Problem 40 - Algebra A Combined Function

Question

The function \(f(x)=\frac{100,000 x}{100-x}\) models the cost in dollars for removing \(x\) percent of the pollutants from a bayou in which a nearby company dumped creosol. a. Find the cost of removing \(20 \%\) of the pollutants from the bayou. (Hint: Find \(f(20) .)\) b. Find the cost of removing \(60 \%\) of the pollutants and then \(80 \%\) of the pollutants. c. Find \(f(90)\), then \(f(95)\), and then \(f(99)\). What happens to the cost as \(x\) approaches \(100 \%\) ?

Step-by-Step Solution

Answer

a. \$25,000; b. \$150,000 for 60%, \$400,000 for 80%; c. Costs escalate rapidly toward infinity as x approaches 100%.

Step 1: Understanding the function

The function \( f(x) = \frac{100,000x}{100-x} \) is designed to model the cost of removing a certain percentage \( x \) of pollutants from a bayou. Our task is to calculate specific values of \( f(x) \) for given percentages \( x \) and analyze the results.

Step 2: Evaluating f(20)

To find the cost of removing \( 20\% \) of the pollutants, substitute \( x = 20 \) into the function: \[ f(20) = \frac{100,000 \times 20}{100 - 20} = \frac{2,000,000}{80} = 25,000. \] Therefore, the cost of removing \( 20\% \) of the pollutants is \$25,000.

Step 3: Evaluating f(60)

For removing \( 60\% \) of the pollutants, substitute \( x = 60 \) into the function:\[ f(60) = \frac{100,000 \times 60}{100 - 60} = \frac{6,000,000}{40} = 150,000. \]Thus, the cost of removing \( 60\% \) of the pollutants is \$150,000.

Step 4: Evaluating f(80)

To find the cost of removing \( 80\% \) of the pollutants, substitute \( x = 80 \) into the function: \[ f(80) = \frac{100,000 \times 80}{100 - 80} = \frac{8,000,000}{20} = 400,000. \]Therefore, the cost of removing \( 80\% \) of the pollutants is \$400,000.

Step 5: Evaluating f(90)

For \( x = 90 \), substitute into the function:\[ f(90) = \frac{100,000 \times 90}{100 - 90} = \frac{9,000,000}{10} = 900,000. \]The cost of removing \( 90\% \) of the pollutants is \$900,000.

Step 6: Evaluating f(95)

Substitute \( x = 95 \) into the function:\[ f(95) = \frac{100,000 \times 95}{100 - 95} = \frac{9,500,000}{5} = 1,900,000. \]Thus, the cost of removing \( 95\% \) of the pollutants is \$1,900,000.

Step 7: Evaluating f(99)

For \( x = 99 \), substitute into the function:\[ f(99) = \frac{100,000 \times 99}{100 - 99} = \frac{9,900,000}{1} = 9,900,000. \]The cost of removing \( 99\% \) of the pollutants is \$9,900,000.

Step 8: Analyzing the cost as x approaches 100%

As \( x \) approaches \( 100\% \), the denominator \( 100 - x \) approaches zero, which makes the cost \( f(x) \) tend toward infinity. This indicates that removing nearly all the pollutants becomes extremely costly.

Key Concepts

Pollution Removal

Understanding how pollution removal costs escalate is crucial. The cost function for pollution removal is given by \( f(x) = \frac{100,000x}{100-x} \), where \( x \) is the percentage of pollutants being removed. This function shows how the cost dynamically increases as more pollutants are removed from an environment like a bayou.
This model reflects real-world challenges of diminishing returns in environmental cleanup. In simpler terms, it becomes more expensive per unit of pollution as you remove higher percentages. This is because the easy-to-remove pollutants are tackled first, leaving behind the hard-to-remove ones.

• For example, removing 20% of pollutants costs less than 60% or 80%.
• As the function illustrates, initial costs like \( f(20) \) are quite manageable at $25,000.

This aligns with the fact that only significant investments can tackle near-total cleanup efforts.

Function Evaluation

The process of function evaluation involves substituting specific values into the given function to find costs at different removal percentages. You start by inserting values for \( x \) to compute \( f(x) \).
For example:

• For \( x = 20 \), you calculate \( f(20) = \frac{2,000,000}{80} = 25,000 \), revealing a cost of \(25,000.
• Similarly, for \( x = 60 \), you have \( f(60) = \frac{6,000,000}{40} = 150,000 \), indicating a larger cost of \)150,000.

By evaluating the function for different percentages, one gets a clearer picture of how costs increase progressively. Such insights help in planning and budgeting environmental projects effectively.
This approach of substituting values for evaluation is crucial to operationalizing theoretical models into practice.

Asymptotic Behavior

Asymptotic behavior in functions refers to how they behave as the input reaches certain limits. For the cost function \( f(x) = \frac{100,000x}{100-x} \), observe how costs rise dramatically as \( x \) nears 100%.
Let’s look at this behavior:

• At \( x = 90 \), cost shoots up to \(900,000, and even more significantly at \( x = 95 \) to \)1,900,000.
• Near the limit, \( f(99) = 9,900,000 \), as the denominator "100 - x" approaches zero.

This tells us that costs increase almost uncontrollably as 100% removal is approached, demonstrating an asymptotic increase. It suggests that practical limits exist when targeting 100% cleanup. This is important to understand the impracticality of total pollutant removal without incurring infinite costs.

Environmental Modeling

The cost function \( f(x) = \frac{100,000x}{100-x} \) serves as an example of environmental modeling. This means it helps simulate real-world environmental situations mathematically. This model plays a role in depicting cost scenarios for pollution mitigation in natural habitats.
Models enable policy makers and engineers to:

• Anticipate costs for different levels of pollutant removal.
• Plan budgets for environmental restoration projects.
• Determine feasible targets for pollution cleanup.

Environmental models like this are a backbone for comprehensive decision-making frameworks. They help balance costs with ecological benefits.
Overall, incorporating such models aids in predicting challenges and potential costs before implementation and fosters a deeper understanding of environmental economics.