Question

Project on to minimise the cost of the food, meeting the dietary requirements of the staple food of the adolescent students of your school. Task to be done (1) select two food items constituting one cereal and one pulse.(2) find the minimum cost of the selected cereal and pulse from market. (3) formulate the converponding linear programming problem. (4) solve the problem graphically.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The linear programming problem is - $Z_{minimize}$=100x+20y, Constrains are- $220x+100y\geq 60$. $630x+760y\geq 1500$ and x,y$\geq 0$.

Given,

There are some dietary requirements of the staple food of the adolescent students of some schools.

To Find,

(1) 2 food items including one pulse and one cereal.

(2)The minimum cost of one pulse and one cereal.

(3) The formulated version of the corresponding linear programming problem.

(4)The problem graph,

Solution,

We can solve this mathematical linear programming problem using the method.

At first let's assume that the 2 food items be, Wheat and arhar dal or pigeon peas.

The rate of Wheat is Rs. 20 per kg and the rate of Arhar dal is Rs. 100 per kg.

So, we need to minimize the 100x+20y.

We know protein in arhar dal is 220gm / kg and in wheat 100gm/kg.

Carbohydrate in arhar dal is 630gm / kg and in wheat 760gm/kg.

Suppose the dietary requirements of adolescent students are as below,

• The requirement for protein is 60gm
• Carbohydrate is 1500gm.

So we can formulate our corresponding linear programming problem as follows,

$Z_{minimize}$=100x+20y

Constrains are-

$220x+100y\geq 60$

$630x+760y\geq 1500$ and x,y$\geq 0$.

Hence, the answers are as follows- The two food items including one pulse and one cereal are Wheat and Arhar dal/ pigeon peas. The arhar dal is Rs.100/kg and wheat is Rs. 20/kg. The linear programming problem is - $Z_{minimize}$=100x+20y, Constrains are- $220x+100y\geq 60$. $630x+760y\geq 1500$ and x,y$\geq 0$.

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