Problem 16 - Applied Calculus

Question

The cost, \(C\) (in dollars), to produce \(g\) gallons of a chemical can be expressed as \(C=f(g)\). Using units, explain the meaning of the following statements in terms of the chemical: (a) \(f(200)=1300\) (b) \(f^{\prime}(200)=6\)

Step-by-Step Solution

Answer

(a) 200 gallons cost $1300; (b) Each additional gallon at 200 gallons costs $6 more.

Step 1: Understanding statement in part (a)

The expression \(f(200) = 1300\) indicates that the cost to produce 200 gallons of the chemical is 1300 dollars. In this context, \(f(g)\) represents the cost function in terms of the gallons \(g\). Therefore, when 200 gallons are produced, the total cost becomes 1300 dollars.

Step 2: Understanding statement in part (b)

The expression \(f^{\prime}(200) = 6\) indicates the rate of change of the cost with respect to the number of gallons produced when 200 gallons are being produced. Specifically, it means that producing one additional gallon, beyond the 200th gallon, will increase the cost by 6 dollars. \(f^{\prime}(g)\) represents the derivative of the cost function with respect to \(g\), showing how costs change as the volume produced changes.

Key Concepts

Derivative Interpretation

In calculus, the derivative of a function gives us vital information about how the function behaves and changes. Specifically, it tells us about the rate at which the function's value is changing. When we talk about the derivative of a cost function, like the one in the exercise, we're interested in understanding how the cost changes as production increases or decreases. The derivative is not just a number; it provides deep insights into the behavior of the function.
The expression \( f^{\prime}(200) = 6 \) implies the derivative of the cost function \( f(g) \) at \( g = 200 \). This derivative tells us how the cost changes when we increase production from 200 gallons. Here, a derivative of 6 means that for every additional gallon produced after 200 gallons, the cost increases by 6 dollars. This information could help a business in decision-making, like deciding the most cost-effective amount of product to manufacture.

Rate of Change

The concept of a rate of change is crucial in understanding how one quantity varies with another. In this case, the rate of change tells us how the cost varies with the quantity of chemical produced. When we say \( f^{\prime}(200) = 6 \), this number represents the rate of change of the cost when producing 200 gallons.
This rate of change is a constant value at \( g = 200 \), suggesting that each additional gallon has an identical impact on the cost. It shows us the cost-efficiency of producing more chemicals. If this rate of change were to decrease, it would mean producing additional gallons costs less, potentially allowing for more profit or lower prices. Understanding the rate of change is vital in economics and business strategy, as it directly influences production and pricing decisions.

Units in Calculus

In calculus, units are important as they give context and meaning to our calculations and results. Without units, numbers can become meaningless and misleading. For example, \( f(g) \) measures cost in terms of dollars, while \( g \), the input of the function, is measured in gallons. Thus, \( f(g) \) describes the total cost in dollars for producing \( g \) gallons of chemicals.
Similarly, when we look at the derivative \( f^{\prime}(g) \), the units become incredibly instructive in interpreting the results. In our example, \( f^{\prime}(200) = 6 \), the derivative's units are dollars per gallon. This unit tells us that the cost increases by 6 dollars for each additional gallon produced at the point of 200 gallons. Without understanding the units, we risk misinterpreting this critical information. Therefore, always pay attention to and use the appropriate units in calculus problems to ensure accurate and meaningful solutions.