Question

A firm manufactures two types of products

A

and

B

and sells them at a profit of Rs

5

per unit of type

A

and
Rs

3

per unit of type

B

. Each product is processed on two machines

M_{1}

and

M_{2}

. One unit of type $A$ requires
one minute of processing time on

M_{1}

and two minutes of processing time on

M_{2}

, whereas one unit of
type

B

requires one minute of processing time on

M_{1}

and one minute on

M_{2}

. Machines

M_{1}

and

M_{2}

are
respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of
product should the firm produce a day in order to maximize the profit. Solve the problem graphically.

Accepted Answer

Let required num ber of product $A$ and $B$ be $x$ and $y$ respectively.
since, profit on each product $A$ and $B$ are Rs $5$ and $R s$ 3 respectively, so, profits on $M_{2}$ product $A$ and $y$ product $B$ are Rs $M_{2}$ and $R s$ $3 y$ respectively Let $Z$ be to tal profit so
$z = 5 x + 3 y$

since each unit of product $A$ and $B$ require one min. each on m achine $M_{1}$ , $50$ ,
$x$ unit of product $A$ and $y$ units of product $B$ require $x$ and $y$ min. respectivley on machine $M_{1}$ but $M_{1}$ can work at most $5 \times 60 = 300 \min ., 50$
$x + y \leq 300$

since each unit of product $A$ and $B$ require $2$ and one min.respectively on machine $M_{2}, so, x$ unit of product $A$ and $y$ units of product $B$ require $2 x$ and $y$ min. respectivley on machine $M_{2} but M_{2}$ can work at $m$ ost $6 \times 60 = 360 \min ., so$
$2 x + y \leq 360$ (second constraint)

Hence, mathem atical form ulation of LPP is find $x$ and $y$ which
maximize $Z = 5 x + 3 y$

Subject to constriants,

$x + y \leq 300$

$\Rightarrow 2 x + y \leq 360$

$x, y \geq 0$ $[$ since production can not be less than zero $]$

Region $x + y \leq 300 :$ line $x + y = 300$ meets axes at $A_{1} (300, 0), B_{1} (0, 300)$ respectively. Region containing origin represents $x + y \leq 300$ as $(0, 0)$ satisfies $x + y = 300$ .
Region $2 x + y \leq 360 :$ line $2 x + y = 360$ meets axes at $A_{2} (180, 0), B_{2} (0, 360)$ respectively. Region containing origin represents $2 x + y \leq 360$ as $(0, 0)$ satisfies $2 x + y \leq 360$
Region $x, y \geq 0 :$ it represent first quandrant
Shaded region $O A_{2} P B_{1}$ represents feasible region. Point $P (60, 240)$ is obtained by solving

$x + y = 300 and 2 x + y = 360$

The value of $z = 5 x + 3 y$ at

$0 (0, 0) = 5 (0) + 3 (0) = 0$

$A_{2} (180, 0) = 5 (180) + 3 (0) = 900$

$p (60, 240) = 5 (60) + 3 (240) = 1020$

$B_{1} (0, 300) = 5 (0) + 3 (300) = 900$ '

maximum $z = 1020 at x = 60, y = 240$

Num ber of product $A = 60$ , product $B = 240$ maximum profit $=$ Rs $1020$

From \ To	Cost (in ₹)
	A	B	C
$P$	$160$	$100$	$150$
$Q$	$100$	$120$	$(0, 0)$

Types of Toys	Machines
	$I$	$I I$	$I I I$
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

R D Sharma Solutions for Chapter: Linear Programming, Exercise 4: EXERCISE

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Linear Programming, Exercise 4: EXERCISE

Questions from R D Sharma Solutions for Chapter: Linear Programming, Exercise 4: EXERCISE with Hints & Solutions

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