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  • Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.1 question 10

Provide solution for RD Sharma maths class 12 chapter Linear Programming exercise 29.1 question 10

Answers (1)

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Answer: Z = 2000x + 30000y

Sub to

\begin{aligned} &5 x+2 y \leq 180 \\ &3 x+3 y \leq 135 \\ &x \geq 0, y \geq 0 \end{aligned}

Hint:         Let x and y  in given condition.

Given:      Profit of ?3000 on each truck and ?2000 on each automobile.

Solution:   Let x  Total number of trucks.

y : Total number of automobiles.

Objective function:

Z  = 2000x  +  30000y

                  Sub to

\begin{aligned} &5 x+2 y \leq 180 \ldots \ldots \ldots \ldots (1) \\ &3 x+3 y \leq 135 \ldots \ldots . . . . . . .(2) \end{aligned}

From equation (1) and (2), the

\begin{aligned} 5 x+2 y &=180 \\ & \Rightarrow \frac{x}{36}+\frac{y}{40}=1 \quad.......(3)\\ 3 x+3 y &=135 \end{aligned}

\begin{aligned} &\Rightarrow \frac{x}{135 / 3}+\frac{y}{135 / 3}=1 \ldots (4) \\ &\Rightarrow \frac{x}{45}+\frac{y}{45}=1 \end{aligned}

Solve (3) and (4) , we get

      P(35,15)

The regionA,Bare feasible region

\begin{array}{|c|c|} \hline \text { Corner point } & \text { Value of } \mathrm{Z}=2000 \mathrm{x}+30000 \mathrm{y} \\ \hline 0,0 & 0 \\ \hline 0,36 & 10,80,000 \\ \hline 15,30 & 9,30,000 \\ \hline 45,0 & 90,000 \\ \hline \end{array}

Thus  0  automobiles  and  36  trucks  give  max.  profit  of  Rs  10,  80,  000

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