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 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?

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 (1) is the objective function.
And constraints are:
x + 3y ≤ 12 (2)
3x + y ≤ 12 (3)
x ≥ 0 (4)
y ≥ 0 (5)
On plotting the graph of the above constraints or inequalities (2), (3), (4), and (5), we get a shaded region as the feasible region having corner points A, O, B, and C.
For the coordinate of 'C', we have two equations:
x + 3y = 12 (6)
3x + y = 12 (7)
On solving, we get x = 3 and y = 3.
Hence, the coordinates of C are (3, 3).
Now, the value of z is evaluated at each corner point as shown in the graph.
Therefore, the maximum profit is Rs. 73.5 when 3 packages of nuts and 3 packages of bolts are produced.