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  • Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 27

Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 27

Answers (1)

Answer:

Maximum Revenue is Rs.1260 obtained when 3 units of x and 8 units of y were produced is as follows.

Hint:

By using the mathematical formulation of the given Linear programming is Max Z= ax+by

Given:

A producer has 30 and 17 units of labour and capital respectively which he can use to produce two types of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and I unit of capital is required to produces one unit of y

Solution:

Let x1 and y1 units of goods x and y were produced respectively.

Number of units of goods cannot be negative.

Therefore, x_{1}, y_{1} \geq 0

To produce one unit of x, 3 units of Capital is required and 1 unit of capital is required to produce one unit of y.

                3 x_{1}+y_{1} \leq 17

If x and y are priced at Rs.100 and Rs.120 per unit respectively. Therefore, cost of x1 and y1 units of goods x and y is Rs.100 x1 and Rs.120 y1

Total revenue = Z = 100 x1 +120 y1 which B to be maximized.

Thus the mathematical formulation of the given linear programming problem is

                               Max Z = 100 x1 +120 y1

Subject to

                               2 x_{1}+3 y_{1} \leq 30 \\

                               3 x_{1}+y_{1} \leq 17 \\

                                \begin{aligned} & &x, y \geq 0 \end{aligned}

First, we will convert in equation into equations as follows 2 \mathrm{x}_{1}+3 \mathrm{y}_{1}=30,3 \mathrm{x}_{1}+\mathrm{y}_{1}=17, \mathrm{x}=0 \& \mathrm{y}=0

Region represented by 2 \mathrm{x}_{1}+3 \mathrm{y}_{1} \leq 30 : the line 2 x_{1}+3 y_{1}=30   meets axes at A1(15,0) and B1(0,10)respectively.

By joining these points we obtain the line  2 x_{1}+3 y_{1}=30 . Clearly (0,0) satisfies the 2 x_{1}+3 y_{1}=30 . So

the region which contains origin represents the solution set of the in equation 2 x_{1}+3 y_{1} \leq 30 .

Region represented by 3 x_{1}+y_{1} \leq 17 : the line 3 x_{1}+y_{1}=17   meets axes at C\left(\frac{17}{3}, 0\right) \& D(0,17) respectively. By joining these points we obtain the line  3 x_{1}+y_{1}=17 . Clearly (0,0) satisfies the in equation 3 x_{1}+y_{1} \leq 17 . So the region which contains origin represents the solution set of the in equation  3 x_{1}+y_{1} \leq 17 .

Region represented by x_{1} \geq 0 \& y_{1} \geq 0 : Since, every point in the first quadrant satisfies these in equations. So, the first quadrant is the region represented by the in equation x \geq 0 \& y \geq 0

The feasible region determined by the system of constraints 2 x_{1}+3 y_{1} \leq 30,3 x+y \leq 17,x \geq 0 \& y \geq 0 as follows.

                                                                          

The corner points are B(0,10), E(3,8) and C(17/3,0)

The values of Z at these corner points.

                                               

Corner Points

z=100 x_{1}+120 y_{1}

 

B

1200

E

1260

C

\frac{1700}{3} 

The maximum value of Z is 1260 which is attained at E(3,8).

Thus, the maximum revenue is Rs.1260 obtained when 3 units of x and 8 units of y were produces are as follows.

Posted by

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