A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs.12 and Rs. 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

Let the company manufacture x dolls of type A and y dolls of type B.
Let Z be the profit.
We are to maximize
Z = 12x + 16y
subject to the constraints

Now we draw the graph of x + y = 1200
For x = 0, y = 1200
For y = 0, x = 1200
∴ line meets OX in A(1200, 0) and OY in L(0, 1200) x - 2 y = 0 is a straight line passing through (0, 0) and (2. 1).
Again we draw the graph of x - 3 y = 600
For x = 0, - 3 y = 600 or y = - 200
For y = 0, x = 600
∴ line meets OX in B(600, 0) and OY in M(0, - 200)
Since feasible region satisfies all the constraints.

∴ OBCD is the feasible region.
The corner points are O(0, 0), B(600, 0), C(1050, 150), D(800, 400).
At O(0, 0), Z = 12 × 0 + 16 × 0 = 0 + 0 = 0
At B(600, 0), Z = 12 × 600 + 16 × 0 = 7200 + 0 = 7200
At C(1050, 150), Z = 12 × 1050 + 16 × 150 = 12600 + 2400 = 15000
At D(800, 400), Z = 12 × 800 + 16 × 400 = 9600 + 6400 = 16000
∴ maximum profit Rs. 16000 is when 800 dolls of type A and 400 dolls of type B are manufactured and sold.