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  • Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 23

Need solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 23

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Answer: min value of Z is \frac{43}{14}  and max value is \frac{183}{13}

Hint: plot the points on the graph

Given: maximize and minimize 2x+y

Solution: first, we willconvert the given equations into equations we obtain the following equation;

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

The line x+3 y=6   meets the coordinates axis at A(0, 0) and B(0, 2). Join the points to obtain the line .clearly (0, 0) does not satisfy the equation x+3 y=6 x+3 y=6  so the region in x-y plane that does not contain the origin represents the solution of the given equation.

The line 3 x+4 y=24  meets the coordinates axis at E (8, 0) and F (0, 6). Join these points to obtain the line 3 x+4 y=24 . Clearly (0, 0) satisfies the equation 3 x+4 y \leq 24 . So the region in uy plane that contains the origin represents the solution set of the given equation.

The line -3 x+4 y=6  meets the coordinates at n(-2, 0) and H(0, 3). Join these points to obtain the line -3 x+4 y=6  . Clearly (0,0) satisfies the equation -3 x+4 y \leq 6  . So the region in xy plane that contains the region represents the solution set for the given equation.

The line5x+y=5  meets the coordinates axis at I(1, 0) and J(0, 5). Join these points to obtain the line 5 x+y=5  . Clearly (0, 0) does not satisfy the equation 5 x+y \leq 5

Corner points are

P\left(\frac{4}{13}, \frac{45}{13}\right), Q\left(\frac{9}{2}, \frac{1}{2}\right), R\left(\frac{9}{14}, \frac{25}{14}\right), S\left(\frac{4}{3}, 5\right), T\left(\frac{84}{13}, \frac{15}{13}\right)

    point     value
    \left(\frac{4}{13}, \frac{45}{13}\right)     \frac{57}{13}
    \left(\frac{9}{2}, \frac{1}{2}\right)     \frac{19}{2}
    \left(\frac{9}{14}, \frac{25}{14}\right)     \frac{43}{14}
    \left(\frac{4}{3}, 5\right)     \frac{23}{3}
    \left(\frac{84}{13}, \frac{15}{13}\right)     \frac{183}{3}

 

Maximum value=\frac{183}{13}

Minimum value  =\frac{43}{14}

Posted by

infoexpert26

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