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  • Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 15

Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 15

Answers (1)

Answer:

The optimal value of z is 55.

Hint:

Plot the points on the graph.

Given:

z=3x+5y

Solution:

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

x+2y=20, x+y=15, y=5, x=0 and y=0.

The line x+2y=20  meets the coordinate axes at A(20,0) and B(0,10)  respectively. By joining these points are obtain the line x+2y=20.

Clearly,( 0,0)  satisfies the equation x+2y\leq 20 . So, the region in xy-plane that contain the origin represents the solution set of the given equation.

The line  x+y=15  meets the coordinate axes at C(15,0) and D(0,15)  respectively. By joining these points are obtain the line x+y=15

Clearly, ( 0,0)  satisfies the equation  x+y\leq 15 . So, the region in xy-plane that contain the origin represents the solution set of the given equation.

Y=5 is the line passing through (0,5) parallel to the x-axis. The region below the line y=5 will satisfy the given equation.

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

The lines are drawn using a suitable scale.

The corner points of the feasible region are

0(0,0), C(15,0), E(10,5) \text { and } F(0,5)

The value of z at these corner points are as follows

\begin{array}{|c|c|} \hline \text { Corner Points } & z=3 x+5 y \\ \hline 0(0,0) & 3 \times 0+5 \times 0=0 \\ \hline C(15,0) & 3 \times 15+5 \times 0=45 \\ \hline E(10,5) & 3 \times 10+5 \times 5=55 \\ \hline F(0,5) & 3 \times 0+5 \times 5=25 \\ \hline \end{array}

We see that the maximum value of objective function z is 55  at the point E(10,5 )

Thus, the optimal of z is 55.

 

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