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.
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.
10x+5y≥20
Since, each lb of food I contains 5 units of protein.Therefore, x lbs of food I contains 5x units of protein.
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.
5x+4y ≥ 20
Since, each lb of food I contains 2 units of calories.Therefore, x lbs of food I contains 2x units of calories.
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 2x + 6y units of calories.
But, the minimum requirement is 13 lbs of calories.
2x+6y≥13
Finally, the quantities of food I and food II are non negative values.
So, x, y ≥ 0x, y ≥ 0
Hence, the required LPP is as follows:
Min Z = 60x + 100y
subject to
10x+5y≥20
5x+4y≥20
2x+6y≥13
x, y≥0
Answered by Aaryan | 1 year agoMinimize Z = 2x + 4y
Subject to
x+y≥8
x+4y≥12
x≥3, y≥2
Maximize Z = 7x + 10y
Subject to
x+y≤30000
y≤12000
x≥6000
x≥y
x, y≥0
Maximize Z = 10x + 6y
Subject to
3x+y≤122x+5y≤34 x, y≥0
Maximize Z = 15x + 10y
Subject to
3x+2y≤802x+3y≤70 x, y≥0