Question

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

Types of Toys	Machines
	I	II	III
A	12	18	6
B	6	0	9

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Answer 1

Let x toys of type A and y toys of type B were manufactured.
Number of toys cannot be negative.
Therefore, x ≥ 0 and y ≥ 0

Types of Toys	Machines
	I	II	III
A	12	18	6
B	6	0	9

Each machine is available for a maximum of 6 hours per day i.e. 360 minutes.

According to question the constraints are

$$\begin{gathered} 12x+6y\le 360 \\ 18x\le 360 \\ 6x+9y\le 360 \end{gathered}$$

If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5.Therefore, profit on x toys of type A and y toys of type B is Rs 7.50x and Rs 5y respectively.

Total profit = Z = $7.5x+5y$

The mathematical formulation of the given problem is

Max Z = $7.5x+5y$
subject to

$$\begin{gathered} 12x+6y\le 360 \\ 18x+0y\le 360 \\ 6x+9y\le 360 \\ x\ge 0,y\ge 0 \end{gathered}$$

First we will convert inequations into equations as follows:
12x + 6y = 360, 18x + 0y = 360, 6x + 9y = 360, x = 0 and y = 0

Region represented by 12x + 6y ≤ 360:
The line 12x + 6y = 360 meets the coordinate axes at A₁(30, 0) and B₁(0, 60) respectively. By joining these points we obtain the line 12x + 6y = 360. Clearly (0,0) satisfies the 12x + 6y = 360. So, the region which contains the origin represents the solution set of the inequation 12x + 6y ≤ 360.

Region represented by 18x + 0y ≤ 360:
The line 18x + 0y = 360 or 18x = 360 or x = 20 is the line that passes through C₁(20, 0) and is parallel to the Y axis.

Region represented by 6x + 9y ≤ 360:
The line 6x + 9y = 360 meets the coordinate axes at D₁(60, 0) and E₁(0, 40) respectively. By joining these points we obtain the line 6x + 9y = 360. Clearly (0,0) satisfies the inequation 6x + 9y = 360. So,the region which contains the origin represents the solution set of the inequation 6x + 9y = 360.

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 12x + 6y ≤ 360, 18x + 0y ≤ 360, 6x + 9y ≤ 360, x ≥ 0 and y ≥ 0 are as follows:

The corner points are O(0, 0), E₁(0, 40), F₁(15, 30), G₁(20, 20) and C₁(20, 0).

The values of Z at the corner points are

Corner points	Z = $7.5x+5y$
O	0
E1	200
F1	262.5
G1	250
C1	150

The maximum value of Z is 262.5 which is at F₁(15, 30).

Thus, for maximum profit is Rs 262.5, 15 units of toy A and 30 units of toy B should be manufactured.

Hence, it is proved that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.