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.

Asked by Sakshi | 1 year ago |  39

##### 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.

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 ago

### Related Questions

#### Minimize Z = 2x + 4y

Minimize Z = 2x + 4y

Subject to

x+y≥8

x+4y≥12

x≥3, y≥2

#### Maximize Z = 7x + 10y

Maximize Z = 7x + 10y

Subject to

x+y≤30000

y≤12000

x≥6000

x≥y

x, y≥0

#### Maximize Z = 3x + 4y

Maximize Z = 3x + 4y

Subject to

2x+2y≤802x+4y≤120

#### Maximize Z = 10x + 6y

Maximize Z = 10x + 6y

Subject to

3x+y≤122x+5y≤34  x, y≥0