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  • Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 6

Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 6

Answers (1)

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Answer:

The optimal value of z is 400.

Hint:

Plot the points on the graph.

Given:

z=15x+10y

Solution:

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

3x+2y=80, 2x+3y=70, x=0  and y=0

Region presented by 3x+2y\leq 80 . The line 3x+2y=80  meets the coordinate axes at A(\frac{80}{3},0)  and B(0,40) respectively. By joining these points are obtain the line 3x+2y\leq 80 . So, the region

containing the origin represents the solution set of the equation 3x+2y\leq 80.

Region presented by 2x+3y\leq 70. The line 2x+3y\leq 70  meets the coordinate axes at C(35,0) and D(0,\frac{70}{3})  respectively. By joining these points are obtain the line 2x+3y\leq 70.

Clearly, (0,0) satisfies the equation 2x+3y\leq 70. So, the region in xy plane which does not contain the origin represents the solution set of the equation 2x+3y\leq 70.

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+2y\leq 80, 2x+3y\leq 70, x\geq 0  and y\geq 0 are as follows

The corner points of the feasible region are

E(20,10) \text { and } D\left(0, \frac{70}{3}\right)

The value of z at these corner points are as follows

    Corner Points     z=15 x+10 y
    O(0,0)     15 \times 0+10 \times 0=0
    A\left(\frac{80}{3}, 0\right)         15 \times \frac{80}{3}+10 \times 0=400
    E(20,10)         15 \times 20+10 \times 10=400
    D\left(0, \frac{70}{3}\right)     15 \times 0+10 \times \frac{70}{3}=\frac{700}{3}

 

Therefore, the maximum value of z is 400 at the point   A\left(\frac{80}{3}, 0\right) \text { and } E(20,10)

Thus, the optimal of z is 400.

 

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