Name: Linear Programming: Manufacturing Problem
Uploaded: 2019-07-19
Duration: 7 min 3 s
Description: In this video we are going to learn linear programming manufacturing problem comprehensively with the help of real life examples. Check it out!

Question

A manufacturer produces nuts and bolts for industrial machinery. It takes

1

hour of work on machine

A

and

3

hours on machine

B

to produce a package of nuts which it takes

3

hours on machine

A

and

1

hour on machine

B

to produce a package of bolts. He earns a profit of

₹ 2.5

per package of nuts and

₹ 1

per package of bolts. How many packages of each should he produce each day so as to maximize his profit, if he operates his machines for at most

12

hours a day? Formulate this mathematically and then solve it.

Accepted Answer

The given information can be put in tabular form as below:

Machines Time required to produce Nut Time required to produce bolt Max. machine hours available.

$A$ $1$ $3$ $12$

$B$ $3$ $1$ $12$

Profit (in $₹$ ) $2.50$ $1.00$

Let the manufacture produce $x$ packages of nuts and $y$ packages of bolts each day.

Since, machine $A$ takes one hour to produce one package of nuts and three hours to produce one package of bolts. So, total time required by machine $A$ to produce $x$ packages of nuts and $y$ packages of bolts is $(x + 3 y)$ hours. But machine $A$ operates for atmost $12$ hours.

$∴ x + 3 y \leq 12$

Similarly the total time required by machine $B$ to produce $x$ packages of nuts and $y$ packages of bolts is $(3 x + y)$ hours. But machine $B$ operates for atmost $12$ hours.

$∴ 3 x + y \leq 12$

Since the profit on one package of nuts is $₹ 2.50$ and on one package of bolts the profit is $₹ 1$ . So, profit on $x$ packages of nuts and $y$ packages of bolts is of $₹ (2.50 x + y)$ . Let $Z$ denote the total profit. Then, $Z = 2.50 x + y$ . Clearly, $x \geq 0 and y \geq 0$ .

Thus, the above LPP can be stated mathematically as follows:

Maximize $Z = 2.50 x + y$

Subject to

$x + 3 y \leq 12$

$3 x + y \leq 12$

and $x, y \geq 0$

To solve this LPP we first convert inequations into equations to obtain the following equations.

$x + 3 y = 12, 3 x + y = 12, x = 0, y = 0$ .

The line $x + 3 y = 12$ meets the coordinate axes at $A_{1} (12, 0) and B_{1} (0, 4)$ . Join these two points to obtain the line represented by $x + 3 y = 12$ . The region represented by $x + 3 y \leq 12$ is region containing origin as $x = 0$ , $y = 0$ satisfies $x + 3 y \leq 12$ .

The line $3 x + y = 12$ meets the coordinate axes at $A_{2} (4, 0) and B_{2} (0, 12)$ . Join these two points to obtain the line represented by $3 x + y = 12$ . Since $x = 0, y = 0$ satisfies $3 x + y \leq 12$ . So, region containing origin and below line $3 x + y = 12$ represents the region represented by $3 x + y \leq 12$ .

Clearly, $x \geq 0 and y \geq 0$ represent all points in first quadrant.

Thus, shaded region ${O A}_{2} {P B}_{1}$ in figure represnts the feasible region of given LPP. The coordinates of the corner points of feasible region ${O A}_{2} {P B}_{1}$ are $O (0, 0), A_{2} (4, 0)$ $, P (3, 3) and B_{1} (0, 4)$ . These points are obtained by solving corresponding intersecting lines simultaneously.

The value of the objective function at corner points of feasible region are given in the table below:

Point $(x, y)$ Value of the objective function $Z = 250 x + y$

$O (0, 0)$ $Z = 2.50 \times 0 + 1 \times 0 = 0$

$A_{2} (4, 0)$ $Z = 2.50 \times 4 + 1 \times 0 = 10$

$P (3, 3)$ $Z = 2.50 \times 3 + 1 \times 3 = 10.50$

$B_{1} (0, 4)$ $Z = 2.50 \times 0 + 1 \times 4 = 4$

So, $Z$ is maximum at $x = 3, y = 3$ and maximum value of $Z$ is $10.50$ .

Hence, optimal production strategy for manufacturer will be to manufacture three packages each of nuts and bolts daily and in this case his maximum profit will be $₹ 10.50$ .

	Machines
	I	II	III
$\begin{matrix} A \\ B \end{matrix}$	$\begin{matrix} 12 \\ 6 \end{matrix}$	$\begin{matrix} 18 \\ 0 \end{matrix}$	$\begin{matrix} 6 \\ 9 \end{matrix}$

	Area occupied by the machine	Labour force for each machine	Daily output in units
Machine $A$ Machine $Z = 2.50 \times 4 + 1 \times 0 = 10$	$1000 sq . m$ $1200 sq . m$	$12$ men $8$ men	$60$ $40$

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