Problem 34 - Applied Calculus

Question

The amount of waste a company produces, \(W,\) in tons per week, is approximated by \(W=3.75 e^{-0.008 t},\) where \(t\) is in weeks since January \(1,2005 .\) Waste removal for the company costs \(\$ 15 /\) ton. How much did the company pay for waste removal during the year \(2005 ?\)

Step-by-Step Solution

Answer

The company paid approximately $2384.55 for waste removal in 2005.

Step 1: Identify the Function Components

The given formula for waste produced is \( W = 3.75 e^{-0.008t} \). Here, \( W \) represents the waste in tons, and \( t \) is the time in weeks since January 1, 2005. We need to find the waste produced throughout the year 2005.

Step 2: Determine Time Interval for 2005

The year 2005 consists of 52 weeks. Therefore, we calculate the total waste produced from \( t = 0 \) to \( t = 52 \).

Step 3: Set Up the Integral for Waste Produced

To find the total waste produced from \( t = 0 \) to \( t = 52 \), we'll integrate the function \( W(t) = 3.75 e^{-0.008t} \) with respect to \( t \) over this interval. The integral is given by: \[\int_0^{52} 3.75 e^{-0.008t} \, dt\]

Step 4: Calculate the Integral

To solve the integral, we perform the integration: \[= \left[-\frac{3.75}{0.008} e^{-0.008t}\right]_0^{52}\]Plug in the limits of integration:\[= \left(-\frac{3.75}{0.008} e^{-0.008 \times 52}\right) - \left(-\frac{3.75}{0.008} e^{-0.008 \times 0}\right)\]

Step 5: Simplify and Evaluate the Expression

Calculate the expression: \[= \left(-468.75 e^{-0.416}\right) + 468.75\]Compute the numerical values:\[\approx -468.75 \times 0.65924 + 468.75 \approx 158.97 \text{ tons}\]

Step 6: Calculate the Total Cost for Waste Removal

The cost of waste removal is \( \\(15 \) per ton. Multiply the total tons by this cost:\[158.97 \times 15 = 2384.55\]Therefore, the total cost is approximately \( \\)2384.55 \).

Key Concepts

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. In our example, the waste produced by the company decreases over time, modeled by the formula: \[W = 3.75 e^{-0.008t}\] Here:

• \(W\) is the waste in tons.
• \(t\) is time in weeks.
• The coefficient of \(t\), \(-0.008\), represents the decay rate.

Exponential decay is common in various natural and financial processes, where values reduce steadily over time. The base of the exponential function \(e\) (approximately 2.718) is crucial as it allows calculations of continuations like this model.

Cost Analysis

Cost analysis in this context refers to determining the monetary expenditure required for waste removal. Once the amount of waste produced over a given period is calculated, multiplying this total by the cost per unit gives us the overall cost. In our exercise:

• The company produced approximately 158.97 tons of waste in 2005.
• Cost of removing waste is \( \\(15\) per ton.

Calculating the payment involves a simple multiplication:\[158.97 \times 15 = 2384.55\]This results in an expenditure of approximately \( \\)2384.55\) for waste management. Cost analysis helps businesses monitor and control expenses effectively, ensuring financial efficiency.

Definite Integral

The definite integral is a fundamental concept in calculus used to calculate the accumulation of quantities, such as area under a curve. In practical scenarios, like our case of waste calculation, integration helps in finding the total waste over a given interval, i.e., from week 0 to week 52.For the waste function:\[\int_0^{52} 3.75 e^{-0.008t} \, dt\]This operation gives the total waste produced during 2005. To solve the integral:

• The integrated function results in \[-\frac{3.75}{0.008} e^{-0.008t}\]
• Evaluate this at the boundaries \(t=0\) and \(t=52\).

When evaluated, the integral computes the reduction and accumulation over the specified time. Understanding definite integrals helps in predicting cumulative outcomes in real-life situations, significantly contributing to decision-making.