Problem 40 - Calculus and its applications

Question

Nevin Patio Contractors determines that its marginal-profit function is given by $$ P^{\prime}(x)=1000 x^{2} e^{-0.2 x} $$ Find the total profit given that \(P(0)=-\$ 2000\).

Step-by-Step Solution

Answer

Integrate the marginal-profit function and use the initial condition to find total profit.

Step 1: Understanding the Problem

The marginal-profit function is provided as \(P'(x) = 1000x^2 e^{-0.2x}\). The problem asks for the total profit, which can be found by integrating the marginal-profit function, given an initial condition \(P(0) = -2000\).

Step 2: Integration of the Marginal-Profit Function

To find the profit function \(P(x)\), we integrate the marginal-profit function \(P'(x) = 1000x^2 e^{-0.2x}\). This requires finding the integral:\[\int 1000x^2 e^{-0.2x} \, dx\]

Step 3: Using Integration by Parts

We apply integration by parts where \(u = x^2\) and \(dv = 1000e^{-0.2x}\). Differentiating and integrating these gives \(du = 2x \, dx\) and \(v = -5000 e^{-0.2x}\) after integrating \(-0.2\). The integration by parts formula \(\int u \, dv = uv - \int v \, du\) is used.

Step 4: Solving the Integral

Substituting into the integration by parts formula:\[\int 1000x^2 e^{-0.2x} \, dx = -5000 x^2 e^{-0.2x} + \int 5000 (2x) e^{-0.2x} \, dx\]The second integral now needs integration by parts again with \(2x\) and \(e^{-0.2x}\). This continues until all parts are integrated.

Step 5: Finding the Profit Function

After solving all parts, the profit function \(P(x)\) includes results from the previous integrals plus the constant of integration, \(C\).

Step 6: Determining the Constant of Integration

Use the initial condition \(P(0) = -2000\) to solve for \(C\). Substitute \(x = 0\) and solve the equation derived in previous steps to find \(C\).

Step 7: Finding the Total Profit Function

Substitute \(C\) back into the total profit function \(P(x)\) derived from integration, incorporating all components and simplifications to write the final function.

Step 8: Evaluation of Total Profit

The function \(P(x)\) now represents the total profit with the initial condition applied. Interpretation of this function allows for specific evaluations at any \(x\).

Key Concepts

Marginal Profit

Marginal profit is a critical concept in calculus applications, particularly in business and economics. It represents the change in profit for producing one more unit of a good or service. In mathematical terms, marginal profit is often represented as the derivative of the profit function, denoted as \(P'(x)\). Here, \(x\) is the quantity of goods produced or sold. In the problem we are discussing, the marginal-profit function is given as \( P^{\prime}(x) = 1000 x^{2} e^{-0.2 x} \). This function can tell us how the profit will change as we produce or sell additional units, allowing us to make more informed business decisions.

Integration by Parts

Integration by parts is a valuable technique for solving integral problems, especially when dealing with products of functions. It's derived from the product rule of differentiation and is stated as:

• \( \int u \, dv = uv - \int v \, du \)

To use this technique effectively, identify two parts of the integrand as \(u\) and \(dv\), then differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\).In our exercise, to integrate the marginal-profit function \(P'(x) = 1000x^2 e^{-0.2x}\), we set \( u = x^2 \) and \( dv = 1000 e^{-0.2x} \, dx \).This yields \( du = 2x \, dx \) and \( v = -5000 e^{-0.2x} \) after integrating. The technique often requires repeating these steps until the integral is fully solved. Applying integration by parts simplifies complex integrations, making them manageable and helping to find a solution to our larger problem.

Profit Function

A profit function is an essential tool for businesses to understand their financial performance. It relates to revenue and cost functions and is generally the difference between total revenue and total cost.In calculus applications, the profit function \(P(x)\) can be found by integrating the marginal profit function \(P'(x)\). This is exactly what's needed in our exercise to determine the total profit over a range of units produced or sold.Once integrated, you obtain the profit function, which also includes a constant of integration, \(C\).To find \(C\), use known conditions, such as an initial profit, which in our problem is \(P(0) = -2000\).Substituting this into the integrated function allows us to solve for \(C\), providing a complete picture of the profit over time or production levels. The final profit function \(P(x)\) can then be evaluated at any point \(x\) to determine specific profit values for different quantities produced.