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  • Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 13 maths textbook solution

Answers (1)

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Answer:

The optimal value of z is 24.

Hint:

Plot the points on the graph.

Given:

z=4x+3y

Solution:

We need to maximize z=4x+3y

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

3x+4y=24, 8x+6y=48, x=5, y=6, x=0 and y=0. 

The line 3x+4y=24  meets the coordinate axes at A(8,0) and B(0,6)  respectively. By joining these points are obtain the line 3x+4y=24 .

Clearly, (0,0)  satisfies the equation 3x+4y\leq 24 . So, the region in xy-plane that contain the origin represents the solution set of the given equation.

The line 8x+6y=48   meets the coordinate axes at C(6,0) and D(0,8)  respectively. By joining these points are obtain the line 8x+6y=48 .

Clearly, (0,0)  satisfies the equation 8x+6y\leq 48 . So, the region in xy-plane which does not contain the origin represents the solution set of the given equation.

X=5 is the line passing through x=5 parallel to y-axis.

Y=6 is the line passing through y=6 parallel to the x-axis.

Region presented by x\geq 0   and y\geq 0 . Since every point in the first quadrant satisfies these equations. So, the first quadrant is the region represented by the given equation.

The lines are drawn using a suitable scale.

The corner points of the feasible region are

0(0,0), G(5,0), F\left(5, \frac{4}{3}\right), E\left(\frac{24}{7}, \frac{24}{7}\right) \text { and } B(0,6)

The value of z at these corner points are as follows

    Coner Points     z=4 x+3 y
    0(0,0)     4 \times 0+3 \times 0=0
    h(5,0)     4 \times 5+3 \times 0=20
    F\left(5, \frac{4}{3}\right)     4 \times 5+3 \times \frac{4}{3}=24
    E\left(\frac{24}{7}, \frac{24}{7}\right)     4 \times \frac{24}{7}+3 \times \frac{24}{7}=24
    B(0,6)     4 \times 0+3 \times 6=18

We see that the maximum value of objective function z is 24  at the point F\left(5, \frac{4}{3}\right)  and E\left(\frac{24}{7}, \frac{24}{7}\right)  Thus, the optimal of z is 24.

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