Problem 60 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A city's main well was recently found to be contaminated with trichloroethylene, a cancer-causing chemical, as a result of an abandoned chemical dump leaching chemicals into the water. A proposal submitted to the city's council members indicates that the cost, measured in millions of dollars, of removing \(x \%\) of the toxic pollutant is given by $$ C(x)=\frac{0.5 x}{100-x} $$ Find \(C^{\prime}(80), C^{\prime}(90), C^{\prime}(95)\), and \(C^{\prime}(99)\). What does your result tell you about the cost of removing all of the pollutant?

Step-by-Step Solution

Answer

\(C'(80) = \frac{1}{8}\), \(C'(90) = \frac{1}{2}\), \(C'(95) = 2\), and \(C'(99) = 50\). The marginal cost of removing the pollutant increases as we try to remove more of it. This indicates that removing all of the pollutant will be extremely expensive due to the high costs associated with removing the last few percentages of the pollutant.

Step 1: Find the derivative of \(C(x)\).

To find the derivative of the given function, we can first notice that \(C(x)\) is a rational function. We can use the quotient rule to find the derivative: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] In this case, \(u(x) = 0.5x\) and \(v(x) = 100 - x\). First, let's find the derivatives of \(u(x)\) and \(v(x)\): \[ \frac{du}{dx} = 0.5 \] \[ \frac{dv}{dx} = -1 \] Now we can apply the quotient rule: \[ C'(x) = \frac{(100-x)(0.5) - (0.5x)(-1)}{(100-x)^{2}}. \] Let's simplify the expression: \[ C'(x) = \frac{50 - 0.5x + 0.5x}{(100-x)^{2}}. \] So, \[ C'(x) = \frac{50}{(100-x)^{2}}. \]

Step 2: Compute \(C'(80)\), \(C'(90)\), \(C'(95)\), and \(C'(99)\)

Now that we have the derivative, we can find its value at the required points: \[ C'(80) = \frac{50}{(100-80)^{2}} = \frac{50}{20^{2}} = \frac{1}{8}\] \[ C'(90) = \frac{50}{(100-90)^{2}} = \frac{50}{10^{2}} = \frac{1}{2}\] \[ C'(95) = \frac{50}{(100-95)^{2}} = \frac{50}{5^{2}} = 2\] \[ C'(99) = \frac{50}{(100-99)^{2}} = \frac{50}{1^{2}} = 50\]

Step 3: Interpret the results

The derivative \(C'(x)\) represents the marginal cost of removing the pollutant, that is, the additional cost of removing one more percent of the pollutant. When we evaluate the derivative at the given points, we find that as we get closer to removing all of the pollutant (i.e., higher values of \(x\)), the marginal cost increases. This means that the cost of removing the last few percentages of the pollutant is much higher than the cost of removing the first few percentages. In other words, it becomes more difficult and expensive to remove the pollutant as the percentage of remaining pollutant decreases, and if we would like to remove 100% of the pollutant, the cost could be extremely high.

Key Concepts

Rational Functions

A rational function is a fraction where both the numerator and the denominator are polynomials. It's a concept that often appears in various fields of mathematics, including calculus and algebra. Rational functions are important because they can model many types of real-world situations, such as population growth and cost analysis.

In our exercise, the function \(C(x) = \frac{0.5x}{100-x}\) represents the cost of removing \(x\%\) of the toxic pollutant. Here, the numerator is a simple polynomial, \(0.5x\), and the denominator, \(100-x\), is another simple polynomial. The behavior of this function is crucial in understanding how costs escalate as more pollutants are removed.

• Numerator: Reflects the direct proportionality of costs to the percentage \(x\) of pollutant removed.
• Denominator: Shows that as the percentage \(x\) approaches 100, the value of the function increases significantly. This indicates higher costs.

This escalation can be explored further using derivatives, helping us understand how costs change as more pollutants are removed.

Marginal Cost

Marginal cost is an economic concept used to express how much additional cost is incurred when producing one more unit of a good or service. In environmental mathematics, it also measures the additional cost of removing an additional percentage of a pollutant.

In the context of our exercise, the derivative \(C'(x)\) gives us the marginal cost function. When evaluated at various points, it reveals the increasing difficulty and expense of making the city's water free of trichloroethylene.

• For example, at 80\% removed, the marginal cost is \(\frac{1}{8}\) million dollars.
• As we increase to 90\% and 95\%, the marginal costs rise to \(\frac{1}{2}\) and 2 million dollars, respectively.
• At 99\%, the marginal cost surges to 50 million dollars, showing a sharp increase as we approach full removal.

This demonstrates that in environmental cleanups, the last few percentages of pollutant removal are often the most costly. Understanding these costs is crucial for city planners and policy-makers when allocating resources for environmental rehabilitation.

Quotient Rule

The quotient rule is a key technique in calculus for finding the derivative of a function that is the ratio of two differentiable functions. It is particularly useful when dealing with rational functions like \(C(x) = \frac{0.5x}{100-x}\).

The quotient rule is expressed as:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]where \(u(x)\) is the numerator and \(v(x)\) is the denominator.

• In our case, \(u(x) = 0.5x\) and \(v(x) = 100 - x\).
• The derivative of the numerator \(\frac{du}{dx} = 0.5\) and the derivative of the denominator \(\frac{dv}{dx} = -1\).

Applying the rule, we simplify to determine \(C'(x) = \frac{50}{(100-x)^2}\), showing how cost relates to pollutant removal percentage. This technique provides a disciplined way to analyze changes and trends in mathematical models.

Environmental Mathematics

Environmental mathematics applies mathematical principles to understand and solve environmental issues. In our scenario, it involves analyzing the cost model for removing pollutants from water—a task critical to public health and safety.

Understanding this concept can help in:

• Modelling environmental problems: Math can quantify the effect of pollutants or the effectiveness of cleanup strategies.
• Predicting costs: Your results show how cleanup expenses escalate dramatically near total removal, implying that decision-makers must plan carefully.
• Finding solutions: Mathematics helps identify cost-effective strategies for managing or mitigating pollution.

For city councils and environmental bodies, such mathematical insights are essential in forming policies that balance environmental safety with economic practicality. As pollutants become harder to remove, strategic planning helps mitigate the financial impact while striving for cleaner environments.