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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.
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Solution
Let x units of type A and y units of type B cups were made.
Quantities cannot be negative.Therefore, \[x, y \geq 0\]
As we are given,
| Machines | |||
| I | II | III | |
| A B |
12 6 |
18 0 |
6 9 |
Therefore, the constraints are
\[12x + 6y \leq 360\]
\[18x + 0y \leq 360\]
\[6x + 9y \leq 360\]
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 formulation of the given linear programmimg problem is
Max Z = \[0 . 75x + 0 . 50y\] subject to
\[12x + 6y \leq 360\]
\[18x + 0y \leq 360\]
\[6x + 9y \leq 360\]
First we will convert inequations into equations as follows :
12x + 6y = 360, 18x = 360,
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
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,
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.
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