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Systems and Nutrition based on aglebra equations

Gather nutritional information for two different foods to devise a meal plan that provides optimal nutritional value.

To maintain health, people need to consume a recommended daily amount of many different vitamins and minerals, as established by government health agencies. To achieve those levels, people have their choice of many types of food to consume. For the sake of simplicity, assume that you are a poor college student who can only afford two items: broccoli and corn. Also assume that the only nutrients you absolutely need to maintain proper health are calcium and iron.

Gather from credible sources the following data:

-The amount of calcium and iron (in milligrams) found in one 10 oz package of broccoli
-The amount of calcium and iron (in milligrams) found in a 1½ cup of frozen corn
-The recommended daily values (in milligrams) of calcium and iron

After collecting the data, it should be clear that one 10 oz package of broccoli and one 1½ cup of frozen corn will not allow a person to achieve the minimum levels of calcium and iron for one day. To that end, you need to determine how much of each food item you need in order reach those levels. Construct and solve a linear system of two equations that would make that determination.

Naturally, people need more than two minerals to survive. Likewise, people eat more than just broccoli and corn. Find four more essential vitamins and minerals and their recommended daily values (in milligrams). Likewise, find four more foods that contain each of the six vitamins and minerals you have listed. Determine that the amount of each nutrient found in one serving of each food.

Using that information, construct but do not solve a linear system of equations that determines how much of each item a person would need to consume to achieve the recommended daily values of each nutrient.

Solution Preview

Located on attachment worksheet 2
b=10oz of broccoli
c=half-cup of corn

calcium equation
159b + 3.3c = 1000
b=10oz of broccoli
c=half-cup of corn

calcium equation
159b + 3.3c = 1000

Iron equation
2.3b + 0.3c = 18

modify the equations
1590b+33c=10000 multiply by 10
23b+3c=180 multiply by 10

modify to eliminate c
3(1590b)+99c=3(10000)
33(23b)+99c=33(180)

Subtract the second equation from the first equation and solve for ...

Solution Summary

An algebraic approach to gather nutritional information for two different foods to devise a meal plan that provides optimal nutritional value.

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