Problem 31 - Calculus and its applications

Question

Stellar Lawn Care introduces a new kind of lawn seeder. It finds that its marginal supply for the seeder satisfies $$ S^{\prime}(x)=\frac{100 x}{(20-x)^{2}}, \quad 0 \leq x \leq 19 $$ where \(S(x)\) is the quantity purchased when the price is \(x\) thousand dollars per seeder. Find \(S(x)\), given that the company sells 2000 seeders when the price is 19 thousand dollars.

Step-by-Step Solution

Answer

The supply function is \( S(x) = -\frac{2000}{20-x} - 100 \ln |20-x| + 4000 \).

Step 1: Understand the Problem

We are given the marginal supply function \( S'(x) = \frac{100x}{(20-x)^2} \), which represents the rate of change of the quantity supplied with respect to price. The task is to find the supply function \( S(x) \), given that \( S(19) = 2000 \).

Step 2: Set Up the Integral

To find \( S(x) \), we need to integrate the marginal supply function. Set up the integral: \[ S(x) = \int S'(x)\, dx = \int \frac{100x}{(20-x)^2}\, dx. \]

Step 3: Perform u-Substitution

Let \( u = 20 - x \). Then \( du = -dx \), or \( dx = -du \). Replace \( x \) in terms of \( u \), giving: \[ x = 20 - u. \] Substitute into the integral to get: \[ S(x) = \int \frac{100(20-u)}{u^2}(-du) = \int \frac{100(20-u)}{u^2}(-du). \]

Step 4: Simplify and Integrate

Simplify the integrand: \[ -\int \left(\frac{2000}{u^2} - \frac{100u}{u^2}\right) du = -\int \left(\frac{2000}{u^2} - \frac{100}{u}\right) du. \]Separate and integrate each term:\[ -\int \frac{2000}{u^2} du = 2000\cdot \frac{1}{u}, \]\[ -\int \frac{100}{u} du = -100 \ln |u|. \]

Step 5: Write General Supply Function

Combining the integrals, the general supply function becomes:\[ S(x) = -\frac{2000}{20-x} - 100 \ln |20-x| + C. \]

Step 6: Use Initial Condition

Plug in the given condition \( S(19) = 2000 \):\[ 2000 = -\frac{2000}{20-19} - 100 \ln |20-19| + C. \]Calculate values to solve for \( C \):\[ 2000 = -2000 - 100 \cdot 0 + C \]\[ C = 4000. \]

Step 7: Rewrite the Supply Function

Substitute \( C \) back into the supply function to get:\[ S(x) = -\frac{2000}{20-x} - 100 \ln |20-x| + 4000. \]

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in calculus that focuses on finding the integral of a function. Integration, the primary operation in integral calculus, is essentially the reverse process of differentiation. It allows us to find quantities like areas under curves, accumulated quantities, and in exercises like these, the total supply function over a range of prices.
In the context of supply functions, integration helps us determine the actual supply function from its rate of change, known as the marginal supply function. The indefinite integral combines antiderivatives and a constant of integration, denoted as C. This constant is vital since it accounts for initial conditions, similar to the initial value conditions given in problems.
To integrate a marginal supply function, we typically use techniques such as u-substitution, integration by parts, or partial fraction decomposition, depending on the structure of the function to be integrated.

Marginal Supply Function

A marginal supply function describes the rate at which the quantity supplied changes concerning the price. It is represented as the derivative of the supply function, denoted as \( S'(x) \). Understanding the relationship between marginal and total supply functions is crucial in economic contexts.
In this exercise, the marginal supply function \( S'(x) = \frac{100x}{(20-x)^2} \) tells us how the number of seeders adjusts as prices shift in thousands of dollars. Calculating the total supply \( S(x) \) involves integrating this marginal function over a specified range corresponding to various price points.
The marginal supply function is valuable in determining how sensitive the quantity supplied is to changes in price, offering insights into supply elasticity and its economic implications.

Definite Integral

The definite integral computes the total value of a function over a given interval. Unlike indefinite integration, which seeks a general form with a constant, the definite integral gives a numerical result.
In the problem of finding \( S(x) \), initially, we compute an indefinite integral, but the final step often involves evaluating the expression for specific values, akin to a definite integral.
The concept of a definite integral is central to understanding accumulated changes. For example, in economics, it can represent the total increase or decrease in supply over a price interval, providing a clear picture of supply dynamics over a set price range.

u-Substitution

U-substitution is a technique used to simplify the integration process, especially when dealing with complex functions. It involves changing variables to transform the integrand into a simpler form.
In this problem, to simplify the integration of the marginal supply function \( \frac{100x}{(20-x)^2} \), we use u-substitution. By letting \( u = 20 - x \), the derivative \( du = -dx \) allows us to rewrite the integral in terms of \( u \).
This process shifts the focus from the original variable to the substitution variable, often simplifying the integrand significantly. After integrating with respect to \( u \), the results are converted back into terms of the original variable, completing the integration and yielding a more easily interpretable supply function. This technique is powerful in integral calculus as it can drastically simplify what initially appears to be a challenging integral.