Question

1.     A firm manufactures two products; the net profit on product 1 is Rupees 3 per unit and Rupees 5 per unit on product 2. The manufacturing process is such that each product has to be processed in two departments D1 and D2. Each unit of product1 requires processing for 1 minute at D1 and 3 minutes at D2; each unit of product 2 requires processing for 2 minutes at D1 and 2 minutes at D2. Machine time available per day is 860 minutes at D1 and 1200 minutes at D2. How much of product 1 and 2 should be produced every day so that total profit is maximum. Make the mathematical model for the given problem.

Answer 1

Let $x_1$ and $x_2$ be levels of production of two products, then

$z(x_1,x_2)=3\cdot x_1+5\cdot x_2$  is profit function and it should be maximized so $3\cdot x_1+5\cdot x_2\rarr max$

For manufacturing  $x_1$unis of product 1 and $x_2$ units of product 2 we need in

using $x_1+2\cdot x_2$ minutes on depatment D1 and $3\cdot x_1+2\cdot x_2$ minutes on department D2 that have availiable machine time 800 and 1200 minutes per day correspondingly therefore we have two restrictions for the plan

$x_1+2\cdot x_2\le800;\\ 3\cdot x_1+2\cdot x_2\le1200;\\$

We also must take in account positiveness restrictions :

$x_1\ge0,x_2\ge0$

Thus we have math model of production:

$3\cdot x_1+5\cdot x_2\rarr max$

$x_1+2\cdot x_2\le800;\\ 3\cdot x_1+2\cdot x_2\le1200;\\$

$x_1\ge0,x_2\ge0$

Standard form convenient for simplex method is

z->max;

$z-3\cdot x_1-5\cdot x_2=0;$

$x_1+2\cdot x_2+1\cdot t+0\cdot s=800;\\ 3\cdot x_1+2\cdot x_2+0\cdot t+1\cdot s=1200;\\ x_1,x_2,t,s\ge0$

To solve we use simplex method:

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} z & x_1 & x_2 &t&s&b\\ \hline 1 & -3 & -5 &0&0&0\\ \hdashline 0 & 1 & ! 2&1&0&800\\ \hdashline 0 & 3 & 2&0&1&1200\\ \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} z & x_1 & x_2 &t&s&b\\ \hline 1 & -1/2 & 0 &5/2&0&2000\\ \hdashline 0 & 1/2 & 1&1/2&0&400\\ \hdashline 0 & ! 2 & 0&-1&1&400\\ \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} z & x_1 & x_2 &t&s&b\\ \hline 1 & 0 & 0 &9/4&1&2100\\ \hdashline 0 & 0 & 1&3/4&-1&300\\ \hdashline 0 & 1 & 0&-1/2&1/2&200\\ \end{array}$

Answer:

$x_1=200,x_2=300,z_{max}=2100$