RD Sharma Class 12 Exercise 29.5 Linear Programing Solutions Math's

Lovekush kumar sainiUpdated on 25 Jan 2022, 02:53 PM IST

Practice makes perfect. This saying is especially true for maths which needs a lot of practice to perfect their calculations and problem-solving skills of students. This is why students should always use the RD Sharma class 12th exercise 29.5 solution, which comes highly recommended by numerous students and teachers in India. The book contains some high-quality answers which will help you solve all your NCERT questions without any problems. The RD Sharma class 12 chapter 29 exercise 29.5 is easily available online and comes with updated syllabus.

RD Sharma solutions Linear Programming 29.5 can be a student's best guide for any exam preparations like school exams, boards, and JEE mains. Exercise 29.5 has two questions and both the questions based on the Transportation Problem that cover some concepts. They are Minimum cost, an optimization problem. The RD Sharma class 12th exercise 29.5 will provide you with answers so that you can solve these questions.

RD Sharma Class 12 Solutions Chapter29Linear Programming - Other Exercise

Linear Programming Excercise: 29.5

Linear Programming Exercise 29.5 Question 1

Answer:
Minimum transportation cost $= 150$
From $A: 10,50,40 \text { quintals to } D, E, F$
From $B: 50,0,0 \text { quintals to } D, E, F$
Hint:
Draw the graph and find the value of $z$
Given:

	Transportation	Cost per quintal
From	$A$	$B$
To	$A$	$B$
$D$ $E$ $F$	$6.00$ $3.00$ $2.50$	$4.00$ $2.00$ $3.00$

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Let the supply of wheat is $x$ quintal from to $A\: to\: D$ and $Y$ quintal from $A$ to $E$ .Then wheat supply will be $(100-x-y)$ quintal from $A$ to $F$.
Similarly, $(60-x),(50-y),(x+y-60)$ quintals of wheat will be supplied from $B\: to\: D, E, F$ respectively.

Where, N represents the Quintals
Now minimum transportation cost,
$\begin{aligned} &z=6 x+3 y+2.50(100-x-y)+4(600-x)+2(50-y)+3(x+y-60) \\ & \end{aligned}$
$z=2.50 x+1.50 y+410$
And constant $x \geq 0, y \geq 0$
$\begin{aligned} &100-x-y \geq 0 \Rightarrow x+y \leq 100 \\ & \end{aligned}$
$60+x \geq 0 \Rightarrow x \leq 60 \\$
$50-y \geq 0 \Rightarrow y \leq 50 \\$
$x+y-60 \geq 0 \Rightarrow x+y \geq 60$
First, we draw the graph of the lines,
$x+y=100, x=60, y=50, x+y=60$

Now, we find the feasible region by constants
$x+y \leq 100, x \leq 60, y \leq 50, x+y \geq 60, x \geq 0, y \geq 0$
And shade it. Its vertices are
$A(10,50), B(60,0), C(60,40), D(50,50)$ at which we find $z$

Vertices	coordinates	$z=2.50 x+1.50 y+410$
$A$ $B$ $C$ $D$	$\left ( 10,50 \right )$ $\left ( 60,0 \right )$ $\left ( 60,40 \right )$ $\left ( 50,50 \right )$	$510-minimum$ $560$ $620$ $610$

Therefore, minimum transportation cost $Rs.510$. For this $10,50,40$ quintals will supply from $A \: to \: D, E, F \text { and } 50,0,0$ quintals will supply $B \: to\: D, E, F$.

Linear Programming Exercise 29.5 Question 2

Answer:
Minimum cost $= 400$ and packets sent $x=10, y=0$
From $A: 10,0,50 \text { quintals to } P Q R$
From $B: 30,40,0 \text { quintals to } P Q R$
Hint:
Pulling coordinates in $z=3 x+4 y+370$
Given:

Transportation cost per packet (in Rs)
From	A	B
To
$P$ $Q$ $R$	$5$ $4$ $3$	$425$

Solution:
According to question,

So, $x \geq 0, y \geq 0$
$60-x-y \geq 0,40-x \geq 0,40-y \geq 0 \text { and }-10+x+y \geq 0$
Total transportation cost $z$
$z=5 x+4 y+3(60-x-y)+4(40-x)+2(40-y)+5(x+y-10)$
$\begin{aligned} &z=3 x+4 y+370 \\ & \end{aligned}$
$\qquad x+y \leq 60$
$x \leq 40 y \leq 40 x+y \geq 10$
Now, we draw the graph of the lines,
$x+y=60$

$x$	$0$	$60$
$y$	$60$	$0$

$x+y=10$

$x$	$0$	$10$
$y$	$10$	$0$

Coordinate $A(0,40), B(20,40), C(40,20), D(40,0), E(10,0), F(0,10)$
Putting coordinates in $z=3 x+4 y+370$

Vertices	coordinates	$Z=3x+4y+370$
$A$ $B$ $C$ $D$ $E$ $F$	$(0,40)$ $(20,40)$ $(40,20)$ $(40,0)$ $(10,0)$ $(0,10)$	$530$ $590$ $570$ $490$ $400 ->Minimum$ $410$

$\therefore x=10 \text { and } y=0$
Minimum cost $= 400$

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