Problem 93 - Contemporary Precalculus

Question

The cost-benefit equation \(\frac{18 x}{100-x}=D\) relates the cost \(D\) (in thousands of dollars) needed to remove \(x\) percent of a polIutant from the emissions of a factory. Find the percent of the pollutant removed when the following amounts are spent. (a) \(\$ 50,000[\text { Here } D=50]\) (b) \(\$ 100,000\) (c) \(\$ 200,000\)

Step-by-Step Solution

Answer

Answer: (a) For $50,000, approximately 73.53% of the pollutant is removed. (b) For $100,000, approximately 84.75% of the pollutant is removed. (c) For $200,000, approximately 91.74% of the pollutant is removed.

Step 1: Rewrite the given equation

Rewrite the given equation as: \(\frac{18x}{100-x}=D\)

Step 2: Solve for x when D= 50 (a)

Given that D = 50 in this case, we need to solve the equation \(\frac{18x}{100-x}=50\) First, cross-multiply to get rid of the fraction: \(18x = 50(100-x)\). Expand the equation: \(18x = 5000 - 50x\). Now solve for x: Add \(50x\) to both sides to get \(68x = 5000\) Finally, divide by 68: \(x = \frac{5000}{68} \approx 73.53\%\) So, to spend $50,000, the factory can remove approximately 73.53\% of the pollutant.

Step 3: Solve for x when D= 100 (b)

Given that D = 100 in this case, we need to solve the equation \(\frac{18x}{100-x}=100\) First, cross-multiply to get rid of the fraction: \(18x = 100(100-x)\). Expand the equation: \(18x = 10000 - 100x\). Now solve for x: Add \(100x\) to both sides to get \(118x = 10000\) Finally, divide by 118: \(x = \frac{10000}{118} \approx 84.75\%\) So, to spend $100,000, the factory can remove approximately 84.75\% of the pollutant.

Step 4: Solve for x when D= 200 (c)

Given that D = 200 in this case, we need to solve the equation \(\frac{18x}{100-x}=200\) First, cross-multiply to get rid of the fraction: \(18x = 200(100-x)\). Expand the equation: \(18x = 20000 - 200x\). Now solve for x: Add \(200x\) to both sides to get \(218x = 20000\) Finally, divide by 218: \(x = \frac{20000}{218} \approx 91.74\%\) So, to spend $200,000, the factory can remove approximately 91.74\% of the pollutant. In conclusion, for the amounts spent, the factory can remove the following percentage of pollutants: (a) \(50,000: \approx 73.53\%\) (b) \(100,000: \approx 84.75\%\) (c) \(200,000: \approx 91.74\%\)

Key Concepts

Cost-Benefit Equation

Understanding a cost-benefit equation is crucial in assessing the financial side of pollution reduction. The formula provided is: \[\frac{18x}{100-x}=D\]Here, \(D\) represents the cost in thousands of dollars to remove \(x\) percent of a pollutant. This kind of equation helps us determine how much it will cost to remove different percentages of pollutants from emissions.
In such equations:

• The numerator \(18x\) shows the relationship between cost and the percentage of pollutant we wish to remove.
• The denominator \(100-x\) represents the reduction from a baseline of 100% pollutant presence.

These equations are practical as they give a clear picture of financial investments relative to environmental improvements, making them suitable for real-world applications like factory emissions management.

Pollutant Removal Calculation

Calculating how much of a pollutant can be removed by spending a certain amount involves solving the cost-benefit equation for \(x\), the percentage of pollutant reduction. Let's break it down using the steps:To solve for \(x\):1. Start with the equation: \(\frac{18x}{100-x} = D\)2. Cross-multiply to eliminate the fraction: \(18x = D(100-x)\)
3. Distribute \(D\), and align terms to bring all \(x\) terms to one side.This rearranging transforms the equation into a simpler format that can be easily solved. The target is to express \(x\) as a simple division once you've consolidated terms. By breaking it down, you find out the pollutant reduction at different spending levels, e.g., \(73.53\%\) at \(50,000, \(84.75\%\) at \)100,000, and \(91.74\%\) at $200,000.
This clear calculation is helpful for businesses to budget their environmental improvement strategies effectively.

Solving Linear Equations

Solving linear equations often involves isolating the variable you're interested in, which in this context is \(x\), the percentage of pollutant removed. To solve equations like:\[18x = D(100-x)\]Follow these steps:

• First, distribute \(D\) to achieve \(18x = D100 - Dx\).
• Next, bring all terms involving \(x\) to one side, for instance, \(18x + Dx = D100\).
• Factor out \(x\) from the terms on one side to simplify: \(x(18 + D) = D100\).
• Lastly, isolate \(x\) by dividing by the coefficient: \(x = \frac{D100}{18 + D}\).

These consistent steps allow you to solve for \(x\) in various scenarios provided by different \(D\) values, clarifying the relationship between costs and pollutant removal. Through practice, solving such equations becomes straightforward, enhancing one's ability to tackle more complex algebraic challenges similarly.