Question

A company produces two types of leather belts, say type $A$ and $B$ . Belt $A$ is a superior quality and belt $B$ is of a lower quality. Profits on each type of belt are $Rs 2$ and $Rs 1.50$ per belt, respectively. Each belt of type $A$ requires twice as much time as required by a belt of type $B$ . If all belts were of type $B$ , the company could produce $1000$ belts per day. But the supply of leather is sufficient only for $800$ belts per day (both $A$ and $B$ combined). Belt $A$ requires a fancy buckle and only $400$ fancy buckles are available for this per day. For belt of type $B$ , only $700$ buckles are available per day.

How should the company manufacture the two types of belts in order to have a maximum overall profit?

Accepted Answer

Let the company produces $A$ belts of type $A$ and $y$ belts of type $B$ .

Number of belts cannot be negative.

Therefore,

$x, y \geq 0$

It is given that leather is sufficient only for $800$ belts per day (both $A$ and $B$ combined). Therefore,

$x + y \leq 800$

It is given that the rate of production of belts of type $B$ is $1000$ per day. Hence, the time taken to produce $B$ belts of type $B$ is $\frac{y}{1000}$

And, since each belt of type $A$ requires twice as much time as a belt of type $B$ , the rate of production of belts of type $A$ is $500$ per day and therefore, time taken to produce $x$ belts of type $A$ is $\frac{x}{500}$ Thus, we have

$\frac{x}{500} + \frac{y}{1000} \leq 1$

$\Rightarrow 2 x + y \leq 1000$

Belt $A$ requires a fancy buckle and only $400$ fancy buckles are available for this per day. $x \leq 400$

For belt of type $B$ , only $700$ buckles are available per day.

$y \leq 700$

Profits on each type of belt are $Rs 2$ and $Rs 1.50$ per belt, respectively. Therefore, profit gained on $x$ belts of type $A$ and $Rs 1.50$ belts of type $B$ is $Rs 2 x$ and $Rs 1.50 y$ respectively.

Hence, the total profit would be $Rs (2 x + 1.50 y)$ .

Let $Z$ denote the total profit.

$∴ z = 2 x + 1.5 y$

Thus, the mathematical formulation of the given linearprogrammingproblem is Maximize $Z = 2 x + 1.5 y$ subject to the constraints

$x + y \leq 800$

$2 x + y \leq 1000$

$x \leq 400$

$y \leq 700$

$x, y \geq 0$

The above inequalities will form a feasible region depicted below.

The corner points are $F, M, G, H, E & O$ . We will ignore $O$ since the profit is zero. Also at $M (100, 700)$ the profit is $2 \times 100 + 1.5 \times 700 = 1250$

The values of $Z$ at other corner points are as follows

Corner point $Z = 2 x + 1.5 y$

$F (0, 700)$ $1050$

$G$ $1300$

$H (400, 200)$ $1100$

$E (400, 0)$ $800$

The maximum value of $Z$ is $1300$ which is attained at $G (200, 600)$ .

Thus, the maximum profit is $Rs 1300$ and the company manufactures $200$ belts of type $A$ and $600$ belt of type $B$ .

	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 $B$	$1000 sq . m$ $1200 sq . m$	$12$ men $8$ men	$60$ $40$

Important Questions on Linear Programming

Learn and Prepare for any exam you want !