Question

A​ nutritionist, working for​ NASA, must meet certain minimum nutritional requirements and yet keep the weight of the food at a minimum. He is considering a combination of two​ foods, which are packaged in tubes. Each tube of food A contains 4 units of​ protein, 2 units of​ carbohydrates, and 2 units of fat and weighs 3 pounds. Each tube of food B contains 3 units of​ protein, 6 units of​ carbohydrates, and 1 unit of fat and weighs 3 pounds. The requirement calls for 48 units of​ protein, 42 units of​ carbohydrates, and 20 units of fat. How many tubes of each food should be supplied to the​ astronauts?

Answer 1

Answer:

9 tubes of A and 4 tubes of B.

Step-by-step explanation:

Let the number of tubes of A and B that should be supplied be represented by letters, X and Y.

For protein,

X tubes contain 4X units of protein

Y tubes contain 3Y units of protein

4X + 3Y ≥ 48

For Carbohydrates,

X tubes contain 2X units of Carbohydrates

Y tubes contain 6Y units of Carbohydrates

2X + 6Y ≥ 42

For fats,

X tubes contain 2X units of fats

Y tubes contain 1Y units of fats

2X + Y ≥ 20

And X ≥ 0, Y ≥ 0

The objective functions put together

4X + 3Y ≥ 48

2X + 6Y ≥ 42

2X + Y ≥ 20

X ≥ 0

Y ≥ 0

So, plotting the graph to obtain optimum points.

The optimal points obtained from the graph include

(6, 8), (9,4), (0,20), (21, 0)

Putting all the optimal points into the first 3 equations

(6,8) gives 48, 50 and 20 units just like in the objective functions

(9,4) gives 48, 42 and 22

(0, 20) gives 60, 120, and 20

(21, 0) gives 84, 42, and 42.

It is evident that (9,4) is the most optimal solution.