Problem 43 - Applied Calculus

Question

In an effort to reduce its inventory, a warehouse runs a sale on its least popular Blu-ray discs. The sales rate (discs sold per day) on day \(t\) of the sale is predicted to be \(50 / t\) (for \(t \geq 1),\) where \(t=1\) corresponds to the beginning of the sale, at which time none of the inventory of 200 discs had been sold. a. Find a formula for the total number of discs sold up to day \(t\) b. Will the store have sold its inventory of 200 discs by day \(t=30 ?\)

Step-by-Step Solution

Answer

The store will not sell its inventory by day 30, as only 164 discs are sold.

Step 1: Understand the sales rate

The sales rate given is \( \frac{50}{t} \) discs per day. This means on each day \( t \), the number of discs sold that day decreases as \( t \) increases.

Step 2: Establish the integral for cumulative sales

To find the total number of discs sold up to day \( t \), we integrate the sales rate function from day 1 to day \( t \). The integral we need to solve is \( \int_1^t \frac{50}{x} \, dx \).

Step 3: Calculate the integral

The integral \( \int \frac{50}{x} \, dx \) gives \( 50 \ln|x| + C \). Evaluating from 1 to \( t \), we have \([50 \ln|t| - 50 \ln|1|] = 50 \ln(t) \) since \( \ln(1) = 0 \).

Step 4: Total number of discs sold

Thus, the total number of discs sold up to day \( t \) is given by the formula \( S(t) = 50 \ln(t) \). This function gives us the cumulative number of discs sold by the time day \( t \) ends.

Step 5: Check sales by day 30

Evaluate \( S(t) = 50 \ln(30) \) to determine how many discs are sold by day 30. Calculate \( 50 \ln(30) \approx 164.4 \) discs.

Step 6: Compare with inventory

Since \( 164.4 \) is less than \( 200 \), the warehouse will not have sold all its inventory by day 30.

Key Concepts

Sales Rate

The sales rate is the number of items sold per unit of time, which in this scenario is per day. For our Blu-ray disc example, the sales rate is given by the function \( \frac{50}{t} \). This function is crucial because it tells us how sales change over time.

In particular, in this case, the sales rate decreases over time. Here's why: as \( t \) gets larger, the denominator of \( \frac{50}{t} \) increases, making the entire fraction and thus the sales rate smaller. This is an important property as it suggests that the most significant number of sales occur at the start of the sale (e.g., day 1) and gradually decline afterwards.

Understanding the sales rate helps businesses predict inventory needs and plan future promotions. It tells the business owner how fast they are clearing their stock during the sale.

Cumulative Sales

Cumulative sales represent the total number of items sold over a given time period up to a certain day. For our problem, we want to determine how many Blu-ray discs have been sold by a specific day, \( t \).

To find out this total number, we don't just rely on the sales rate for a single day, but instead sum up or accumulate the sales over multiple days. The mathematical tool needed here is integration. By integrating the sales rate over the desired period, we can calculate the cumulative sales.

For example, the cumulative sales by day \( t \) can be calculated using the integral \( \int_1^t \frac{50}{x} \, dx \). This tells us how many discs were sold from day 1 through day \( t \). In these types of problems, cumulative sales information is essential for managing stock and assessing sales performance over time.

Integration of Functions

Integration is a fundamental concept in calculus that helps determine quantities when given a rate. In the context of sales, if we know the rate at which products are sold, integration allows us to find the total number sold over a period.

In our exercise, the integration of \( \frac{50}{x} \) from day 1 to day \( t \) is necessary to find the cumulative sales function, \( S(t) = 50 \ln(t) \). This function gives the total discs sold by the end of day \( t \). Integration, here, transforms a variable rate into a total quantity, capturing the total impact of the sales rate over time.

Understanding integration is crucial in various fields, not just sales. Whether calculating distance from speed, or volumes from density, integration enables us to compute a total from an ever-changing amount. This makes it an indispensable tool in both mathematics and real-world applications.