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  • Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 8 maths

Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 8 maths

Answers (1)

Answer:

The optimal value of z is 140.

Hint:

Plot the points on the graph.

Given:

z=3x+4y

Solution:

We have to maximize z=3x+4y

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

2x+2y=80, 2x+4y=120

Region presented by 2x+2y\leq 80. The line2x+2y=80 meets the coordinate axes at A(40,0) and B(0,40 ) respectively. By joining these points are obtain the line 2x+2y=80. Clearly, (0,0) satisfies the equation 2x+2y\leq 80. So, the region contain the origin represents the solution set of the equation2x+2y\leq 80.

Region presented by 2x+4y\leq 120. The line 2x+4y=120 meets the coordinate axes at C(60,0) and D(0,30 ) respectively. By joining these points are obtain the line 2x+4y\leq 120 . Clearly, (0,0) does not satisfies the equation 2x+4y\leq 120. So, the region in xy plane which does not contain the origin represents the solution set of the equation 3x+2y\geq 34.

The feasible region determined by the system of constraints,
2x+2y\leq 80, 2x+4y\leq 120 are as follows

The corner points of the feasible region are

O(0,0), A(40,0), E(20,20) \text { and } D(0,30)

The value of z at these corner points are as follows

\begin{array}{|c|c|} \hline \text { Corner Points } & z=3 x+4 y \\ \hline O(0,0) & 3 \times 0+4 \times 0=0 \\ \hline A(40,0) & 3 \times 40+4 \times 0=120 \\ \hline E(20,20) & 3 \times 20+4 \times 20=140 \\ \hline D(0,30) & 3 \times 0+4 \times 30=120 \\ \hline \end{array}

We see that the maximum value of objective function z is 140 at the point E(20,20). Which is at x=20 and y=20 is the optimal solution of the given LPP.

Thus, the optimal of z is 140.

 

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