Problem 25 - Calculus an Applied Approach

Question

Sales The profit for a product is increasing at a rate of \(\$ 5600\) per week. The demand and cost functions for the product are given by \(p=6000-25 x\) and \(C=2400 x+5200\). Find the rate of change of sales with respect to time when the weekly sales are \(x=44\) units.

Step-by-Step Solution

Answer

The rate of change of sales with respect to time when the weekly sales are 44 units is \$8000 per week.

Step 1: Express the Profit Function

First, we express the profit in terms of the given functions: \(P = R - C\), where \(R\) is revenue, \(C\) is cost and \(P\) is profit. As we're given that the Cost \(C = 2400x + 5200\) and Revenue \(R = x*p = x*(6000 - 25x)\), we substitute these into the profit function to get: \(P = x*(6000 - 25x) - (2400x + 5200)\).

Step 2: Differentiate the Profit Function

Now, to find the rate of change of this profit function, we differentiate it with respect to time \(x\)(weekly). So, \(P'= R' - C'\). Doing this, we get: \(P' = (6000 - 50x) - 2400\).

Step 3: Find the Rate of Change

Now we substitute the given values of \(x = 44\) and \(P' = 5600\). So, we plug those values into our differentiated profit function obtained in Step 2 and solve for \( R'\), which gives us rate of change of sales. Solving we get, \(5600 = R' - 2400\), on solving we get \(R' = 8000\) per week.

Key Concepts

Profit Function

The profit function is a mathematical representation that helps us understand how profit behaves regarding sales and costs. Essentially, profit is defined as the difference between revenue and cost. It's expressed as:

• **Profit (P)** = **Revenue (R)** - **Cost (C)**

In the given exercise, revenue and cost functions are provided, allowing us to construct the profit function. Understanding these components is crucial as it directly influences business decisions regarding pricing, production, and sales strategies. By finding relationships and trends through such a profit function, businesses can optimize their operations.
If sales increase or costs fluctuate, these changes are reflected in the profit function, making it a vital tool for financial analysis. For example, in this scenario, calculating the derived equations helps predict how changes in the sales quantity affect the overall profit. Understanding this interplay assists stakeholders in making informed strategic choices.

Derivative in Calculus

Derivatives in calculus provide us with a powerful tool to analyze how a quantity changes over time. In the context of this exercise, the derivative is used to find the rate at which profit changes as a function of time. The derivative of a function essentially tells us its rate of change.
Mathematically, if we have a function that defines a relationship, such as profit involving sales and cost, differentiating it concerning time gives us insight into how changes in one variable, like sales, influence another variable, such as profit.
For our exercise, differentiating the profit function provides the rate of change of profits with respect to sales. This step involves calculating the derivative of both revenue and cost functions separately, and then finding their difference, known as the profit derivative. It’s an application of calculus to real-world problems, illustrating its utility in economic and business contexts.

Revenue and Cost Functions

Understanding revenue and cost functions is essential when analyzing financial questions like those in this exercise. The revenue function gives us insight into the total income generated from sales, often represented as price times quantity. Mathematically, if the price depends on the quantity sold, this relationship is expressed within the revenue function.

• **Revenue (R)** = **Price (p) x Sales (x)**

In the problem, the price is expressed as a function of sales: \( p = 6000 - 25x \). Thus, revenue becomes \( R = x(6000 - 25x) \).
Similarly, the cost function represents the total costs involved in producing a product. These costs can be fixed or variable, and having a function allows for precise calculations when assessing overall financial performance.

• **Cost (C)** = Total production cost, in this case, \( C = 2400x + 5200 \)

By calculating these functions, weeks determine not only current profitability but also predict future insights if either sales or costs were to change, making them foundational to business operations.