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  • Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 2 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 2 Maths Textbook Solution.

Answers (1)

Answer:

Maximum Profit = Rs 40 (A = 4, B = 4)

Hint:

Let  x_{1} and y are required quantity of x and y.

Given:

He makes a profit of ?6.00 on item A and ?4.00 on item B.

Solution:

 So, profits of one X item =s of type A and Y item of type B are 6x and Rs 4y respectively.

Maximum  z=6x+4y  (Where x and y are required quantity)

Given machine I works 1 hour and 2 hours on item A and B respectively., so x number of item A and y number of item B need x hour and 2y hours on machine  I respectively, but machine I works at most 12 hours so,

x+2y\leq 12( first constraint)

Given, machine II works 2 hours and 1 hour on item A and B respectively, but machine II works maximum 12 hours, So

2x+y\leq 12 (second constraint)

Given, machine III works 1 hour and  \frac{5}{4}  hour on one item A and B respectively, but machine III works maximum 5 hours, So

x+54y\geq 5

\Rightarrow 4x+5y\geq 20   (third constraint)

The required LPP

z=6x+4y

Subject to constraint

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

2 x+y 1 \leq 2 \\

4 x+5 y \geq 20

x, y\geq 0  since the number of item A and B not be less than zero

The corresponding line is

\begin{aligned} &x+2 y=12 \Rightarrow \frac{x}{12}+\frac{y}{6}=1 \\ & \end{aligned}

2 x+y=12 \Rightarrow \frac{x}{6}+\frac{y}{12}=1 \\

4 x+5 y=20 \Rightarrow \frac{x}{5}+\frac{y}{4}=1

Shaded region  A_{2}A_{3}PB_{3}B_{1}   represent feasible region

Points

z=6x+4y

A_{2}\left ( 6,0 \right )

 36

A_{3}\left ( 5,0 \right )

 \\30

B_{3}\left (0 ,4 \right )

 16

B_{2}\left (0 ,6 \right )

 \\24

P\left ( 4,4 \right )

 40

 

Hence z is maximum at x = 4 and y = 4

Required number of product A = 4, B = 4

Maximum profit=Rs 40

Posted by

infoexpert27

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