Question

A company is increasing the production of a product at the rate of 25 units per week. The demand and cost functions for the product are given by \(p=50-0.01 x\) and \(C=4000+40 x-0.02 x^{2} .\) Find the rate of change of the profit with respect to time when the weekly sales are \(x=800\) units. Use a graphing utility to graph the profit function, and use the zoom and trace features of the graphing utility to verify your result.

Short answer

The rate of change of the profit with respect to time when the weekly sales are 800 units is -750 units per week.

Step-by-step solution

Calculate the revenue function

Since revenue is calculated as the product of price and quantity, the revenue function \(R(x)\) can be calculated as \(p \cdot x = x(50-0.01x) = 50x-0.01x^{2}\).

Get the profit function

The profit function \(P(x)\) is given by the difference between the revenue function and the cost function. So, \(P(x) = R(x) - C(x) = (50x-0.01x^{2}) - (4000+40x-0.02x^{2}) = 10x-0.02x^{2} - 4000\).

Derive the profit function with respect to x

The derivative of the profit function with respect to x, \(P'(x)\), can be obtained by differentiating the profit function with respect to x, yielding \(P'(x) = 10 - 0.04x\).

Compute the rate of change of profit

By the chain rule, the rate of change of the profit with respect to time is given by \(dp/dt = P'(x) \cdot dx/dt\). Provided that \(dx/dt = 25\) (the rate at which production is increasing), the rate of change of the profit when the weekly sales are 800 units is \(dp/dt = P'(800) \cdot 25 = (10 - 0.04\cdot800) \cdot 25 = -750\) units per week.

Key concepts

Revenue Function

Understanding how a business performs financially starts with the revenue function, which determines the total income a company can generate based on the sales of its goods or services. In mathematical terms, the revenue function, often represented as \( R(x) \), is calculated by multiplying the price per unit, \( p \), by the quantity of units sold, \( x \). In this specific problem, the price per unit varies with the number of units sold, following the demand function \( p = 50 - 0.01x \). Therefore, the revenue function is expressed as \( R(x) = x(50 - 0.01x) \), which simplifies to \( R(x) = 50x - 0.01x^2 \).

Using this function, we can estimate the company's income at different levels of sales. It is a crucial element in calculating the profit, as it provides the basic input into the profit function equation.

Profit Function

The heart of any business analysis is the profit function. Profit is what remains after deducting all costs from the revenue. To establish the profit function, \( P(x) \), we take the revenue function and subtract the cost function, \( C(x) \). In our example, the cost function is given by \( C = 4000 + 40x - 0.02x^2 \). Subtracting this from our revenue function gives the profit function equation as \( P(x) = (50x - 0.01x^2) - (4000 + 40x - 0.02x^2) \), which simplifies to \( P(x) = 10x - 0.02x^2 - 4000 \).

This equation represents the company's profit depending on how many units are sold, \( x \). It is a vital tool for businesses to predict financial outcomes and make strategic decisions based on different sales volumes.

Derivative of Profit Function

To find how profit changes with respect to the number of units sold, we turn to calculus and calculate the derivative of the profit function. The derivative, \( P'(x) \), indicates the rate of change of profit with respect to the number of units, \( x \). For our profit function, \( P'(x) = 10 - 0.04x \), which we get by applying basic differentiation rules to each term in the profit function.

In practical terms, the derivative tells us how small changes in the number of units sold will affect the profit. If we want to know the rate of change of profit with respect to time, we need to consider how the number of units sold changes over time. This information is crucial when making production and sales decisions.

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus that describes the relationship between two or more functions that are composed of one another. In the context of our problem, we need to use the chain rule to find out how profit is changing over time (\( dp/dt \)), not just over the number of units (\( dp/dx \)).

To do this, we take the derivative of the profit with respect to \( x \), which we already have as \( P'(x) = 10 - 0.04x \), and then multiply that by the derivative of \( x \) with respect to time, \( dx/dt \), to get \( dp/dt = P'(x) \cdot dx/dt \). Since \( dx/dt = 25 \) units per week, we can find the rate of change of the profit with respect to time by plugging in the value of \( x \) at 800 units. The result is the rate of change of profit at a specific moment in time, which gives us insight into the company's financial performance dynamics.