Answer to Question #151319 in Operations Research for K.manivarma

Question

2A firm manufactures two products A and B on which the profits earned per unit are Rs. 3 and Rs.4 respectively. Each product is processed on two machines M and M Product Arequires one minute of processing time on M, and two minutes on M, while requires one minute on M, and one minute on M, Machine M is available for not more than 7 hours 30 minutes while machine M is available for 10 hours during any working day. Find the number of units of products A and B to be manufactured to get maximum profit. Formulate the above as a LPP and solve by graphical method.

Answer 1

Let $x$ units of firm A and $y$ units of firm B were manufactured: $x\geq0, y\geq0.$

Given information can be tabulated as

$$\def\arraystretch{1.5} \begin{array}{c:c:c} & Time\ of\ M_1 & Time\ of\ M_2 \\ \hline Product\ A\ (x) & 1\min & 2\min \\ Product\ B\ (y) & 1\min & 1\min \\ \hdashline Availability & 450\min & 600\min \end{array}$$

The constrants are

$$x+y\leq450, 2x+y\leq600$$

Total profit $Z=3x+4y$ which is to be maximized.

The objective function: $Max\ Z=3x+4y$

Subjective Constraints

$x+y\leq450$

$x+2y\leq600$

$x\geq0, y\geq0$

Horizontal $(x)$axis: Product A , Vertical $(y)$ axis: Product B

Constraint No. 1: $x+y\leq450$ Converting into equality: 

$x+y=450$

The two points which make the constraint line are: $(0,450)$ and $(450, 0)$

Constraint No. 2: $x+2y\leq600$ Converting into equality: 

$x+2y=600$

The two points which make the constraint line are: $(0,300)$ and $(600, 0)$

Each constraint will be represented by a single straight line on the graph. There are two constraints, hence there will be two straight lines.

The feasible region determined by the system of constraints

$$\begin{matrix} x+y\leq450 \\ x+2y\leq600\\ x\geq 0 \\ y\geq0 \end{matrix}$$

is as follows

The corner points are $O(0,0), D(0,300),E(300,150),A(450,0).$

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

$$\begin{matrix} Corner\ point & Z=3x+4y \\ O & 0 \\ D & 1200\\ E & 1500\\ A & 1350 \end{matrix}$$

The maximum value of $Z$ is $Rs1500$ which is attained at $E(300,150).$

The maximum profit is $Rs1500$ obtained when 300 units of firm A and 150 products of firm B are manufactured.