Question

A firm is engaged in producing two models $X_{1.}{X}_2$ performing only three operations assembling, pianting and testing.The relevant data are as follows.
Total number of hours available each week for assemblling $600$ hours, painting $100$ hours and testing $30$ hours.The firm wishes to determine its weekly products mix so as to maximize revenue formulate I.P.P model for maximize the revenue

Answer 1

Let the no. of models of types $X_1,X_2$ be x and y respectively.

According to the table,

Total no. of hours of required for assembling $=1\times x+1.5\times y=x+1.5y$

Total no. of hours of required for painting $=0.2\times x+0.2\times y=0.2(x+y)$

Total no. of hours of required for testing $=0\times x+0.1\times y=0.1y$

According to question,

Maximum no. of hours available for assembling$=600$hrs

$\therefore x+1.5y\le 600\Rightarrow 2x+3y\le \mathrm{1200....}\left(1\right)$

Maximum no. of hours available for painting$=100$hrs

$\therefore 0.2(x+y)\le 100\Rightarrow x+y\le \mathrm{500....}\left(2\right)$

Maximum no. of hours available for testing$=30$hrs

$\therefore 0.1y\le 30\Rightarrow y\le \mathrm{300....}\left(3\right)$

The objective function is the revenue $Z=50x+80y$ which we have to maximize against the constraints (1),(2) and (3)

Using the graphical method of LPP we first need to plot the equation of constraints and find the feasible region.

According to the graph and inequations of the constraints, the feasible region is the area formed by the line x+y=500

So, the corner points are $(0,0),(0,250),(250,0)$

Value of Z at (0,0): $Z=50\times 0+80\times 0=0$

Value of Z at (0,250): $Z=50\times 0+80\times 250=20000$

Value of Z at (250,0): $Z=50\times 250+80\times 0=12500$

So Z is maximum at $x=0$ and $y=250$