Problem 64 - Calculus and its applications

Question

Melanie's Crafts estimates that its sales are growing continuously at a rate given by $$S^{\prime}(t)=20 e^{t}$$ where \(S^{\prime}(t)\) is in dollars per day, on day \(t\). a) Find the accumulated sales for the first 5 days. b) Find the sales from the 2 nd day through the 5 th day. (This is the integral from 1 to \(5 .)\)

Step-by-Step Solution

Answer

a) Approximately \(2948.2\) dollars; b) Approximately \(2913.8\) dollars.

Step 1: Understanding the Problem

We are given the sales growth rate \(S'(t) = 20 e^t\), which tells us how sales are increasing per day in dollars. We need to find the accumulated sales over a specific period by integrating this rate over the given time intervals.

Step 2: Set Up the Integral for Part (a)

To find the accumulated sales for the first 5 days, we calculate the definite integral of \(S'(t) = 20 e^t\) from \(t = 0\) to \(t = 5\). This means we need to evaluate \(\int_0^5 20 e^t \, dt\).

Step 3: Evaluate the Integral (Part a)

The integral \(\int 20 e^t \, dt\) is \(20 e^t + C\), where \(C\) is the constant of integration. To evaluate from \(0\) to \(5\), compute:\[[20 e^t]_0^5 = 20 e^5 - 20 e^0 = 20 e^5 - 20.\]

Step 4: Compute Numerical Value for Part (a)

Calculate the actual accumulated sales: \(20 e^5 - 20 = 20(148.41) - 20 = 2968.2 - 20 = 2948.2\). So, the accumulated sales for the first 5 days is approximately \(2948.2\) dollars.

Step 5: Set Up the Integral for Part (b)

For sales from the 2nd to the 5th day, we calculate the definite integral from \(t = 1\) to \(t = 5\): \(\int_1^5 20 e^t \, dt\).

Step 6: Evaluate the Integral (Part b)

Using the same antiderivative \(20 e^t\), evaluate the integral:\[[20 e^t]_1^5 = 20 e^5 - 20 e^1.\]

Step 7: Compute Numerical Value for Part (b)

Calculate the sales for the time period:\(20 e^5 - 20 e = 20(148.41) - 20(2.72) = 2968.2 - 54.4 = 2913.8\).Thus, the sales from the 2nd to the 5th day is approximately \(2913.8\) dollars.

Key Concepts

Definite Integral

The concept of a definite integral is essential in calculus for calculating the accumulated quantity over an interval. In layman's terms, a definite integral computes the total or accumulated value of a function between two points, known as the limits of integration. It is like adding up tiny slices to find the complete picture. This measurement becomes extremely helpful when analyzing changes over time. For instance, in the exercise above, integrating the sales growth rate function gives you total sales over a certain number of days. The notation for a definite integral is \( \int_a^b f(t) \, dt \), where \( a \) and \( b \) are the lower and upper limits of the time interval, and \( f(t) \) is the function being integrated.
To solve the problem, you evaluate \( \int_0^5 20 e^t \, dt \) to obtain the accumulated sales for the first 5 days. Similarly, for the period from day 2 to day 5, evaluate \( \int_1^5 20 e^t \, dt \).
Remember, the results from definite integrals give you a specific number because they remove the need for a constant of integration, which appears in indefinite integrals.

Exponential Growth

Exponential growth is a type of growth where the rate of increase in a quantity is proportional to its current value. It means the more you have, the faster it grows. This is contrasted with linear growth, where the increase is steady and constant.
In the exercise, Melanie's Crafts experiences sales growth described by \( S'(t) = 20 e^t \). Here, \( e \) is the base of the natural logarithm, approximately equal to 2.718. It is a constant often used to describe continuous growth processes. During exponential growth, even minor initial values can lead to huge numbers over time, which is why it can be so impactful for businesses.
Exponential functions play a significant role in various fields beyond sales, such as population growth, radioactive decay, and interest computations.

Accumulated Sales

Accumulated sales refer to the total sales accumulated over a specific period. It's about finding out how much has been sold in total, from one point in time to another.
The exercise requires integrating the growth rate over time to find out how much revenue is gathered within the given period. For instance, calculating sales accumulation from day 0 to day 5 means finding the "area under the curve" of the sales rate function from 0 to 5.
The significance of understanding accumulated sales lies in budgeting, forecasting, and gauging a company's performance over time. It helps in making informed business decisions, such as analyzing market trends and optimizing future sales strategies.