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  • Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 24 maths

Explain solution RD Sharma class 12 chapter Linear Programming exercise 29.2 question 24 maths

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Answer: the optimal value of Z is \frac{29}{3}

Hint: plot the points on the graph

Given: minimum value of 3x+5y

Solution: first, we will convert the given equations into equations, we obtain the following equations;

-2 x+y=4, x+y=3, x-2 y=2, x=0, y=0

The line -2 x+y=4   must meet the coordinates axis at A(-2, 0) and B(0, 4). Join the points to obtain the line -2 x+y=4 .

Clearly, (0, 0) satisfies the equation -2 x+y \leq 4 . So the region in xy-plane that contains the origin represents the solution set of the given equation.

The line x-2 y=2  meets the coordinates axis at E(2, 0) and F(0, -1). Join these points to obtain the line x-y plane that contains the origin represents the solution set of the given equation.

Region represented by x \geq 0 \text { and } y \geq 0 . Since every point in the first quadrant satisfies these equations, so the first quadrant is the region represented by the equations. These lines are shown using a suitable scale.

The corner points of the feasible region are B(0, 4), D(0, 3) and U(\frac{0}{3}, \frac{1}{3}) . The value of z at these corners are as follows;

    Corner point     z=3 x+5 y
    B(0,4)     3(0)+5(4)=20
    D(0,3)     3(0)+5(3)=15
    u\left(\frac{8}{3}, \frac{1}{3}\right)     3\left(\frac{8}{3}\right)+5\left(\frac{1}{3}\right)=\frac{29}{3}

We see that the minimum value of objective function z is \frac{29}{3}

 

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