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

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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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