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  • Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 50

Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 50

Answers (1)

Answer: Therefore, the maximum profit is Rs.262.50

Hint:

Use properties of LPP

Given:

Types of toys

Machines

                                 

            I                           II                                III

A

           12                         18                                6

B

           16                          0                                 9

Solution:

Suppose the manufacturer makes x toys of type A and y types of toy B.

Since each toy of type A requires 12 minutes on machine I and each toy of type B require 6 minutes on machine I, therefore, x toys of type A and y toys of type B require (12x + 6y) minutes on machine I.But, machines I is available for at most 6 hours.

                                                              \begin{aligned} &12 x+6 y \leq 360 \\ &2 x+y \leq 60 \end{aligned}

Similarly, each toy of type A requires 18 minutes on machine II and each toy of type B require 0 minutes on machine II, therefore, x toys of type A and y toys of type B require (18x + 0y) minutes on machine II. But, machines II is available for at most 6 hours.

                                                              \begin{aligned} &18 x+0 y \leq 360 \\ &\Rightarrow x \leq 20 \end{aligned}

Also, each toy of type A requires 6 minutes on machine III and each toy of type B require 9 minutes on machine III, therefore, x toys of type A and y toys of type B require (6x + 9y) minutes on machine III. But, machines III is available for at most 6 hours.

                                                              \begin{aligned} &6 x+9 y \leq 360 \\ &\Rightarrow 2 x+3 y \leq 120 \end{aligned}

The profit on each toy of type A is Rs.7.50 and each toy of type B is Rs.5. Therefore, the total profit from x toys of type A and y toys of type B is Rs.(7.50x+5y)

Thus the given linear programming problem is

Maximize Z=7.5x+5y

Subject to the constraints:

                                                       \begin{aligned} &2 x+y \leq 60 \\ &x \leq 20 \\ &2 x+3 y \leq 120 \\ &x, y \geq 0 \end{aligned}

The feasible region determined by the given constraints can be diagrammatically represented as.

The coordinates of the corner points of the feasible region are O(0,0), A(20,0), B(20,20), C(15,30) and D(0,40).

The value of the object function at these points is given in the following table.

Corner Points

Z=7.5x+5y

(0,0)

7.5 \times 0+5 \times 0=0

(20,0)

7.5 \times 20+5 \times 0=150

(20,20)

7.5 \times 20+5 \times 20=250

(15,30)

262.5

(maximum)

(0,40)

7.5 \times 0+5 \times 40=200

 

The maximum value of Z is 262.5 at x = 15, y = 30.

Hence, 15 toys of type A and 30 toys of type B should be manufactured in  a day to get maximum profit.

The maximum profit is Rs.262.5

Posted by

infoexpert27

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