Question 1142242: It takes 10 units of carbohydrates and 4 units of protein to satisfy Jacob's minimum weekly requirements. The meat contains 2 units of carbohydrates and 2 units of protein per pound. The cheese contains 3 units of carbohydrates and 1 unit of protein per pound. The meat costs $3.20 per pound and the cheese costs $4.70 per pound. How many pounds of each are needed to fulfill the minimum requirements at minimum cost? What is Jacob's minimum cost?
Answer by ikleyn(52794)   (Show Source):
You can put this solution on YOUR website! .
From the condition, you have these two equations in two unknowns
2M + 3C = 10 units of carbohydrates (1)
2M + 1C = 4 units of protein (2)
where M is the amount of Meat in pounds and C is the amount of Cheese in pounds.
To solve the system, use the Elimination method. For it, subtract eq(2) from eq(1). You will get
3C - 1C = 10 - 4
2C = 6
C = 6/2 = 3 pounds of cheese.
Then from eq(2) you get
2M + 3 = 4,
which implies
2M = 4 - 1 = 1 =====> M = 0.5 pounds of meat.
So, Jacob's minimum need is 0.5 of a pound of meat and 3 pounds of cheese weekly, ACCORDING TO THE SOLUTION. ANSWER 1
The cost of this food is
3.20*0.5 + 4.70*3 dollars = 15.70 dollars. ANSWER 2
Solved.
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Again: the solution is in two steps.
First step is to find the amounts of ingredients required.
The second step is to calculate the total cost of the food.
Although the word "mimimum" occurs many times in the text of the problem, this problem is not
a "minimization type" of problems. It is on "solving systems of linear equations".
It is important to distinct between these two different types of problems.
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If you want to see other similar solved problems, look into the lessons
- Counting calories and grams of fat in combined food
- Solving word problems by reducing to systems of linear equations in three unknowns, Problem 3
in this site.
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