begin{array}{|c|c|} \hline \text { Corner Points } & z=18 x+10 y \\ \hline B(0,20) & 18 \times 0+10 \times 20=200 \\ \hline C(15,0) & 18 \times 15+10 \times 0=270 \\ \hline E(3,8) & 18 \times 3+10 \times 8=134 \\ \hline \end{array}

The maximum value of z is 134 at the point E(3,8) . Hence, x=3  and y=8 is the optimal solution of the given LPP.

Thus, the optimal of z is 134.

 

 

 

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