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To Maintain His Health Person Must Fulfil Certain Minimum Daily Requirements for Several Kinds of Nutrients .What Combination of Two Food Items Will Satisfy Daily Requirement and Entail Least Cost? - Mathematics

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Sum

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.

\[\therefore\] \[10x + 5y \geq 20\]
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.
\[\therefore\] \[5x + 4y \geq 20\]
Since, each lb of food I contains 2 units of calories.Therefore, x lbs of food I contains 2units 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.
\[\therefore 2x + 6y \geq 13\]
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  
\[10x + 5y \geq 20\]
\[5x + 4y \geq 20\]
\[2x + 6y \geq 13\]
\[x, y \geq 0\]
Concept: Linear Programming Problem and Its Mathematical Formulation
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.1 | Q 7 | Page 15
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