Problem 61 - Calculus and its applications

Question

Kitchens-to-Please Contracting determines that the marginal cost, in dollars per square foot, of installing \(x\) square feet of kitchen countertop is given by $$C^{\prime}(x)=4 x^{1 / 3}$$ a) Find the cost of installing \(50 \mathrm{ft}^{2}\) of countertop. b) Find the cost of installing an extra \(14 \mathrm{ft}^{2}\) of countertop after \(50 \mathrm{ft}^{2}\) have already been installed.

Step-by-Step Solution

Answer

The cost of installing 50 ft² is approximately $549.04, and the additional cost of 14 ft² after 50 ft² is calculated similarly for 64 ft².

Step 1: Understanding the Marginal Cost

The provided marginal cost function is \( C'(x) = 4x^{1/3} \), which represents the rate of change of the cost with respect to the area \( x \). To find the cost, we need to integrate this marginal cost function.

Step 2: Set Up the Integral for Total Cost

Integrate the marginal cost function to find the total cost function \( C(x) \). The integral of \( 4x^{1/3} \) with respect to \( x \) is set up as follows:\[C(x) = \int 4x^{1/3} \, dx\]

Step 3: Solve the Integral

Perform the integration:\[C(x) = \int 4x^{1/3} \, dx = \frac{4}{4/3}x^{4/3} + C = 3x^{4/3} + C\]where \( C \) is the constant of integration.

Step 4: Determine the Constant of Integration (Optional)

Since we are interested in cost calculations starting from zero square feet, we can consider the constant of integration as zero in this context, simplifying our task.

Step 5: Calculate Cost for 50 Square Feet

Calculate \( C(50) \) to find the total cost for installing 50 square feet:\[C(50) = 3(50)^{4/3}\]Calculate \( 50^{4/3} \) and then multiply by 3.

Step 6: Approximate Cube Root and Power

Calculate \( 50^{1/3} \) or cube root of 50, approximately 3.68. Then raise to the power of 4.\[50^{4/3} \approx (3.68)^4 \approx 183.014\]

Step 7: Calculate Total Cost for 50 Square Feet

Substitute back to find the total cost:\[C(50) = 3 \times 183.014 = 549.042\]So, the cost to install 50 square feet is approximately $549.04.

Step 8: Determine Extra Cost for Additional 14 Square Feet

Now find the cost after 64 square feet (50 + 14). Calculate \( C(64) \):\[C(64) = 3(64^{4/3})\]

Key Concepts

Marginal Cost

Marginal cost is a fundamental concept in economics and calculus that helps us understand how much an additional unit of a good or service will cost to produce. In the context of this exercise, the marginal cost is represented by the function \( C'(x) = 4x^{1/3} \). This function tells us how the cost changes as the area \( x \) of the kitchen countertop increases.

To effectively use this concept, it's important to realize that the marginal cost is essentially the derivative of the total cost function. This derivative provides us with the cost rate for each additional square foot of countertop. In simple terms, just like checking the speedometer of a car gives the speed at a particular moment, the marginal cost tells us the additional cost at a specific quantity of installation.

Understanding the marginal cost function allows us to explore how costs increase in a non-linear fashion, particularly when the rate of increase isn't constant and changes with the number of units produced or installed.

Integration in Calculus

Integration in calculus is the process of finding the function that describes the area under a curve. In simpler terms, if a derivative function gives us a slope, integration allows us to "reverse" this process to find the original function.

In our exercise, we have the marginal cost function \( C'(x) = 4x^{1/3} \), which we need to integrate to find the total cost function \( C(x) \). The process involves setting up an integral:

• \[C(x) = \int 4x^{1/3} \, dx\]

Performing this integration enables us to add up all the marginal costs across the installation space. The solution comes to the integrated function: \[C(x) = 3x^{4/3} + C\], where \( C \) is the constant of integration.

The integration here captures the accumulation of costs as more square feet are installed, illustrating how calculus can provide powerful tools for solving real-world problems involving rates and accumulated quantities.

Application of Derivatives

Applying derivatives is crucial when solving problems related to changing rates, such as in this exercise. Understanding how derivatives play a role in determining marginal cost helps in making informed decisions, like cost-efficient production levels.

Here, the derivative, \( C'(x) = 4x^{1/3} \), illustrates how the cost changes with each additional square foot of countertop. The power of derivatives in calculus is their ability to provide instant slopes or rates, which in business contexts translate into understanding efficiencies and predicting costs.

Using derivatives to find the marginal cost allows individuals or companies to perform precise cost calculations, plan budgets, and forecast financial outcomes more accurately. In this case, it's utilized to figure out the exact cost when installing specific amounts of countertop square footage by evaluating points along the derivative and then using integration to compute total costs.

Mathematical Problem Solving

Solving mathematical problems like this one involves a series of structured steps to ensure accuracy and obtain meaningful results. When approaching such problems, it is important to systematically break down complex equations into simpler components.

First, identify the given information and what needs to be found. For the problem at hand:

• Identify the marginal cost function \( C'(x) = 4x^{1/3} \)
• Set up integrals to find the total cost \( C(x) \)
• Perform necessary calculations to determine specific costs, like for 50 square feet and then 64 square feet.

It's essential to maintain clarity in each step. Use of consistent variables helps track calculations and ensures that results are accurate and retraceable. Mathematical problem solving also enhances analytical skills, enabling one to derive solutions not just in academic settings, but in practical, real-world scenarios as well.