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Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.5 Question 2 Maths Textbook Solution.

Answers (1)

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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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