Problem 95 - Intermediate Algebra

Question

Suppose the cost (in dollars) of removing \(p \%\) of the pollution in a river is given by the rational function \(f(p)=\frac{50,000 p}{100-p}\) where \(0 \leq p<100\) Find the cost of removing each percent of pollution. a. \(50 \%\) b. \(80 \%\)

Step-by-Step Solution

Answer

Removing 50% costs \(\$50,000\), and removing 80% costs \(\$200,000\).

Step 1: Understand the problem

We need to find the cost of removing a certain percentage of pollution from a river using the function given by \(f(p) = \frac{50,000 p}{100-p}\). We need to find the cost when \(p = 50\%\) and \(p = 80\%\).

Step 2: Substitute \(p = 50\) into the function

Plug \(p = 50\) into the function. The function becomes: \(f(50) = \frac{50000 \times 50}{100 - 50}\). This simplifies to \(f(50) = \frac{2500000}{50}\).

Step 3: Simplify the expression for \(f(50)\)

Simplify \(\frac{2500000}{50}\) by dividing 2500000 by 50, which gives \(50000\). So, \(f(50) = 50000\).

Step 4: Substitute \(p = 80\) into the function

Now, plug \(p = 80\) into the function. The function becomes: \(f(80) = \frac{50000 \times 80}{100 - 80}\). This simplifies to \(f(80) = \frac{4000000}{20}\).

Step 5: Simplify the expression for \(f(80)\)

Simplify \(\frac{4000000}{20}\) by dividing 4000000 by 20, which gives \(200000\). So, \(f(80) = 200000\).

Step 6: Summary of Costs

The cost of removing 50% of the pollution is \(\\(50,000\) and the cost of removing 80% of the pollution is \(\\)200,000\).

Key Concepts

Understanding Pollution Removal Cost

The cost function, \(f(p) = \frac{50,000 p}{100-p}\), models how removing a percentage of pollution from a river relates to the financial cost involved. This function is called a rational function, and it helps predict costs for different percentages of pollution removal easily.

• As the percentage \(p\) increases, the denominator \(100 - p\) decreases, which leads to a higher cost.
• The formula highlights the complexity of pollution removal, where costs can escalate significantly as you aim to remove near-total pollution.

For example, to remove 50% of pollution, the cost is \(f(50)\) which stems from inserting \(p = 50\) into the function. The simplification involves calculating \(\frac{50,000 \times 50}{50}\) yielding a cost of \(50,000\) dollars. Similarly, for 80%, inserting \(p = 80\) turns into \(\frac{50,000 \times 80}{20}\), resulting in \(200,000\) dollars. These calculations display how significantly costs climb with increasing percentages of pollution removed.

Role of Percentage Substitution

Substituting a percentage into a function means simply replacing the variable in the expression with a given value. For those new to these concepts, think of substituting as plugging in numbers to find specifics.Substitution is the crux of solving problems involving variable percentages, such as pollution removal. It helps transform the general model into specific numerical insights.

• At \(p = 50\), the function becomes \(f(50) = \frac{50,000 \times 50}{100-50}\). This reveals how using specific values provides exact answers, simplifying the otherwise complex function.
• Using \(p = 80\), the function becomes \(f(80) = \frac{50,000 \times 80}{100-80}\), which helps estimate costs when changes in the percentage of pollution removal occur.

This simple practice of substitution can solve many real-world problems by applying carefully defined mathematical models.

Simplifying Rational Expressions

Simplifying rational expressions means reducing fractions to their simplest form. It involves dividing both the numerator and the denominator by their greatest common divisor. This process makes complex calculations manageable.In the given exercise, you simplified the expressions \(\frac{2500000}{50}\) and \(\frac{4000000}{20}\). It's a straightforward division:

• Dividing 2,500,000 by 50 gives 50,000 which makes the expression clear and easy to work with.
• When you divide 4,000,000 by 20, you get 200,000, simplifying the expression effectively.

Understanding the steps of simplification aids in breaking down seemingly daunting problems into basic operations. It becomes especially effective when dealing with larger numbers in rational functions, making them intuitive and less intimidating.