Problem 29 - Algebra Form and Function

Question

A student takes a part-time job to earn $$\$ 2400$$ for summer travel. The number of hours, \(h,\) the student has to work is inversely proportional to the wage, \(w\), in dollars per hour, and is given by $$ h=\frac{2400}{w} $$ (a) How many hours does the student have to work if the job pays $$\$ 4$$ an hour? What if it pays $$\$ 10$$ an hour? (b) How do the number of hours change as the wage goes up from $$\$ 4$$ an hour to $$\$ 10$$ an hour? Explain your answer in algebraic and practical terms. (c) Is the wage, \(w\), inversely proportional to the number hours, \(h\) ? Express \(w\) as a function of \(h\).

Step-by-Step Solution

Answer

Answer: As the wage rate increases from 4 to 10 per hour, the number of hours needed to earn 2400 decreases.

Step 1: (Calculate hours at \(4 and \)10 per hour)

We are given the formula, \(h = \frac{2400}{w}\). To find \(h\) at \(w = 4\), let's plug in the value of \(w\) in the formula. $$ h = \frac{2400}{4} $$ Similarly, to find \(h\) at \(w = 10\), plug the value of \(w\) into the formula. $$ h = \frac{2400}{10} $$

Step 2: (Evaluate the expressions for \(h\))

Now let's evaluate the expressions for \(h\). $$ h = \frac{2400}{4} = 600 \\ h = \frac{2400}{10} = 240 $$ The student has to work 600 hours when the wage rate is \(4 per hour and 240 hours when the wage rate is \)10 per hour.

Step 3: (Discuss the change in hours as the wage rate changes)

As we can observe, when the wage rate increases from \(4 to \)10 per hour, the number of hours required to earn \(2400 decreases. Algebraically speaking, the relationship between \)h\( and \)w\( is given by \)h = \frac{2400}{w}\(. This implies that when \)w\( increases, \)h$ decreases, as they are inversely proportional. Practically, this means that the student needs to work fewer hours to earn the same amount when they receive a higher wage rate.

Step 4: (Express \(w\) as a function of \(h\))

To find \(w\) as a function of \(h\), let's rearrange the given equation: $$ h = \frac{2400}{w} $$ Multiplying both sides by \(w\), we have: $$ w \cdot h = 2400 $$ Now, dividing both sides of the equation by \(h\), we obtain: $$ w = \frac{2400}{h} $$ So, the wage, \(w\) is inversely proportional to the number of hours \(h\).

Key Concepts

Algebraic Function

In mathematics, an algebraic function is a function that relates input quantities or variables to output values through algebraic expressions. An example of this is the equation used in our problem: \[ h = \frac{2400}{w} \] This equation represents an inverse proportionality between the number of hours, \(h\), the student works, and the wage per hour, \(w\).
This means that when one value increases, the other decreases. The constant relationship of \(2400\) shows the total amount the student aims to earn. As the wage \(w\) increases, \(h\) decreases, meaning fewer hours are needed to reach the target of \$2400.

• The constant \(2400\) represents the fixed amount of money to be earned.
• \(h\) and \(w\) are inversely proportional; doubling \(w\) will halve \(h\).

This equation illustrates how algebraically capturing constraints helps solve real-world problems.

Wage Calculation

Understanding wage calculation through algebraic functions can simplify how one plans financially. In the exercise, the student needs to calculate the hours required to earn \\(2400. By applying the function: \[ h = \frac{2400}{w} \] for different wages, one can find the corresponding hours needed.
Here's how it works practically, using the exercise examples:

• If the wage \(w\) is \\)4: \( h = \frac{2400}{4} = 600 \) hours
• If the wage \(w\) is \\(10: \( h = \frac{2400}{10} = 240 \) hours

It's evident that as \(w\) rises, \(h\) goes down, illustrating the power of efficient earning.
For every \\) increase in \(w\), \(h\) is reduced, saving time. This understanding is not only beneficial for immediate calculations but helps in making informed decisions on job selections.

Algebraic Expression

An algebraic expression utilizes numbers, variables, and arithmetic operations to represent relationships. The expression: \[ h = \frac{2400}{w} \] is a clear example of how algebra can articulate inverse relationships.
In simpler terms, this formula tells us that the earnings goal (\$2400) remains constant as you play with the numbers \(h\) and \(w\). This mathematical communication is key to understanding real-world dynamics such as economic exchanges.

• Variables \(h\) and \(w\) allow the relationship to be flexibly analyzed.
• The expression shows how real scenarios can be converted into solvable math problems.

Mastering algebraic expressions like these equips you with the ability to model and solve various practical problems mathematically.