Question

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?

Answer 1

Let x dolls of type A and y dolls of type B are manufactured.
Number of dolls cannot be negative.
$\therefore x,y\ge 0$

Demand for dolls of type B is at most half of that for dolls of type A.

$y\le \frac{x}{2}$

The combined production level should not exceed 1200 dolls per week.

$x+y\le 1200$

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.

$\therefore x-3y\le 600$

The company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B. Therefore, profit gained from x dolls of type A and y dolls of type B is Rs 12x and Rs 16y respectively.

Total profit = Z = 12x + 16y

Thus, the mathematical formulation of the given LPP is

Maximize Z = 12x + 16y

subject to
$y\le \frac{x}{2}$
$x+y\le 1200$
$x-3y\le 600$
$x,y\ge 0$

First we will convert inequations into equations as follows:
x + y = 50, 2x + y = 80, x = 0 and y = 0

Region represented by $y\le \frac{x}{2}$:
The line $y=\frac{x}{2}$ is the line that passes through (0, 0). The region below the line $y=\frac{x}{2}$ will satisfy the inequation $y\le \frac{x}{2}$. Like if we take a point (400, 400) above the line $y=\frac{x}{2}$. Here, 400 > 200 which is not satisfying the inequation $y\le \frac{x}{2}$. Therefore, the region below the line $y=\frac{x}{2}$ will satisfy the inequation $y\le \frac{x}{2}$.

Region represented by x + y ≤ 1200:
The line x + y = 1200 meets the coordinate axes at A(1200, 0) and $B\left(0,1200\right)$ respectively. By joining these points we obtain the line
x + y = 1200. Clearly (0,0) satisfies the inequation x + y ≤ 1200. So,the region which contains the origin represents the solution set of the inequation x + y ≤ 1200.

Region represented by x − 3y ≤ 600:
The line x − 3y = 600 meets the coordinate axes at C(600, 0) and $D\left(0,-200\right)$ respectively. By joining these points we obtain the line
x − 3y = 600 . Clearly (0,0) satisfies the inequation x − 3y ≤ 600. So,the region which contains the origin represents the solution set of the inequation x − 3y ≤ 600.


Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints $y\le \frac{x}{2}$ , x + y ≤ 1200, x − 3y ≤ 600, x ≥ 0 and y ≥ 0 are as follows.


The corner points are O(0, 0), F(800, 400), E(1050, 150), C(600, 0).

The value of the objective function at the corner points

Corner points	Z = 12x + 16y
O(0, 0)	0
F(800, 400)	16000
E(1050, 150)	15000
C(600, 0)	7200

The maximum value of Z is 16000 which is attained at F(800, 400).
Thus, for maximum profit = Rs 16000, 800 dolls of type A and 400 dolls of type B should be manufactured.