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  • Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 4 maths

Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 4 maths

Answers (1)

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Answer:

The optimal value of z is \frac{1700}{3}

Hint:

Plot the points on the graph.

Given:

z=50x+30y

Solution:

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

2x+y=18, \; \; 3x+2y=34

Region presented by 2x+y\geq 18 . The line 2x+y=18  meets the coordinate axes at A(9,0 ) and B(0,18) respectively. By joining these points are obtain the line 2x+y=18 . Clearly, (0,0)  does not

satisfies the equation 2x+y\geq 18 . So, the region in xy plane which does not contain the origin represents the solution set of the equation 2x+y\geq 18.

Region presented by 3x+2y\geq 34 . The line 3x+2y=34  meets the coordinate axes at C(\frac{34}{3},0) and D(0,17)respectively. By joining these points are obtain the line 3x+2y=34.Clearly, (0,0) does not

satisfies the equation 3x+2y\geq 34. So, the region in xy plane which does not contain the origin represents the solution set of the equation 3x+2y\geq 34.

The corner points of the feasible region are

A(9,0), C\left(\frac{34}{3}, 0\right) \text { and } E(2,14)

The value of z at these corner points are as follows

    Corner Points     z=50 x+30 y
    A(9,0)     50 \times 9+30 \times 0=450
    C\left(\frac{34}{3}, 0\right)     50 \times \frac{34}{3}+30 \times 0=\frac{1700}{3}
    E(2,14)     50 \times 2+30 \times 14=520

Therefore, the maximum value of z is \frac{1700}{3} at the point \left ( \frac{34}{3},0 \right )

Hence, x=\frac{34}{3}  and y=0  is the optimal solution of the given LPP.

Thus, the optimal of z is \frac{1700}{3}

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