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  • Need solution for RD Sharma Maths Class 12 Chapter 29 Linear Programmig Excercise 29. 4 Question 28

Need solution for RD Sharma Maths Class 12 Chapter 29 Linear Programmig Excercise 29.4 Question 28

Answers (1)

Answer:

Maximum Value of Z is Ra.1020 which is attained at \mathrm{B}_{1}(60,240)  . Thus the maximum profit is Rs.1020 obtained when 60 units of product A and 240 units of product B were manufactured.

Hint:

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

Given:

Firm manufacturer’s two types of Products A and B and sells them at a profit of Rs.5 per unit of Type A and Rs.3 per unit of Type B.

Solution:

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

Number of products cannot be negative.

Therefore, x, y \geq 0

According to question, the given information can be tabulated as                                      

 

Time on M1(Minutes)

Time on M2(Minutes)

Product A(x)

1

2

Product B(y)

1

1

Availability

300

360

The constraints are

                                       \begin{aligned} &x+y \leq 300 \\ &2 x+y \leq 360 \end{aligned}

Firm manufactures two types of Products A and B and sells them at a profit of Rs.5 per unit of type A and Rs. 3 per unit of the type B. Therefore x unit of product A and y units of product B costs Rs.5x and Rs.3y respectively.

Total Profit = Z = 5x + 3y which is to be maximized

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

                                       Max Z = 5x + 3y

Subject to

                                       \begin{aligned} &x+y \leq 300 \\ &2 x+y \leq 360 \\ &x, y \geq 0 \end{aligned}

First we will convert in equations as follows x+y=300,2 x+y=360, x=0 \& y=0

Region represented by x+y \leq 300 : the line x+y=300  meets the coordinate axes at A1(300,0) and B1(0,300)respectively.

By joining these points we obtain the line  x+y=300  . Clearly (0,0) satisfies the  x+y=300 . So

the region which contains origin represents the solution set of the in equation x+y \leq 300 .

Region represented by 2 x+y \leq 360   : the line meets the coordinate axes at C1(180,0) and D1(0,360) respectively.

By joining these points we obtain the line 2 x+y=360  . Clearly (0,0) satisfies the 2 x+y \leq 360 . So

the region which contains origin represents the solution set of the in equation  2 x+y \leq 360 .

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  x+y \leq 300,2 x+y \leq 360_{\ell} x \geq 0 \& y \geq 0  are as follows.

 

                                                                                                Scale: On x-axis: 1 Big Division = 100 units

                                                                                                            On y- axis: 1 Big Division = 100 units

 

  

The corner points are Q(0,0), B_{1}(0,300), E_{1}(60,240) \text { and } C_{1}(180,0)

The values of Z at these corner points are as follows

Corner Points

z=5 x+3 y

O

0

B_{1}

900

E_{1}

1020

C_{1} 

900

The maximum value of X is Rs. 1020 which B attained at B1(60,240)

Thus, the maximum profit is Rs.1020 obtained when 60 units of Product A and 240 units of Product B were manufactured.

 

Posted by

infoexpert27

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