Problem 1 - Applied Calculus

Question

If the marginal cost function \(C^{\prime}(q)\) is measured in dollars per ton, and \(q\) gives the quantity in tons, what are the units of measurement for \(\int_{800}^{900} C^{\prime}(q) d q ?\) What does this integral represent?

Step-by-Step Solution

Answer

The units are dollars; it represents the total cost change between 800 and 900 tons.

Step 1: Understanding the Integral

The integral \(\int_{800}^{900} C^{\prime}(q) dq\) is used to find the total change in cost when the quantity \(q\) increases from 800 tons to 900 tons. This means we need to integrate the marginal cost function over this interval.

Step 2: Analyzing Units

The marginal cost \(C^{\prime}(q)\) is given in dollars per ton. The units of \(dq\), which represents a small change in quantity, are in tons. Therefore, the units of \(C^{\prime}(q) dq\) equation become (dollars/ton) \(\times\) (tons) = dollars.

Step 3: Constructing the Total Cost

Integrating \(C^{\prime}(q)\) from 800 to 900 gives us the total cost in dollars. Specifically, it gives us the total additional cost incurred when production changes from 800 tons to 900 tons. This value represents the aggregate of marginal costs over this production interval.

Key Concepts

Integral Calculus

Integral calculus helps us understand the accumulation of quantities. In this context, it is used to find the total additional cost of producing goods. When we look at the integral \[\int_{800}^{900} C^{\prime}(q) dq\]we are calculating the area under the curve of the marginal cost from 800 to 900. This area represents the total change in cost when the quantity increases from 800 tons to 900 tons.
Imagine plotting the marginal cost per ton as a continuous curve, where the integral gives you the sum of all these tiny costs. Each tiny slice of this curve contributes to the total cost.
In simple terms: integration in this scenario allows us to see how costs add up as production grows. It converts a series of small, per-unit costs into a cumulative sum, representing the total additional costs over a given interval of production.

Units of Measurement

Understanding units is crucial in solving calculus problems. Units help us interpret what mathematical results mean in real-world terms.
In the exercise, the marginal cost function \(C^{\prime}(q)\) is in dollars per ton. This "per ton" part is because it tells us the cost to produce just one more ton of material.
When we integrate this function from 800 to 900 tons, we consider a small change in quantity, represented by \(dq\), which has the unit "tons."

• The multiplier for \(C^{\prime}(q)\) is "tons," because we are adding up those costs per ton.
• This means the units of \(C^{\prime}(q) dq\) become dollars, as "per ton" cancels out with "tons."

Ultimately, the result of the integral is in dollars, which reflects the total additional cost incurred between these production amounts.

Cost Function

A cost function describes how costs evolve as production changes. The marginal cost function \(C^{\prime}(q)\) represents the cost of producing an additional unit, or in this context, an additional ton.
The original function, \(C(q)\), represents the cumulative cost at a certain production level. The derivative, \(C^{\prime}(q)\), gives us the rate at which costs are changing—hence "marginal."
When we use integration to return from \(C^{\prime}(q)\) to \(C(q)\), we add up all the marginal changes between two production levels.

• This accumulated result shows the total costs added when increasing production from 800 tons to 900 tons.
• The integral is a powerful tool as it helps us reverse the process of differentiation, providing a complete picture of cost accumulation over specific intervals.

This understanding aids businesses in planning and optimizing production, by showing how incremental changes map onto total costs.