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

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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.

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