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Question
To maintain his health a person must fulfil certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients-calcium, protein and calories and the person's diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:
| Food I (per lb) |
Food II (per lb) |
Minimum daily requirement for the nutrient |
||||
| Calcium | 10 | 5 | 20 | |||
| Protein | 5 | 4 | 20 | |||
| Calories | 2 | 6 | 13 | |||
| Price (Rs) | 60 | 100 |
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.
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Solution
Let the person takes x lbs and y lbs of food I and II respectively that were taken in the diet.
Since, per lb of food I costs Rs 60 and that of food II costs Rs 100.
Therefore, x lbs of food I costs Rs 60x and y lbs of food II costs Rs 100y.
Total cost per day = Rs (60x + 100y)
Let Z denote the total cost per day
Then, Z = 60x + 100y
Total amount of calcium in the diet is \[10x + 5y\]
Since, each lb of food I contains 10 units of calcium.Therefore, x lbs of food I contains 10x units of calcium.
Each lb of food II contains 5 units of calciu.So,y lbs of food II contains 5y units of calcium.
Thus, x lbs of food I and y lbs of food II contains 10x + 5y units of calcium.
But, the minimum requirement is 20 lbs of calcium.
Each lb of food II contains 4 units of protein.So,y lbs of food II contains 4y units of protein.
Thus, x lbs of food I and y lbs of food II contains 5x + 4y units of protein.
But, the minimum requirement is 20 lbs of protein.
Each lb of food II contains units of calories.So,y lbs of food II contains 6y units of calories.
Thus, x lbs of food I and y lbs of food II contains
But, the minimum requirement is 13 lbs of calories.
So,
Min Z = 60x + 100y
subject to
\[5x + 4y \geq 20\]
\[2x + 6y \geq 13\]
\[x, y \geq 0\]
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