RD Sharma Class 12 Exercise 29.5 Linear Programing Solutions Math's - Download PDF Free Online

# RD Sharma Class 12 Exercise 29.5 Linear Programing Solutions Math's - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 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.

## Linear Programming Excercise: 29.5

Linear Programming Exercise 29.5 Question 1

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 $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

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 $= 400$

The class 12 RD Sharma chapter 29 exercise 29.5 solution can be the perfect study guide to help students perfect their skills and improve their performance. There are several reasons why students should use the RD Sharma solutions for exam preparations. Here are some of the reasons why RD Sharma class 12 solutions chapter 29 ex 29.5 is an excellent option for all:-

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The RD Sharma class 12 solutions Linear Programming 29.5 is an excellent book for self-study at home. Students can solve their NCERT questions and compare answers with the text to check their own performance. If they practice the book enough, they can even get standard maths questions in their board exams.

## RD Sharma Chapter wise Solutions

1. Who can use this material?

Class 12 CBSE students Who want to get more information about linear programming can use this material.

2. From where can I access this material?

Students can access this material from Career360’s website.

3. Can I prepare for exams using this material?

As this material covers the entire syllabus, students can prepare from it to score good marks in exams.

4. Does this material follow the CBSE syllabus?

Yes, this material complies with the CBSE syllabus.

5. What is the cost of this material?

Career360 has provided this material for free to help students prepare for their exams.

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