(Manufacturing problem) A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts, while it takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹2.50 per package on nuts and ₹1 per package on bolts. Form a linear programming problem to maximize his profit, if he operates each machine for atmost 12 hours.

Maximize $Z=2.5x+y$ 

Subject to: $x+3y≤12,3x+y≤12, x,y≥0$

The correct answer is Option (3) → Maximize $Z=2.5x+y$, Subject to: $x+3y≤12,3x+y≤12, x,y≥0$

Suppose that $x$ packages of nuts and $y$ packages of bolts are produced. The objective of the manufacturer is to maximize the profit $Z=2.50x + 1.y$

Time required on machine A to produce $x$ packages of nuts and $y$ packages of bolts is $1.x + 3.y = x + 3y$, while time required on machine B is $3.x + 1.y = 3x + y$.

From given data, we can formulate the L.P.P. as:

Maximize $Z = 2.50x + y$ subject to the constraints

$x + 3y ≤ 12$ (Machine A constraint)

$3x + y ≤ 12$ (Machine B constraint)

$x ≥0, y ≥0$ (Non-negativity constraints)