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  • Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 22

Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.2 question 22

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Answer: the optimal value of ? is 13

Hint: plot the points on the graph

Given: to prove and to minimize 3x+2y

Solution: first, we will convert the given equations into equations, we obtain the following equations; 5 \lambda+y=10, \lambda+y=6, \lambda+4 y=12, \lambda=0, y=0 .

Region represented by 5 \lambda+y \geq 10 . The line 5 \lambda+y=10   meets the coordinates axes at A(2, 0) and B(0, 10) respectively. By joining these points we obtain line 5 \lambda+y \geq 10

Clearly (0, 0) does not satisfies the equation 5 \lambda+y \geq 10

Region represented by \lambda+y \geq 6 . The line  \lambda+y=6  meets the coordinates axes at C(6,0) and D(0, 6) respectively. by joining these points we obtain the line 2 \lambda+3 y=30   clearly (0, 0) does not satisfy the equation \lambda+y \geq 6 .  . So the region which does not contain the origin represents the solution set of the equation 2 \lambda+3 \geq 30

Region represented by \lambda+4 y \geq 12 , the line \lambda+4 y=12  meets the coordinates axes at E(12, 0) and F(0, 3) respectively. by joining these points we obtain the line.

\lambda+4 y=12, \mathrm{cl} , clearly (0,0) does not contain the origin represents the solution set of equation \lambda+4 y \geq 12

Region represented by \lambda \geq 0 \text { and } y \geq 0. Since, every point is the first quadrant satisfies these inequalitions. So the first quadrant is the region represented by the equations \lambda \geq 0 \text { and } y \geq 0. The feasible region determined by the system of constraints 5 \lambda+y=10, \lambda+y=6, \lambda+4 y=12, \lambda=0, y=0

Therefore the minimum value of z is 13 at the point c(1, 5). Hence x=1 and  y=5 is the optimal solution of the given LPP. Thus the optimal value of Z is 13

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