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  • Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 7

Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 7

Answers (1)

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Answer:

The optimal value of z is 56.

Hint:

Plot the points on the graph.

Given:

z=10x+6y

Solution:

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

3x+y=12, 2x+5y=34, x=0 and y=0

Region presented by 3x+y\leq 12 . The line 3x+y=12  meets the coordinate axes at A(4,0 ) and B(0,12) respectively. By joining these points are obtain the line 3x+y=12 . So, the region containing the

origin represents the solution set of the equation 3x+2y\leq 80.  Clearly, (0,0) satisfies the equation 3x+y\leq 12 . So, the region contain the origin represents the solution set of the equation 3x+y\leq 12.

Region presented by2x+5y\leq 34 . The line 2x+5y\leq 34  meets the coordinate axes at C(17,0 ) and D(0,\frac{34}{5})  respectively. By joining these points are obtain the line 2x+5y\leq 34 . So, the region containing

the origin represents the solution set of the equation 2x+5y\leq 34. Clearly, (0,0) satisfies the equation 2x+5y\leq 34 . So, the region contain the origin represents the solution set of the equation 2x+5y\leq 34.

Region presented by x\geq 0  and y\geq 0 . 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,

3x+y\leq 12,2x+5y\leq 34, x\geq 0  and y\geq 0 are as follows

The corner points of the feasible region are

A(4,0), O(0,0), E(2,6) \text { and } D\left(0, \frac{34}{5}\right)

The value of z at these corner points are as follows

    Corner Points     z=10 x+6 y
    O(0,0)     10 \times 0+6 \times 0=0
    A(4,0)             10 \times 4+6 \times 0=40
    E(2,6)     10 \times 2+6 \times 6=54
    D\left(0, \frac{34}{5}\right)     10 \times 0+6 \times \frac{34}{5}=\frac{204}{3}

 

Therefore, the maximum value of objective function z is 56 at the point E(2,6) The means at x=2 and y=6.  Thus, the optimal of z is 56.

 

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