Question

A manufacturer makes two types A and B of tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:

	Machines
	I	II	III
A B	12 6	18 0	6 9

Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.

Answer 1

Let x units of type A and y units of type B cups were made.
Quantities cannot be negative.Therefore,
$x,y\ge 0$

As we are given,

	Machines
	I	II	III
A B	12 6	18 0	6 9

Every machine is available for a maximum of 6 hours per day i.e. 360 minutes per day.
Therefore, the constraints are

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

If the profit on each cup A is 75 paise and that on each cup B is 50 paise.

Total profit = Z = $0.75x+0.50y$ which is to be maximised.

Thus, the mathematical formulat​ion of the given linear programmimg problem is

Max Z = $0.75x+0.50y$

subject to

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

First we will convert inequations into equations as follows :
12x + 6y = 360, 18x = 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 meets the coordinate axes at C₁(20, 0) . We obtain the line 18x + 0y = 360.Clearly (0,0) satisfies the inequation 18x + 0y ≤ 360. So,the region which contains the origin represents the solution set of the inequation 18x + 0y ≤ 360.

Region represented by $6x+9y\le 360$:
The line $6x+9y=360$ meets the coordinate axes at E₁(60, 0) and F₁(0, 40) respectively. By joining these points we obtain the line $6x+9y=360$. Clearly (0,0) satisfies the inequation $6x+9y\le 360$ . So,the region which contains the origin represents the solution set of the inequation $6x+9y\le 360$.

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, $6x+9y\le 360$, x ≥ 0, and y ≥ 0 are as follows.



The corner points are O(0, 0) F₁(0, 40), G₁(15, 30), H₁(20, 20) and C₁(20, 0).

The values of Z at these corner points are as follows

Corner point	Z= 0.75x + 0.50y
O	0
F1	20
G1	26.25
H1	25
C1	15

Thus, the maximum profit is Rs 26.25 obtained when 15 units of type A and 30 units of type B cups were made.