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 ₹ 12 and ₹ 16 per doll respectively on dolls *A* and *B*, how many of each should be produced weekly in order to maximise the profit?

#### Solution

Let \[x\] units of doll A and \[y\] units of doll B are manufactured to obtain the maximum profit.

The mathematical formulation of the above problem is as follows:

Maximize \[Z = 12x + 16y\]

Subject to

\[x + y \leq 1200\]

\[y \leq \frac{x}{2}\]

\[x - 3y \leq 600\]

\[x, y \geq 0\]

The shaded region represents the set of feasible solutions.

The coordinates of the corner points of the feasible region are O(0,0), A(800,400), B(1050,150) and C(600,0).

\[= 12\left( 0 \right) + 16\left( 0 \right) = 0\]

The value of Z at A(800,400) \[= 12\left( 800 \right) + 16\left( 400 \right) = 16000\] (Maximum)

The value of Z at B(1050,150) \[= 12\left( 1050 \right) + 16\left( 150 \right) = 15000\]

The value of Z at C(600,0)

\[= 12\left( 600 \right) + 16\left( 0 \right) = 7200\]

Therefore, 800 units of doll A and 400 units of doll B should be produced weekly to get the maximum profit of Rs 16000.