Question

The ZEE Company makes zingos, which it markets at a price of \(p(x)=10-0.001 x\) dollars, where \(x\) is the number produced each month. Its total monthly cost is \(C(x)=200+4 x-0.01 x^{2} .\) At peak production, it can make 300 units. What is its maximum monthly profit and what level of production gives this profit?

Short answer

The maximum monthly profit is approximately $615.91 at a production level of 273 units.

Step-by-step solution

Define Profit Function

Profit is defined as total revenue minus total cost. The total revenue is given by the price function multiplied by the number of units, i.e., \[ R(x) = p(x) \times x = (10 - 0.001x) \times x. \] Therefore, the profit function \( P(x) \) is \[ P(x) = R(x) - C(x) = (10x - 0.001x^2) - (200 + 4x - 0.01x^2). \] Simplifying gives \[ P(x) = -0.011x^2 + 6x - 200. \]

Find the Critical Points

To find the maximum profit, we need to find the critical points by taking the derivative of the profit function and setting it to zero. The derivative of the profit function \( P(x) \) is \[ P'(x) = -0.022x + 6. \] Set this equal to zero: \[ -0.022x + 6 = 0. \] Solving for \( x \), we get \[ x = \frac{6}{0.022} \approx 272.73. \]

Evaluate the Profit at Critical Points and Endpoint

Since peak production is limited to 300 units, we will evaluate the profit at \( x = 272.73 \) and \( x = 300 \). At \( x = 272.73 \), the profit is \[ P(272.73) = -0.011(272.73)^2 + 6 \times 272.73 - 200. \approx 615.91. \] At \( x = 300 \), the profit is \[ P(300) = -0.011(300)^2 + 6 \times 300 - 200. = 400. \]

Determine Maximum Profit

By comparing the profit at \( x = 272.73 \) and \( x = 300 \), we find that the maximum profit of approximately \( 615.91 \) occurs at the production level of 273 units, due to rounding to whole units.

Key concepts

Profit Function

A profit function is a formula that determines the profit a business earns from its operations. It plays a crucial role in decision-making. In simple terms, profit is the total revenue minus total costs. Here, total revenue is the product of price per unit and the number of units sold.
In our exercise, the profit function for ZEE Company is derived from the difference between total revenue and total costs. The total revenue, represented as \( R(x) = (10 - 0.001x) \times x \), is calculated by multiplying the price \( p(x) = 10-0.001x \) by the number of zingos, \( x \). On the other hand, the cost function \( C(x) = 200 + 4x - 0.01x^{2} \) considers both fixed and variable costs associated with production.
By subtracting the cost function from the revenue function, we arrive at the profit function: \( P(x) = -0.011x^2 + 6x - 200 \). This expression gives insights into how profit changes with the number of zingos produced.

Critical Points

Critical points in calculus refer to values of \( x \) where a function's derivative is zero or undefined. These points help in determining maximum or minimum values, which is particularly useful in optimizing business operations like finding maximum profit.
In solving the given problem, once we have the profit function \( P(x) = -0.011x^2 + 6x - 200 \), finding the critical points involves computing the derivative, \( P'(x) \), and solving \( P'(x) = 0 \). This process helps in pinpointing the production level that maximizes profit. The derivative \( P'(x) = -0.022x + 6 \) equals zero at \( x \approx 272.73 \).
This is a critical point where ZEE Company should focus its production to achieve maximum profitability.

Derivative

Derivatives represent the rate of change of a function. They are fundamental in calculus and provide insight into the behavior of functions. In business, derivatives are applied to assess how changes in production affect profit and costs.
For ZEE Company, finding the derivative of the profit function \( P(x) = -0.011x^2 + 6x - 200 \) gives \( P'(x) = -0.022x + 6 \). This indicates how profit changes as the number of units \( x \) varies.
Setting the derivative to zero helps to find the point where profit does not increase further, which is crucial for finding the maximum profit. Understanding the derivative allows the company to make informed production decisions.

Revenue and Cost Functions

Revenue and cost functions are vital in determining how profitable a company's operations are. The revenue function captures how much money is earned from selling a product, while the cost function enumerates expenses associated with production.
In our case, the revenue function \( R(x) = (10 - 0.001x) \times x \) describes how revenue varies with the production of zingos. As more units are produced, the unit price decreases, impacting the total revenue.
The cost function \( C(x) = 200 + 4x - 0.01x^2 \) reflects both fixed costs (like 200, which doesn't change with production level) and variable costs (like \( 4x - 0.01x^2 \), which change with the number of units produced).
Together, these functions allow businesses to analyze and strategize production to maximize profitability by understanding where to balance income and expenditures.