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 $10 x + 5 y$

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.

$∴$ $10 x + 5 y \geq 20$

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.
$∴$ $5 x + 4 y \geq 20$

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.
$∴ 2 x + 6 y \geq 13$

Finally, the quantities of food I and food II are non negative values.

So, $x, y \geq 0$

Hence, the required LPP is as follows:

Min Z = 60x + 100y

subject to

$10 x + 5 y \geq 20 5 x + 4 y \geq 20 2 x + 6 y \geq 13 x, y \geq 0$

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Solution

Food I
(per lb)

Food II
(per lb)

Minimum daily requirement
for the nutrient

Price (Rs)

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