Question

A manufacturer produces nuts and bolts. It takes $1$ hours of work on machine $A$ and $3$ hours on machine $B$ to product a package of nuts. It takes $3$ hours on machine $A$ and $1$ hour on machine $B$ to produce a package of bolts. He earns a profit of Rs $17.50$ per package on nuts and Rs $7$ per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operates his machines for at the most $12$ hours a day? From the above as a linear programming problem and solve it graphically.

Answer 1

Let x be the no. of nuts & y be the bolts.

We have to maximize x & y if the factory worked at capacity of 12hours per day.

Clearly, $x\ge y,y\ge 0$

Since we have only 12 hours of machine, we will use the following constraints:-

$x+3y\le 12$

$3x+y\le 12$

The profit of nuts is Rs.17.5 and bolts is Rs. 7 .

We need to maximize $17.5x+7y$

We can see that the feasible region is bounded and in the first quadrant.

On solving the equations, $x+3y=12$ and $3x+y=12$, we get $x=3$ and $y=3$.

Therefore, feasible points are $(0,0);(0,4);(3,3);(4,4)$

Values of z are($z=17.5x+7y$)

The maximum profit $=73.5$ which is when $3$ packets each of nuts & bolts are produced.