Answer :

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

Max Z = 30x + 25y

Subject to

3x + 3y 18

3x + 2y 15

x, y 0

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,) 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.

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