Answer:
a. The number of pounds of the meat required is 3 pounds and the number of pounds of cheese required is 2 pounds.
b. $ 16.7
Step-by-step explanation:
a. How many pounds of each are needed in order to minimize the cost and still meet the minimum requirements?
Let c represent the carbohydrate units and p the protein units.
For the meat portion M, we have 2 units of carbohydrates and 2 units of protein per pound. So, M = 2c + 2p
For the cheese portion K, we have 3 units of carbohydrates and 1 units of protein per pound. So, K = 3c + p.
Let x be the number of pounds of meat required and y be the number of cheese pounds required. The total number of pounds required is T
So, we have xM + yK = x(2c + 2p) + y(3c + p)
= 2xc + 2xp + 3yc + yp
= 2xc + 3yc + 2xp + yp
= (2x + 3y)c + (2x + y)p
Since the required number of units, R is 12 units of carbohydrates and 8 units of protein, we have R = 12c + 8p
Since T = R, we have
(2x + 3y)c + (2x + y)p = 12c + 8p
Equating coefficients, we have
2x + 3y = 12 (1) and 2x + y = 8 (2)
Subtracting (2) from (1), we have
2x + 3y = 12 (1)
-
2x + y = 8 (2)
2y = 4
y = 4/2
y = 2
Substituting y = 2 into (2), we have
2x + y = 8
2x + 2 = 8
2x = 8 - 2
2x = 6
x = 6/2
x = 3
Since x = 3 and y = 2
The number of pounds of the meat required is 3 pounds and the number of pounds of cheese required is 2 pounds.
What is the minimum cost?
Since meat costs $3.70 per pound and the cheese costs $2.60 per pound and we have 3 pounds of meat and 2 pounds of cheese, the total cost of meat is C = $3.70/pound × 3 pounds = $ 11.1.
The total cost of cheese is C' = $2.60/pound × 2 pounds = $ 5.2.
So, the minimum cost C" = C + C' = $ 11.1 + $ 5.2 = $ 16.7
