Let x and y represent the number of units produced per week of the products A and B respectively. Let Z be the profit corresponding to this rate of production. ThenZ = 40 x + 35 y ...(1)In order to produce these number of units,total consumption of raw material = 2x + 3yand total labour hours needed = 4x + 3yBut total raw material available = 60 kgand total labour hours available = 96∴  we have,                                             ...(2)∵  it is not possible to produce negative number of units∴ x ≥ 0, y ≥ 0    ...(3)∴ firm’s allocation problem can be put in the following mathematical form:Find two real numbers x and y such that2x + 3y ≤ 604x + 3y ≤ 96x ≥ 0, y ≥ 0and for which the objective functionZ = 40x + 35yis maximumConsider a set of rectangular axes OXY in the plane.It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.Let us draw the graph of2x + 3y = 60For x = 0, 3 y = 60 or y = 20For y = 0, 2 x = 60 or x = 30∴ line 2x + 3y = 60 meets OX in A (30, 0) and OY in B (0, 20).Now we draw the graph of 4x + 3y = 96For x = 0, 3y = 96 or y = 32For y = 0, 4x = 96 or x = 24∴ line 4x + 3y = 96 meets OX in C (24, 0) and OY in D (0, 32).Since feasible region is the region which satisfies all the constraints.∴  OCEB is the feasible region.The corner points are O (0, 0), C (24, 0), E (18, 8), B (0, 20).At O (0, 0), Z = 40 (0) 35 (0) = 0 + 0 = 0At C (24, 0), Z = 40 (24) + 35 (0) = 960 + 0 = 960At E (18, 8), Z = 40 (18)+ 35 (8) = 720 + 280 = 1000At B(0, 20), Z = 40(0) + 35 (20) = 0 + 700 = 700Here Rs. 1000 is the maximum values of Z and occurs at E (18, 8)∴  optimal solution is x = 18, y = 8i.e., 18 units of A and 8 units of B.