Let x trunks of first type and y trunks of second type were manufactured. Number of trunks cannot be negative.
Therefore, x, y ≥ 0
According to the question, the given information can be tabulated as
Therefore, the constraints are,
3x + 3y ≤ 18
3x + 2y ≤ 15.
He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. Therefore, profit gained by him from x trunks of first type and y trunks of second type is Rs 30x and Rs 25y respectively.
Total profit Z = 30x + 25y which is to be maximized.
Thus, the mathematical formulation of the given LPP is
Region 3x + 3y ≤ 18: line 3x + 3y = 18 meets axes at A(6, 0), B(0, 6) respectively. Region containing origin represents the solution of the inequation 3x + 3y ≤ 18 as (0, 0) satisfies 3x + 3y ≤ 18.
Region 3x + 2y ≤ 15: line 3x + 2y = 15 meets axes at C(5, 0), D(0, \(\frac{15}{2}\)) respectively. Region containing origin represents the solution of the inequation 3x + 2y ≤ 15 as (0,0) satisfies 3x + 2y ≤ 15.
Region x, y ≥ 0: it represents first quadrant.
The corner points are O(0, 0), B(0, 6), E(3, 3), and C(5, 0).
The values of Z at these corner points are as follows:
The maximum value of Z is 165 which is attained at E(3, 3).
Thus, the maximum profit is of Rs 165 obtained when 3 units of each type of trunk is manufactured.