Let the number of packets of nuts and bolts be x and y respectively.
∴According to the question,
X + 3y ≤ 12, 3x + y ≤ 12, x ≥ 0, y ≥ 0
Maximize Z = 17.50x + 7y
The feasible region determined by X + 3y ≤ 12, 3x + y ≤ 12, x ≥ 0, y ≥ 0 is given by
The corner points of the feasible region are A(0,0), B(0,4), C(3,3)), D(4,0).
The value of Z at the corner point is
The maximum value of Z is 73.50 at (3,3).
The manufacturer should make 3 packets each of nuts and bolts to make maximum profit of Rs.73.50.