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  • Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 48

Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 48

Answers (1)

Answer: The maximum profit of the manufacture is Rs.4000

Hint:

Use properties of LPP

Given:

Item

Number of hours required on machines

                                 

              I                           II                                III

M

             1                           2                                 1

N

             2                            1                             1.25

 

She makes a profit of Rs.600 and Rs.400 on items M and N respectively.

Solution:

Suppose x units of item M and y units of item N are produced to maximize the profit. Since each unit of item M require  hour on machine I and each unit of item N require  hours on Machine I, therefore, the total hours required for producing x units of item M and y units of item N on machine I are (2x +y). But machines I is capable of being operated for at most 12 hours.

                                         2 x+y \leq 12

Similarly, each unit of item M require 2 hours on machine II and each unit of item N require 1 hour on machine II, therefore, total hours required for producing x units of item M and y units of item N on machine II are (x + 2y). But machines II is capable of being operated for at most 12 hours.

                                         x+2 y \leq 12

Also, each unit of item M require 1 hour on machine III and each unit of item N require 1.25 hour on machine III,  therefore, the total hours required for producing x units of item M and y units of item N on machine III are (x + 1.25y). But, machines III must be operated for at least 5 hours.

                                         x+1.25 y \geq 5

The profit from each unit of item M is Rs.600 and each unit of item N is Rs.400, Therefore the total profit from x units of item M and y units of item N is (600x + 400y).

Thus, the given linear programming problem is

                                         Maximize Z = 600x + 400y

Subject to the constraints,

                                         \begin{aligned} &2 x+y \leq 12 \\ &x+2 y \leq 12 \\ &x+1.25 y \geq 5 \\ &x, y \geq 0 \end{aligned}

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

                                       12 x+y=12 \quad x+1.25 y=5

The coordinates of the corner points of the feasible region are A(5,0), B(6,0), C(4,4), D(0,6) and E(0,4)

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

 

Corner Points

Z=600x+400y

(5,0)

3000

(6,0)

 600 \times 6+400 \times 0=3600

(4,4)

600 \times 4+400 \times 4=4000 (maximum)

(0,6)

600 \times 0+400 \times 6=2400

(0,4)

600 \times 0+400 \times 4=1600

 

The maximum value of Z is 4000 at x=4,y=4.

Hence, 4 units of item M and 4 units of item N should be produced to maximize the profit.

The maximum profit of the manufacturer is Rs.4000.

Posted by

infoexpert27

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