Problem 25 - College Algebra and Calculus: An Applied Approach

Question

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 represented by the slope of the line \(x = 80 - 0.8x\) at \(x = 44\).

Step 1: Establish the Profit Function

Profit can be calculated by subtracting the cost from the revenue. In this case, the revenue is the price times the quantity of sales: \(R = p \cdot x = (6000 - 25x) \cdot x\). The cost is given as \(C = 2400x + 5200\). Hence, the profit \(P = R - C = (6000 - 25x) \cdot x - (2400x + 5200)\).

Step 2: Differentiate the Profit Function

Differentiating the profit function with respect to \(x\), we get \(\frac{dP}{dx} = \frac{dR}{dx} - \frac{dC}{dx}\). Here, \(\frac{dR}{dx} = 6000 - 50x\) after applying the power rule and the derivative of a constant. The derivative of the cost with respect to \(x\) is \(\frac{dC}{dx} = 2400\). So, \(\frac{dP}{dx} = 6000 - 50x - 2400\).

Step 3: Substitute the given Profit Rate

We are given that the rate of profit change is $5600 per week, so we set \(\frac{dP}{dx} = 5600\). Substituting our derivative, we get \(6000 - 50x - 2400 = 5600\).

Step 4: Solve for \(x\)

Solving the equation \(6000 - 50x - 2400 = 5600\) for x, we find \(x = 80 - 0.8x\). This equation represents a line whose slope gives the change in sales per unit of time, at any given value of \(x\).

Step 5: Find the rate of change of sales at \(x = 44\)

Substituting \(x = 44\) into the equation from the previous step, we find the rate of change of sales at that point.

Key Concepts

Profit Function

The concept of a profit function is crucial in understanding business operations. Profit essentially measures how much money is being made over the costs. Mathematically, profit is calculated by subtracting the total costs from the total revenues generated. In this exercise, the revenue function is given by the price times the quantity of sales, which is expressed as \( R = p \cdot x = (6000 - 25x) \cdot x \). The cost function is provided as \( C = 2400x + 5200 \). Therefore, the profit function \( P \) can be derived by subtracting the total costs \( C \) from the total revenue \( R \), leading to the equation:

• \( P = (6000 - 25x) \cdot x - (2400x + 5200) \)

This function represents how profit changes based on the number of units \( x \) sold.

Demand Function

The demand function is a relation that shows how the quantity demanded by the market changes with the price of the product. In the problem at hand, the demand function is expressed as \( p = 6000 - 25x \), where \( p \) denotes the price and \( x \) represents the number of units sold. This particular demand function indicates that as the quantity \( x \) increases, the price \( p \) decreases, reflecting a common inverse relationship between demand and price in economics.

Understanding the demand function is essential for setting appropriate pricing strategies. It helps to determine the optimal price point to maximize revenue or profit, considering how the market reacts to changes in pricing. Here, by manipulating the price, businesses can influence consumer demand.

Cost Function

The cost function shows how the total cost incurred varies with the level of production or sales. In this exercise, the cost function is given by \( C = 2400x + 5200 \). This equation suggests that there are two components of cost:

• Variable Costs: These costs change with the level of production, represented here by \( 2400x \). This means for each additional sale, the cost increases by 2400 dollars.
• Fixed Costs: These are constant costs that do not vary with sales or production levels, represented by \( 5200 \) in this case.

The cost function is vital for determining how much needs to be spent to produce a given number of products. It helps in budgeting and setting sales targets that ensure profitability.

Differentiation

Differentiation is a mathematical process that allows us to determine the rate at which one quantity changes with respect to another. In economic problems, this is particularly beneficial for identifying how functions like profit, revenue, or cost behave as certain variables change.

In the given exercise, differentiation is used to find the rate of change of the profit function. By differentiating the profit function \( P = (6000 - 25x) \cdot x - (2400x + 5200) \) with respect to \( x \), it gives us \( \frac{dP}{dx} \), which represents the change in profit for each additional unit sold. The derivative simplifies to \( 6000 - 50x - 2400 \).

• This equation allows us to assess how sensitive profits are to changes in sales volume.
• Determining \( \frac{dP}{dx} = 5600 \) given in the problem further connects this derivative to real-world changes, helping predict sales adjustments.

This use of differentiation aids in making data-driven business decisions by forecasting how changes in one aspect of operations might impact others.