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  • Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 33 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 33 Maths Textbook Solution.

Answers (1)

Answer:The answer of the given question is that maximum profit is Rs.2375, 25 units of A and 125 units of B should be manufactured.

Hint:

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

Given:

Total capacity of 500man-hour. It takes 5 hours to produce unit A and 3 hours to produce unit B.

Solution:

Let x units of Product A and y units of Product B were manufactured.

Clearly, x \geq 0, y \geq 0

It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The two products are produced in a common production process, which has total capacity of 500 man-hours.

            5 x+3 y \leq 500

The maximum number of unit of A that can be sold is 70 and that for B is 125.

           \begin{aligned} &x \leq 70 \\ &y \leq 125 \end{aligned}

If the profit is Rs.20 per unit for the product A and Rs.15 per unit for the product B. Therefore, profit x units of product A and y units of product B is Rs.20x and 15y respectively.

          Total Profit \Rightarrow z=20 x+15 y

The mathematical formulation of the given problem is

            Max z=20 x+15 y

Subject to

          \begin{aligned} &5 x+3 y \leq 500 \\ &x \leq 70 \\ &y \leq 125 \\ &x \geq 0, y \geq 0 \end{aligned}

First we will convert in equation into equations as follows:

        5 x+3 y=500, x=70, y=125, x=0 \& y=0

Region represented by 5 x+3 y \leq 500  : the line 5 x+3 y=500  meets the coordinate axes at A1(100,0) and B1 1\left(0, \frac{500}{3}\right)  respectively.

By joining these points we obtain the line  5 x+3 y=500 . Clearly (0,0) satisfies the  5 x+3 y=500 . So

The region which contains origin represents the solution set of the in equation  5 x+3 y \leq 500 .

Region represented by  x \leq 70 .

The line x = 70 is the line passes through C1(70,0) and is parallel to y-axis. The region to the left of the line x = 70 will satisfy the in equation x \leq 70 .

Region represented by y \leq 125

The line y = 125 is the line passes through D1(0,125) and is parallel to x-axis. The region below the line

y=125 will satisfy the in equation y \leq 125 .

Region represented by x \geq 0 \& y \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  5 x+3 y \leq 500, x \leq 70, y \leq 125,x \geq 0 \& y \geq 0   are as follows:

The corner points are O(0,0), D(0,125), E(25,125), F(70,50) and C(70,0). The values of Z at the corner points are:

 

Corner Points

z=20 x+15 y  

O

0

D_{1}

1875

E_{1}

2375

F_{1}

2150

C_{1}

1400

The maximum value of Z is 2375 which is at E1(25,125)

Thus, the maximum profit is Rs.2375. 25 units of A and 125 units of B should be manufactured.

Posted by

infoexpert27

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