Problem 23 - Algebra for College Students

Question

The cost for burning a 75 -watt bulb is given by the function \(c(h)=0.0045 h\), where \(h\) represents the number of hours that the bulb burns. (a) How much does it cost to burn a 75-watt bulb for 3 hours per night for a 31 -day month? Express your answer to the nearest cent. \(\$ .42\) (b) Graph the function \(c(h)=0.0045 h\). (c) Use the graph in part (b) to approximate the cost of burning a 75 -watt bulb for 225 hours. (d) Use \(c(h)=0.0045 h\) to find the exact cost, to the nearest cent, of burning a 75 -watt bulb for 225 hours. \(\$ 1.01\)

Step-by-Step Solution

Answer

(a) \$0.42. (b) Linear graph. (c) Approx. \$1.01. (d) \$1.01.

Step 1: Calculate Total Hours Spent

First, determine the total number of hours the bulb is used in a 31-day month. If it burns for 3 hours each night, multiply the number of hours per night by the number of days: \( h = 3 \times 31 = 93 \) hours.

Step 2: Calculate Cost for Part (a)

Use the cost function \( c(h) = 0.0045h \) to calculate the cost of running the bulb for 93 hours. Substitute \( h = 93 \) into the function: \( c(93) = 0.0045 \times 93 = 0.4185 \). Rounding to the nearest cent, the cost is \( \$0.42 \).

Step 3: Graph the Function

To graph the function \( c(h) = 0.0045h \), begin by noting that it is a linear function with a slope of 0.0045 and y-intercept of 0. Start with the point (0,0) and use the slope to find other points, such as (100, 0.45) and (200, 0.9), to draw a straight line through these points.

Step 4: Approximate Using Graph

On the graph from part (b), find the point where \( h = 225 \). Use the plotted line to approximately determine \( c(225) \), which should be close to \( 1.0125 \). As a rough estimate, this corresponds to a cost slightly above \( \$1.00 \).

Step 5: Calculate Exact Cost for Part (d)

Use the equation \( c(h) = 0.0045h \) to determine the exact cost when \( h = 225 \). Substitute \( h = 225 \) into the function: \( c(225) = 0.0045 \times 225 = 1.0125 \). Rounding to the nearest cent, the exact cost is \( \$1.01 \).

Key Concepts

Cost Calculation

In our problem, we are interested in calculating the cost of burning a 75-watt bulb. The total cost is determined by the formula \( c(h) = 0.0045h \), where \( h \) represents the number of hours in use. This formula tells us how much it costs to run the bulb for each hour.

To find total cost over a period, multiply the number of hours the bulb is on by the cost per hour rate. For instance, if the bulb burns 3 hours every day for a month of 31 days, the total hours \( h \) would be:

• \( h = 3 \times 31 = 93 \text{ hours} \)

Using the cost formula, plug in the total hours:

• \( c(93) = 0.0045 \times 93 = 0.4185 \)

Rounding this to the nearest cent, gives us a cost of \( \$0.42 \).

To relate this to a practical scenario, consider how power companies often charge you based on how many hours you use certain utilities. This problem is a simplified version of how to calculate such costs based on usage.

Graphing Linear Equations

Graphing the linear function \( c(h) = 0.0045h \) helps visualize how the cost increases with more hours of bulb usage. Observing graphs of functions can make understanding relationships between variables more manageable.

This equation is a straightforward linear equation with:

• a y-intercept of 0, meaning it passes through the origin (0,0), indicating no cost if the light is not used at all.
• a slope of 0.0045, which signifies that for each additional hour of use, the cost increases by \$0.0045.

On graph paper, begin the line at the origin. Use the slope to determine other points. For example, at 100 hours, the cost is \( 0.0045 \times 100 = 0.45 \), plotted at the point (100, 0.45). Similarly, at 200 hours, the cost is \( 0.9 \).

By establishing these points, draw a straight line which represents the function. This visual representation clarifies how costs increase linearly as usage rises and is an intuitive way to approximate or cross-check calculations.

Function Evaluation

Function evaluation is at the heart of many mathematical problems, allowing us to input values and assess specific results. For our schedule, with \( c(h) = 0.0045h \), evaluating this function for different \( h \) values gives us distinct costs.

For part (d), when evaluating at \( h = 225 \), we directly substitute 225 into the function:

• \( c(225) = 0.0045 \times 225 = 1.0125 \)

Rounding, the accurate cost is

• \( \$1.01 \)

This procedure shows how essential it is to accurately plug values into the function and compute them to get the precise output.

Function evaluation is not only useful in mathematical contexts but is pivotal in fields like economic forecasts or any area relying on data-driven decision-making. When you need an exact outcome, simply plug the necessary variable into the formula, and find your result. This neat package of input-output helps reduce errors and guides predictions accurately.