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  • Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 25 maths textbook solution

Please solve RD Sharma class 12 chapter Linear Programming exercise 29.2 question 25 maths textbook solution

Answers (1)

Answer: the maximum value of ? is 1800

Hint: plot the points on the graph

Given: maximize ? =60 x+15 y

Solution: we have to maximize ? =60 x+15 y . First we will convert the given equations into equations, we obtain the following equations x+y=50,3 x+y=90, x=0, y=0

Region represented by  x+y \leq 50  . The line  x+y=50   meets the coordinates axes at A(50, 0) and B(0, 50) respectively. by joining these points we obtain the line 3 x+5 y=15  . Clearly (0, 0) satisfies the equation  x+y \leq 50  , the region containing the origin represents the solution set of the equation x+y \leq 50 .

Region represented by  3 x+y \leq 90 . The line  3 x+y=90   meets the coordinates axes at C(3, 0) and D(0, 90) respectively. by joining these points we obtain the line  3 x+y=90

x \geq 0 \text { and } y \geq 0 . So the region containing the origin represents the solution set of the equation 3 x+y \leq 90 .

Region represented by  x \geq 0 \text { and } y \geq 0  since every point in the first quadrant satisfies these equations, so the first quadrant is the region represented by the inequalities  x \geq 0 \text { and } y \geq 0 .

The feasible region determined by the system of constraints `x+y \leq 50,3 x+y \leq 90, x \geq 0, y \geq 0  .

The corner point of the feasible region are D(0,0), C(30,0), E(20,30) B(0,50)

The value of Z at these corner points are as follows;

\begin{array}{|l|l|} \hline \text { Corner point } & z=60 x+15 y \\ \hline D(0,0) & 60(0)+15(0)=0 \\ C(30,0) & 60(30)+15(0)=1800 \\ E(20,30) & 60(20)+15(30)=1650 \\ B(0,50) & 60(0)+15(50)=750 \\ \hline \end{array}

Therefore, the maximum value of Z is 1800 at the point (30, 0) hence x= 30 and y= 0is the optimal solution pf the given LPP. Thus the optimal value of Z is 1800

 

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