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  • Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 31 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 31 Maths Textbook Solution.

Answers (1)

Answer:.

The answer of the given question B that the maximum profit is Rs.120 obtained when 12 units of articles A and 6 units of articles B were manufactured.

Hint:

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

Given:

The maximum capacity of first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing.

Solution:

Let x units and y units of articles A and B are produced respectively.

Number of articles can’t be negative.

Therefore, x, y \geq 0

The product of each unit of article A requires 4 hours in assembly and that of articles B requires 2 hours in assembly and the maximum capacity of the assembly department in 60 hours a week.

4 x+2 y \leq 60

The product of each unit of article A requires 2 hours in finishing and that of articles B requires 4 hours in assembly and the maximum capacity of the finishing department in 48 hours a week.

2 x+4 y \leq 48

If the profit is Rs.6 for each unit of A and Rs.8 for each unit of B. Therefore, profit gained from x units and y units of articles A and B respectively is Rs.6x and Rs.8y respectively.

Total revenue = Z = 6x + 8y which is to be maximized.

Thus, the mathematical formulation of the given linear programming problem is

                Max Z = 6x + 8y

Subject to

                \begin{aligned} &2 x+4 y \leq 48 \\ &4 x+2 y \leq 60 \\ &x, y \geq 0 \end{aligned}

First, we will convert in in equations into equations as follows:

                2 x+4 y=48,4 x+2 y=60, x=0 \& y=0

Region represented by 2 x+4 y \leq 48 : the line  2 x+4 y=48 meets the coordinate axes at A1(24,0) and B1(0,12)respectively.

By joining these points we obtain the line 2 x+4 y=48 . Clearly (0,0) satisfies the  2 x+4 y=48 . So

The region which contains origin represents the solution set of the in equation 2 x+4 y \leq 48 .

Region represented by 14 x+2 y \leq 60 : the line 4 x+2 y=60  meets the coordinate axes at C1(15,0) and B1(0,30)respectively.

By joining these points we obtain the line 4 x+2 y=60 . Clearly (0,0) satisfies the 14 x+2 y \leq 60 . So

The region which contains origin represents the solution set of the in equation  14 x+2 y \leq 60 .

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+4 y \leq 48,4 x+2 y \leq 60, x \geq 0 \& y \geq 0 are as follows.

The corner points are O(0,0), B(0,12),E1(12,6) and C1(15,0)

The values of Z at these corner points as follows.

Corner Points

z=6 x+8 y

O

0

B1

96

E1

120           

C1

90

The maximum value of Z is 120 which is attained at E1(12,6)

Thus, the maximum profit is Rs. 120 obtained when 12 units of articles A and 6 units of Article B were manufactured.

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