RD Sharma Class 12 Exercise 29.2 Linnear Programing Solutions Maths

RD Sharma Class 12 Exercise 29.2 Linnear Programing Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 02:52 PM IST

Maths for CBSE students is extensive and detailed. To get the best understanding of the subject, students should follow suitable materials that cover all concepts. RD Sharma books are the best option for CBSE students as they are comprehensive and contain numerous topics.

RD Sharma Class 12th Exercise 29.2 covers the chapter Linear Programming. It contains 28 Level 1 sums that are simple and based on fundamentals. Moreover, it covers the graphical method concept in which the equation and constraints are specified. RD Sharma solutions The majority of sums in this exercise are based on the same concept, which means that students can practice a few of them and refer to the solutions for the rest to save time.

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RD Sharma Class 12 Solutions Chapter29Linear Programming - Other Exercise
Linear Programming Excercise: 29.2
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter29Linear Programming - Other Exercise

Linear Programming Excercise: 29.2

Linear Programming exercise 29.2 question 26

Answer: the maximum value of Z is 10
Hint: plot the points on the graph
Given: maximize: $Z=2 x+5 y$
Solution: the given constraints are
$\begin{aligned} &2 x+4 y \leq 8----(1) x+y \leq 4--(3) \\ &3 x+y \leq 6----(2) x \geq 0, y \geq 0 \end{aligned}$
We need to maximize the objective function $Z=2 x+5 y$ . Converting the equations into equations, we obtain the lines $2 x+4 y=8,3 x+y=6, x+y=4, x=0, y=0$
These lines are drawn and the feasible region of the LPP is shaded.
The coordinates of the corner points of the feasible regions are
$O(0,0), A(0,2), B(1.6,1.2), C(2,0)$
The values of the objects function at the points are given in the following table;
$\begin{array}{|l|l|} \hline \text { Points } & \text { Values of the objective function } z=2 x+5 y \\ \hline O(0,0) & 2(0)+5(0)=0 \\ A(0,2) & 2(0)+5(2)=10 \\ B(1.6,1.2) & 2(1.6)+5(1.2)=9.2 \\ C(2,0) & 2(2)+5(0)=4 \\ \hline \end{array}$
Out of these values of Z, the maximum value of Z is 10

Linear Programming exercise 29.2 question 27

Answer: the maximum value of Z is 200
Hint: plot the point on the graph.
Given: maximize $z=20 x+10 y$
Solution: the given constraints are $x+2 y \leq 28,3 x+y \leq 24, x \geq 2, x, y \geq 0$
Converting the inequalities into equations, we obtain the following equations;
$x+2 y=28,3 x+y=24, x=0, y=0$
These equations represents straight lines in Xy palne
The line $x+2 y=28$ must coordinate axes at A(18, 0) and B(0, 14). Join these to obtain the line $x+2 y=28$ .
The line $3 x+y=24$ meets the coordinates axes at $A_{2}(0,0), B_{2}(0,24)$ join these points to obtain the line $3 x+y=24$ . The line $u=2$ is parallel to y-axis passes through the point $A_{2}(2,0)$
Also, $u=0$ is the y-axis and $y=o$ is the x-axis.the feasible region of the LPP is shaded
The point of intersection of lines $x+2 y=28,3 x+y=24$ is D(-1,12)
The point of intersection of line $u=2$ and $x+2 y=28$ R(2,0)
The coordinates of the corner points of the feasible region we have $A_{2}(0,0), B_{2}(0,24)$ the value of the objective function of these points are given in the following table;
$\begin{array}{|l|l|} \hline \text { Points } & \text { Values of the objective z } \\ \hline A_{2}(2,0), & z=20(3)+10(0)=40 \\ A_{2}(8,0) . & z=20(8)+10(0)=160 \\ Q(4,12) & z=20(4)+10(12)=200 \\ R(2,3) & z=20(2)+10(13)=170 \\ \hline \end{array}$
Clearly, the maximum value of Z is 200

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As maths is a subject that requires a lot of practice, Career360 has provided essential questions that are based on fundamentals to help students prepare accordingly. The answers provided by RD Sharma Class 12th Exercise 29.2 material contain step-by-step explanations that are helpful to understand the concepts easily.

This chapter contains hundreds of sums, which is why it is not feasible for students to solve every one of them. Career360 has provided this material to help students cover their syllabus easily while saving time. These solutions are the best medium for preparation as they are simple and easy to understand. Students who face difficulty studying from RD Sharma textbook can refer to this material to simplify their learning.

Students can complete a fixed amount of sums every day to progress in their portion efficiently. This method will help them understand all concepts clearly while saving time for revision. RD Sharma Class 12th Exercise 29.2 material is available on Career360's website for students to access. It is entirely free, and all the solutions are available on the website. This makes it a convenient option for students to prepare for their exams from the comfort of their homes.

RD Sharma Chapter wise Solutions

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