Problem 10 - Intermediate Algebra

Question

One ounce of each of three foods has the vitamin and mineral content shown in the table. How many ounces of each must be used to provide exactly 22 milligrams (mg) of niacin, 12 mg of zinc, and 20 mg of vitamin C? Milligrams per ounce in each food type $$ \begin{array}{|c|c|c|c|} \hline \text { Food } & \text { Niacin } & \text { Zinc } & \text { Vitamin C } \\ \hline \mathrm{A} & 1 \mathrm{mg} & 1 \mathrm{mg} & 2 \mathrm{mg} \\ \mathrm{B} & 2 \mathrm{mg} & 1 \mathrm{mg} & 1 \mathrm{mg} \\ \mathrm{C} & 2 \mathrm{mg} & 1 \mathrm{mg} & 2 \mathrm{mg} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

2 ounces of Food A, 4 ounces of Food B, 6 ounces of Food C.

Step 1: Define the Variables

Let \( x \) be the ounces of Food A, \( y \) be the ounces of Food B, and \( z \) be the ounces of Food C.

Step 2: Formulate Equations for Each Nutrient

Based on the table, create equations for each nutrient:1. **Niacin:** \( x + 2y + 2z = 22 \)2. **Zinc:** \( x + y + z = 12 \)3. **Vitamin C:** \( 2x + y + 2z = 20 \)

Step 3: Use Substitution or Elimination

Use substitution or elimination to solve the system of equations. First, use equations 2 and 3 to eliminate one variable.

Step 4: Subtract Equations

Subtract Equation 2 from Equation 3 to eliminate \( y \):\((2x + y + 2z) - (x + y + z) = 20 - 12\)Simplifying, we get: \(x + z = 8\) (Equation 4).

Step 5: Substitute Equation 4

Substitute \( z = 8 - x \) from Equation 4 into the other equations to find values for \( x \) and \( y \).

Step 6: Solve for \( x \) and \( y \)

Substitute \( z = 8 - x \) into Equation 2, we get:\(x + y + (8 - x) = 12\)which simplifies to \(y + 8 = 12\) so \(y = 4\).

Step 7: Solve for \( z \) and \( x \)

With \( y = 4 \), substitute into Equation 2 to find \( x \):\(x + 4 + (8-x) = 12\), implying \(8 = 8\), so consistent.Now substitute \( y = 4 \) & \( z = 8 - x \) into Equation 1:\(x + 2(4) + 2(8-x) = 22\), simplifying:\(x + 8 + 16 - 2x = 22\), so \(-x + 24 = 22\).Solving gives: \(x = 2\), therefore \(z = 8 - 2 = 6\).

Step 8: Solution

Hence, the ounces needed are \(x = 2\) ounces of Food A, \(y = 4\) ounces of Food B, and \(z = 6\) ounces of Food C.

Key Concepts

Linear Equations

Understanding linear equations is essential when solving systems of equations like the one in this exercise. A linear equation is an equation where every term is either a constant or the product of a constant and a single variable. The general form of a linear equation is \( ax + by + cz = d \), where \( a \), \( b \), \( c \), and \( d \) are constants, and \( x \), \( y \), and \( z \) are variables.
In our exercise, each nutrient requirement is expressed as a linear equation. For instance, the equation for niacin is \( x + 2y + 2z = 22 \). Here, \( x \), \( y \), and \( z \) represent quantities of food types A, B, and C, respectively, each contributing to the total niacin requirement. The goal is to find the right combination of \( x \), \( y \), and \( z \) that satisfies all three linear equations simultaneously.
The power of linear equations lies in their ability to represent complex problems in a simple yet effective format. Whether it's designing a diet plan or optimizing resource allocation, linear equations are a versatile tool used in various applications.

Substitution Method

The substitution method is a useful approach for solving systems of linear equations. This method involves solving one of the equations for one variable and substituting that expression into the other equations. In the exercise, after simplifying the equations, we derived a new equation \( x + z = 8 \) (Equation 4).
Using this equation, you can express one variable in terms of the others, such as \( z = 8 - x \). This expression is then substituted back into the other equations to reduce the number of variables and simplify solving for the remaining unknowns.
Advantages of using the substitution method include simplifying calculations when dealing with clear expressions and small systems of equations. However, it can become complex if the expressions are too complicated, which is why it's often used in combination with other methods like elimination.

Elimination Method

The elimination method, also known as the addition-subtraction method, is another technique to solve systems of linear equations. This method involves aligning equations and adding or subtracting them to eliminate one of the variables.
In our problem, combining and subtracting the relevant equations helped us eliminate a variable early on, simplifying the system. Here, subtracting Equation 2 from Equation 3 resulted in eliminating \( y \), giving us \( x + z = 8 \).
This method is particularly useful for dealing with equations involving a lot of variables, or when the coefficients of one variable are easily canceled out. It can be faster than substitution, especially for larger systems, and can also help highlight dependencies between the equations. By simplifying the equations at an early stage, it often provides a clearer path to finding the solution.

Nutrient Optimization

Nutrient optimization, in the context of this exercise, refers to finding the right mix of foods to meet specific dietary needs. We have to ensure that each food contributes the necessary amount of nutrients, such as niacin, zinc, and vitamin C, while sticking to the exact requirements given (22 mg of niacin, 12 mg of zinc, and 20 mg of vitamin C).
By setting up a system of equations, each representing a nutrient's constraints from the table of food values, we're employing mathematical optimization to solve dietary challenges. The linear equations ensure that the sum of nutrients from each food type equals the required total.
This type of problem is common in diet planning and nutritional science. When resources are limited, optimization helps in making the most effective decisions to balance different nutrition goals, ensuring health benefits while being cost-effective.