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  • Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 5 maths textbook solution

Answers (1)

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Answer:

Thus 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 the 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), h(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

    Corner 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

The maximum value of z is 24.

Posted by

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