Problem 219 - Calculus Volume 1

Question

The cost to remove a toxin from a lake is modeled by the function \(C(p)=75 p /(85-p),\) where \(C\) is the cost (in thousands of dollars) and \(p\) is the amount of toxin in a small lake (measured in parts per billion [ppb]). This model is valid only when the amount of toxin is less than \(85 \mathrm{ppb}\). a. Find the cost to remove 25 ppb, 40 ppb, and 50 ppb of the toxin from the lake. b. Find the inverse function. \(c\). Use part b. to determine how much of the toxin is removed for $$\$ 50,000$$.

Step-by-Step Solution

Answer

a. $31,250, $66,670, $107,140; b. Inverse: \( p(C) = \frac{85C}{75+C} \); c. 34 ppb.

Step 1: Calculate the Cost for 25 ppb

To find the cost to remove 25 ppb of toxin, substitute \( p = 25 \) into the function \( C(p) = \frac{75p}{85-p} \).\[ C(25) = \frac{75 \times 25}{85 - 25} = \frac{1875}{60} = 31.25 \]Thus, the cost is \( 31.25 \) thousand dollars, or \( 31,250 \) dollars.

Step 2: Calculate the Cost for 40 ppb

Substitute \( p = 40 \) into the function \( C(p) = \frac{75p}{85-p} \).\[ C(40) = \frac{75 \times 40}{85 - 40} = \frac{3000}{45} = 66.67 \]Thus, the cost is \( 66.67 \) thousand dollars, or \( 66,670 \) dollars.

Step 3: Calculate the Cost for 50 ppb

Substitute \( p = 50 \) into the function \( C(p) = \frac{75p}{85-p} \).\[ C(50) = \frac{75 \times 50}{85 - 50} = \frac{3750}{35} = 107.14 \]Thus, the cost is \( 107.14 \) thousand dollars, or \( 107,140 \) dollars.

Step 4: Derive the Inverse Function

Start with the original cost function:\[ C = \frac{75p}{85-p} \]To find the inverse, solve for \( p \) in terms of \( C \):Multiply both sides by \( 85 - p \):\[ C(85 - p) = 75p \]Distribute \( C \):\[ 85C - Cp = 75p \]Rearrange to isolate terms involving \( p \):\[ 85C = 75p + Cp \]Factor \( p \) on the right side:\[ 85C = p(75 + C) \]Divide by \( (75 + C) \):\[ p = \frac{85C}{75 + C} \]This is the inverse function, \( p(C) \).

Step 5: Use Inverse to Find Toxin for $50,000

Convert \( 50,000 \) dollars into thousands: \( C = 50 \).Substitute into the inverse function:\[ p = \frac{85 \times 50}{75 + 50} = \frac{4250}{125} = 34 \]The amount of toxin removed is \( 34 \) ppb.

Key Concepts

Cost Function

Understanding a cost function is crucial when analyzing how much it would cost to remove toxins from an environment, like a lake. The cost function provided in this scenario is a formula used to calculate the expense involved to remove a specific concentration of toxins. In this example, the cost function is given by \[ C(p) = \frac{75p}{85-p} \]where \(C\) represents the cost in thousands of dollars, and \(p\) denotes the toxin levels measured in parts per billion (ppb).
This formula is engineered to show how costs escalate as the amount of toxin increases.

• Lower toxin levels result in a smaller denominator, hence a lower cost.
• As the level approaches 85 ppb, the denominator (\(85-p\)) reduces, causing the cost to rise sharply.

This model suggests that the cost is affordable with lower toxin concentrations but becomes significantly more expensive as the concentration approaches an upper limit of 85 ppb. This implies a diminishing return on cost efficiency as the toxin levels increase, making it essential to work within economical toxin removal ranges.

Toxin Removal

Toxin removal refers to the process of decreasing the amount of a harmful substance—in this case, a toxin—from an environment. This exercise focuses on how much it costs to remove specific quantities of toxins from a lake, based on their concentration measured in parts per billion (ppb).

• Toxins often originate from pollutants, which can have detrimental effects on water quality and ecosystem health.
• Removing these toxins is crucial for maintaining the lake's ecological balance and ensuring the safety of water for other uses.

By understanding the cost associated with such environmental remediation efforts, stakeholders can plan effective toxin management strategies. The cost as calculated by the provided function will also consider logistical factors such as the method of removal, the technology used, and the depth and spread of contamination, influencing the overall expenses incurred in the cleaning process.

Parts per Billion (ppb)

Parts per billion (ppb) is a unit of measurement used to describe the concentration of a substance in parts relative to a larger mixture. In environmental science, ppb is often used to convey very small amounts of a chemical in air, water, or soil. The use of ppb in this context mainly helps to measure toxicity levels present in the lake.

• A ppb level indicates the weight ratio of the toxin in one billion parts of the water, thus signifying a minute but potentially impactful presence.
• This measurement allows scientists and policymakers to evaluate acceptable levels of toxins and set regulations accordingly.

Applying ppb measures helps in accurately assessing how dangerous a contaminant concentration is, focusing resources on purification efforts precisely where needed.

Inverse Calculation

The concept of inverse calculation helps to work backwards from the result to discover what input was needed to achieve that result. In the context of the exercise, finding an inverse function allows one to determine the toxin concentration for a given cost. This is particularly useful for budget-constrained scenarios.To find the inverse, the original formula is rearranged to solve for the toxin concentration \(p\) given a specific cost \(C\). Starting with:\[ C = \frac{75p}{85-p} \]The inverse function \( p(C) \) is derived as:\[ p = \frac{85C}{75 + C} \]This inverse function helps calculate what the toxin level would be for, say, a budget of \(50,000. By substituting \)50,000 (\(C = 50\) when represented in thousands of dollars) into the equation, the toxin concentration comes out as 34 ppb.

• Knowing this allows decision-makers to adjust their actions based on financial constraints, offering a more strategic purification approach.
• It provides vital insights into how much toxin removal is feasible for designated budgets.