Problem 31 - Intermediate Algebra

Question

Use linear functions. 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. (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.

Step-by-Step Solution

Answer

(a) $0.42; (c) Approximate from graph; (d) $1.01.

Step 1: Calculate Daily Usage Cost

First, determine how many hours the bulb burns in a day. Since it burns for 3 hours each night, in one day, the bulb burns for 3 hours. To find the cost for one night, substitute 3 for \(h\) in the function \(c(h)=0.0045h\).\[ c(3) = 0.0045 \times 3 = 0.0135 \]Thus, it costs $0.0135 each night to burn the bulb for 3 hours.

Step 2: Calculate Monthly Usage Cost

To find the total cost for 31 days, multiply the daily cost by the number of days in the month: \[ \text{Total monthly cost} = 0.0135 \times 31 = 0.4185 \]Rounding to the nearest cent, the total cost is approximately $0.42.

Step 3: Graph the Function

Begin by identifying key characteristics of the function \(c(h) = 0.0045h\). This is a linear function with a slope of 0.0045 and a y-intercept of 0. It passes through points such as (0,0), (100,0.45), and (200,0.9). Plot these points on a graph and draw the line through them.

Step 4: Approximate Cost Using Graph

Locate 225 hours on the x-axis of the graph, then draw a line vertically up to the function \(c(h)=0.0045h\). From the intersection point, draw a horizontal line to the y-axis to find the approximate cost. Based on the graph, the cost for 225 hours should be close to the actual calculated value in Step 5.

Step 5: Calculate Exact Cost Using Function

Substitute 225 for \(h\) in the function \(c(h)=0.0045h\) to find the exact cost:\[ c(225) = 0.0045 \times 225 = 1.0125 \]Rounding to the nearest cent, the cost is $1.01.

Key Concepts

Cost Calculation

Understanding how costs are calculated using linear functions is an essential skill. In the context of a 75-watt bulb burning for several hours, we have a linear cost function given by \[ c(h) = 0.0045 h \]Here, - \( c(h) \) is the cost in dollars,- \( h \) is the number of hours the bulb burns.For example, to find the cost for a single night when the bulb burns for 3 hours, substitute 3 into \( h \):\[ c(3) = 0.0045 \times 3 = 0.0135 \]Thus, it costs \(0.0135 each night for 3 hours. To project this over a 31-day month, multiply this nightly cost by 31:\[ 0.0135 \times 31 = 0.4185 \]Rounded to the nearest cent, this monthly cost becomes \)0.42. This step-by-step approach helps in accurate cost estimation over different time frames.

Graphing Linear Equations

Visualizing linear equations through graphs provides a compelling way of understanding functions. The function \( c(h) = 0.0045h \) is a linear function because it can be represented by a straight line. It has a slope of 0.0045, describing the rate at which cost increases per hour. The y-intercept is at 0, indicating that when no hours are utilized, no cost incurs.To graph this equation, identify and plot several key points:

• (0,0) - Starting point in the graph
• (100,0.45) - Represents 100 hours at \(0.45
• (200,0.9) - Represents 200 hours at \)0.9

Connect these points with a straight line, illustrating the constant increase in cost with time. This kind of graph allows for easy estimation of the cost for any number of hours by viewing the corresponding y-axis value of any x-coordinate.

Application of Functions

Linear functions have practical applications, such as estimating costs efficiently with given variables. With the bulb's cost function, \( c(h) = 0.0045h \), it can be easy to tailor calculations based on real-world usage scenarios.By using graphical methods, you can locally determine an approximate cost. For example, at 225 hours, position a line from the 225-mark on the x-axis vertically to the linear cost line. Then, horizontally trace from this intersection point toward the y-axis to approximate cost based on the graph. This visualization gives a hands-on sense of cost behavior over time, even without exact calculations.For exact precision beyond approximation, plug the hour value directly into the function as follows:\[ c(225) = 0.0045 \times 225 = 1.0125 \]Rounded, the cost for 225 hours is precisely $1.01.

Algebra

Algebraic techniques form the basis for solving cost functions like \( c(h) = 0.0045 h \). This expression showcases how algebra can be used to compute outcomes by substituting values such as hours (\( h \)) directly into the equation.These functions are especially useful for creating straightforward algebraic models of cost, making them easily adaptable for various situations:

• Allows for consistent cost forecasting over time.
• Provides a simple way to visualize and analyze behavioral trends in expenses.
• Offers accessibility in calculating exact or approximate values efficiently.

Through simple substitution and operations, anyone can apply algebra to model, calculate, or estimate various scenarios, enhancing understanding and decision-making abilities related to costs and other linear relationships.