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  • Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 11

Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 11

Answers (1)

Answer:

The optimal value of z is 27

Hint:

Plot the points on the graph.

Given:

z=5x+3y

Solution:

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

2x+y=10,x+3y=15, x=10, and y=8

Region presented by 2x+y\geq 10 . The line 2x+y=10  meets the coordinate axes at A(5,0 ) and B(0,10)  respectively. By joining these points are obtain the line 2x+y=10Clearly, (0,0)  does not

satisfies the equation 2x+y\geq 10. So, the region in xy plane which does not contain the origin represents the solution set of the equation 2x+y\geq 10.

Region presented by x+3y\geq 15. The line x+3y=15  meets the coordinate axes at C(15,0 )and D(0,5) respectively. By joining these points are obtain the line x+3y=15. Clearly, (0,0) satisfies the

equation x+3y\geq 15 . So, the region in xy plane which does not contain the origin represents the solution set of the equation x+3y\geq 15.

The line x=10 is the line that passes through the point (10,0) and is parallel to y-axis. x\leq 10 is the region to the right the line x=10

The line y=8 is the line that passes through the point (0,8) and is parallel to x-axis.y\leq 8 is the region above the line y=8.

Region presented by x\geq 0 and y\geq 10. Since the every point in the first quadrant satisfies these equations. So, the first quadrant is the region represented by the equation x\geq 0 and y\geq 0.

The feasible region determined by the system of constraints,

2x+y\geq 10, x+3y\geq 15, x\leq 10, y\leq 8, x\geq 0 and y\geq 0 are as follows

 

The value of z at these corner points are as follows

    Corner Points     z=5 x+3 y
    E(3,4)     5 \times 3+3 \times 4=27
    H\left(10, \frac{5}{3}\right)     5 \times 10+3 \times \frac{5}{3}=55
    F(10,8)     5 \times 10+3 \times 8=74
    G(1,8)     5 \times 1+3 \times 8=29

Therefore, the maximum value of objective function z is 27 at the point E(3,4)

Hence, x=3 and y=4 is the optimal solution of the given LPP.

Thus, the optimal of z is 27.

 

Posted by

infoexpert26

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