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  • Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 2

Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 2

Answers (1)

Answer: -Thus, the optimal value of Z is 15.

Hint: -Plot graph.

Given: -Z = 9x + 3y

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

2x + 3y =13, 3x + y = 5, x = 0  and y = 0

Region represent by 2x + 3y \leq 13

The line 2x + 3y = 13  meets the coordinates axes at \mathrm{A}\left(\frac{13}{2}, 0\right) \text { and } \mathrm{B}\left(0, \frac{13}{3}\right) respectively. By joining the points we obtain the line 2x + 3y = 13

Clearly (0, 0) satisfies the equation 2x + 3y \leq 13.  So, the region containing the origin represents the solution set of the inequality 2x + 3y \leq 13.. Region represented by 3x + y \leq 5 . The line 5x + 2y = 10 meets the coordinates axes at   \mathrm{C}\left(\frac{5}{3}, 0\right) \text { and } \mathrm{D}(0,5)   respectively. By joining these points we obtain the line 3x + y = 5.

Clearly (0, 0) satisfies the inequality 3x + y \leq 15 . So, the region containing the origin represents the solution set of the inequality 3x + y \leq 5.

Region represented by x \geq 0 \; and\; y \geq 0;

Since, every point in the first quadrant satisfies these inequalities. So, the first quadrant is the region represented by the inequalities x \geq 0 \; and\; y \geq 0;  The feasible region determined by the system of constants,

2x + 3y \leq 13, 3x + y \leq 5,  x \geq 0 \; and\; y \geq 0; use as follows:

The corner points of the feasible region are (0, 0).

C\left(\frac{5}{3}, 0\right), E\left(\frac{2}{7}, \frac{29}{7}\right) \text { and } B\left(0, \frac{13}{3}\right)

The value of Z at these corner points are as follow

We see that the maximum value of the objective function 2,3, 13 which is at C

 

    Corner Points     z=9x+3y
    O(0,0)     9 \times 0+3 \times 0=0
    C\left(\frac{5}{3}, 0\right)     9 \times \frac{5}{3}+3 \times 0=0
    E\left(\frac{2}{7}, \frac{29}{7}\right)         9 \times \frac{2}{7}+3 \times \frac{29}{7}=0
    B\left(0, \frac{13}{3}\right)     9 \times 0+3 \times \frac{13}{3}=0

\left(\frac{5}{3}, 0\right) \text { and }\left(\frac{2}{7}, \frac{29}{7}\right) . Thus the optimal value of 2 is 15.

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