Problem 49 - Calculus Volume 1

Question

A factory selling cell phones has a marginal cost function \(C(x)=0.01 x^{2}-3 x+229,\) where \(x\) represents the number of cell phones, and a marginal revenue function given by \(R(x)=429-2 x .\) Find the area between the graphs of these curves and \(x=0 .\) What does this area represent?

Step-by-Step Solution

Answer

The area between the curves represents total profit from selling cell phones within the interval \([0, a]\).

Step 1: Identify the Functions

The marginal cost function is given by \(C(x) = 0.01x^2 - 3x + 229\), and the marginal revenue function is \(R(x) = 429 - 2x\). We will integrate these functions to find the total areas under their respective curves.

Step 2: Set Up Integrals for Each Function

To find the area under each curve from \(x = 0\) to some point \(x = a\), we set up the definite integrals:\[ \int_0^a C(x)\, dx \quad \text{and} \quad \int_0^a R(x)\, dx. \]

Step 3: Integrate the Marginal Cost Function

The integral of the cost function is\[ \int C(x)\, dx = \int (0.01x^2 - 3x + 229)\, dx. \]Solving this, we get:\[ 0.01 \frac{x^3}{3} - \frac{3x^2}{2} + 229x + C. \]

Step 4: Integrate the Marginal Revenue Function

The integral of the revenue function is:\[ \int R(x)\, dx = \int (429 - 2x)\, dx. \]Solving this, we get:\[ 429x - \frac{2x^2}{2} + C = 429x - x^2 + C. \]

Step 5: Evaluate the Definite Integrals

To find the definite integrals from \(0\) to \(a\), we evaluate\[ \left[0.01 \frac{a^3}{3} - \frac{3a^2}{2} + 229a \right]_0^a \quad \text{and} \quad \left[429a - a^2 \right]_0^a. \]

Step 6: Find the Area Between the Curves

The area between the curves is given by the absolute difference between the two integrals, evaluated from \(0\) to \(a\):\[ \text{Area} = \left| \int_0^a R(x)\, dx - \int_0^a C(x)\, dx \right|. \]Simplifying, this becomes:\[ \left| \left[(429a - a^2) - (0.01 \frac{a^3}{3} - \frac{3a^2}{2} + 229a)\right]_0^a \right|. \]

Step 7: Analyze the Meaning of the Area

The area between these curves represents the total profit over the interval \([0, a]\). This is because the marginal revenue minus the marginal cost gives the marginal profit, which when integrated gives total profit.

Key Concepts

Area Between Curves

When looking at two curves on a graph, such as a marginal cost function and a marginal revenue function, the area between them tells us something valuable. It represents the difference between what a company earns from selling products and what it costs to make them. To find this area, you need to understand the curves really well.

For example, the marginal cost function can be complex, involving terms like quadratic expressions. It tells you how much it costs to produce each additional unit. The marginal revenue function, on the other hand, shows how much revenue an additional unit brings in.

The area between these two curves is calculated using definite integrals. It helps visualize how profit changes over time or output. This can be done by subtracting the area under the cost curve from the area under the revenue curve. The difference gives the area between the curves, showcasing the profit.

This area doesn't just show profit, it illustrates efficiency and effectiveness in production and pricing strategies. In simple words, if the area under the revenue curve is larger than the area under the cost curve, the business is turning a profit.

Definite Integrals

Definite integrals are powerful tools in calculus used to calculate the exact accumulation of quantities. In the context of business, they help evaluate differences between cost and revenue functions over a specific range.

When integrating a function like the marginal cost from point 0 to some value \(a\), you are summing up all the little changes in cost from producing up to \(a\) units. It's similar for the marginal revenue. By handling these functions with definite integrals, you can pinpoint the area under each curve.

Mathematically, this process involves setting up the integrals for the functions:

• For marginal cost: \( \int_0^a C(x)\, dx \)
• For marginal revenue: \( \int_0^a R(x)\, dx \)

Evaluating these integrals is done by finding antiderivatives, then applying limits from 0 to \(a\). After solving, you gain the total cost and total revenue up to \(a\) units, respectively.

Thus, definite integrals are not just useful in mathematical analysis but provide crucial insights into business dynamics, helping quantify how costs and revenues accumulate with production.

Profit Calculation

Profit calculation is an essential concept in any business, representing the balance between revenues and costs. When we determine profit using calculus, we delve into the finer details of how each unit affects overall earnings.

In this context, profit over an interval is calculated by finding the area between the marginal revenue and marginal cost curves. This correlates with the difference in their respective areas under the curves as defined by definite integrals.

To simplify it:

• Marginal Revenue minus Marginal Cost equals Marginal Profit.
• Marginal Profit integrated over an interval gives total profit.

For instance, if you integrate the difference of the revenue and cost functions from 0 to \(a\), this cumulative profit is represented by \(\left| \int_0^a R(x)\, dx - \int_0^a C(x)\, dx \right|\). The integration illustrates all differentiable profit changes summed over the interval.

Ultimately, understanding these calculations is crucial as it allows businesses to measure how efficiently they are converting inputs into profitable outputs.