Question

A manufacturer produces nuts and bolts. It takes $1$ hour of work on machine A and $3$ hours on machine B to produce 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 $Rs7$ per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operate his machines for at the most $12$ hours a day?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

Let x package nuts and y package bolts are produced
Let z be the profit function, which we have to maximize.
Here, $z=17.50x+7y....\left(1\right)$ is objective function.
And constraints are
$x+3y\le 12$ .... (2)
$3x+y\le 12$ .... (3)
$x\ge 0$ ....(4)
$y\ge 0$ ....(5)
On plotting graph of above constraints or inequalities $\left(2\right),\left(3\right),\left(4\right)$ and $\left(5\right)$ we get shaded region as feasible region having corner points A, O, B and C.
For coordinate of 'c' two equations,
$x+3y=12....\left(6\right)$
$3x+y=12....\left(7\right)$
On solving we get, $x=3$ and $y=3$
Hence coordinate of C are $(3,3)$
Now value of z is evaluated at corner point as shown in the graph.
Therefore, maximum profit is Rs. $73.5$ when $3$ package nuts and $3$ package bolt are produced.