Question

A new animal food, which a company has just developed, consists of two elements. Each pound of element H cost $1 and contains 8 units of carbohydrates, 4 units of proteins and 2 units of fats. Each pound of element I cost $0.80 and contain 3 units of carbohydrates, 5 units of proteins and 6 units of fats. There is no weight specification, but each bag of the food must contain at least 24 units of carbohydrates, 20 units of protein and 12 units of fat. For each element, determine the number of pounds per bag that will yield the most economical mix. What will be the cost of this amount per bag?

Answer 1

Let, in a bag there will be h pounds of H element, and i pounds of I element.

Our optimizing variable is cost.

1 dollar is the cost of 1 pound of element H. 0.8 dollar is the cost of 1 pound of element I.

so, h dollar will be the cost of h pound of element H, 0.8 i dollar will be the cost of 1 pound of element I.

total cost, z = h + 0.8i.

Limiting factors are -

each bag need to contain atleast 24 units of carbohydrate,

each pound of H element needs 8 units of carbohydrate, each pound of I element needs 3 units of carbohydrate.

h pound of element H needs 8h units of carbohydrate, i pounds of element I needs 3i units of carbohydrate.

overall balance on cabohydrate -

8h + 3i >= 24

each bag need to contain atleast 20 units of protein,

each pound of H element needs 4 units of protein, each pound of I element needs 5 units of protein.

h pound of element H needs 4h units of protein, i pounds of element I needs 5i units of protein.

overall balance on protein-

4h + 5i >= 20

each bag need to contain atleast 12 units of fat,

each pound of H element needs 2 units of fat, each pound of I element needs 6 units of fat.

h pound of element H needs 2h units of fat, i pounds of element I needs 6i units of fat.

overall balance on fat-

2h + 6i >= 12

so, our optimizing variable = z = h + 0.8i

limiting constraint equations -

8h + 3i >= 24

4h + 5i >= 20

2h + 6i >= 12

We are solving this problem graphically. For that we first plot this 3 equations considering only equality sign.

Now, considering the inequality our area of investigate is from the lines towards more positive value (opposite of the origin).

Now, for cost minimization, we have to check cost on the lower boundary points of the area i.e point A,B,C,D.

Co ordinates of these points -

A - intersection of 8h+3i = 24 and h = 0.

so, coordinates are - 0 units of H, 24/3 = 8 units of I. We are neglecting this point because we need to feed both elements.

B - intersection of 8h + 3i = 24 and 4h + 5i = 20

or, 8h + 3i - 24 = 2*(4h + 5i - 20)

or, 40-24 = 7i, or i = 16/7, h = (24 - 3*16/7) / 8 = 15/7

z value on point B = 15/7 * 1 + 16/7 * 0.8 = 3.97143

C - intersection of 4h + 5i = 20 and 2h + 6i = 12

or, 4h + 5i - 20 = 2*(2h + 6i - 12)

or, 24 -20 = 7i, or i = 4/7, h = (20 - 5*4/7) / 4 = 30/7

z value on point C = 30/7 + 4/7*0.8 = 4.74286

D - intersection of 2h + 6i = 12 and i = 0

so, coordinates are - 0 units of I, 12/2 = 6 units of H. We are neglecting this point because we need to feed both elements.

We can see cost on point B is minimum.

In a bag 15/7 = 2.14285715 pounds of element H and 16/7 = 2.2857143 pounds of element I loading would provide most economic yield.

In this case cost of each bag would be 3.97143 dollars. Cost of H in each bag = 15/7 = 2.142857 dollar. Cost of I in each bag = 16/7*0.8 = 1.82857 dollar.