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Question
A hospital dietician wishes to find the cheapest combination of two foods, A and B, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of Acontains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?
Solution
Let the dietician wishes to mix x units of food A and y units of food B.
Therefore,
| Thiamine(mg) | Calories | |
| Food A | 0.12 | 100 |
| Food B | 0.1 | 150 |
| Minimum requirement | 0.5 | 600 |
According to the question,
The constraints are
\[0 . 12x + 0 . 1y \geq 0 . 5\]
\[100x + 150y \geq 600\]
It is given that each food costs 10 paise per units
Therefore,
Total cost, Z = \[10x + 10y\]
Thus, the mathematical formulation of the given linear programmimg problem is
\[0 . 12x + 0 . 1y \geq 0 . 5\]
\[100x + 150y \geq 600\]
Region represented by 0.12x +0.1y ≥ 0.5:
The line 0.12x + 0.6y = 20 meets the coordinate axes at\[A_1 \left( \frac{25}{6}, 0 \right)\] and \[B_1 \left( 0, 5 \right)\]respectively. By joining these points we obtain the line 0.12x + 0.6y = 20.Clearly (0,0) does not satisfies the 0.12x + 0.6y = 20. So,the region which does not contains the origin represents the solution set of the inequation 0.12x+0.1y ≥ 0.5.
Region represented by 100x + 150y ≥ 600:
The line 100x + 150y = 600 meets the coordinate axes at\[C_1 \left( 6, 0 \right)\] and \[D_1 \left( 0, 4 \right)\] respectively. By joining these points we obtain the line 100x + 150y = 600. Clearly (0,0) does not satisfies the inequation 100x + 150y ≥ 600. So,the region which does not contains the origin represents the solution set of the inequation 100x + 150y ≥ 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 0.12x +0.1y ≥ 0.5, 100x + 150y ≥ 600, x ≥ 0, and y ≥ 0 are as follows.
The corner points are B1(0, 5),
| Corner point | Z= 10x +10y |
| B1 | 50 |
| E1 | 46.2 |
| C1 | 60 |
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