Question

Let \(C(x)\) represent the cost of producing \(x\) items and \(p(x)\) be the sale price per item if \(x\) items are sold. The profit \(P(x)\) of selling x items is \(P(x)=x p(x)-C(x)\) (revenue minus costs). The average profit per item when \(x\) items are sold is \(P(x) / x\) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that \(x\) items have already been sold. Consider the following cost functions \(C\) and price functions \(p\). a. Find the profit function \(P\). b. Find the average profit function and marginal profit function. c. Find the average profit and marginal profit if \(x=a\) units are sold. d. Interpret the meaning of the values obtained in part \((c)\). $$\begin{aligned} &C(x)=-0.04 x^{2}+100 x+800, p(x)=200-0.1 x,\\\ &\bar{a}=1000 \end{aligned}$$

Short answer

a. The profit function \(P(x)\) is given as: \(P(x) = -0.14x^2 + 300x - 800\). b. The average profit function is: \(\frac{P(x)}{x} = -0.14x + 300 - \frac{800}{x}\), and the marginal profit function is: \(\frac{dP}{dx} = -0.28x + 300\). c. When \(x=a = 1000\) units are sold, the average profit is \(159.2\) and the marginal profit is \(20\). d. The average profit of 159.2 per unit when selling 1000 items means that the company is making a profit of $159.2 for each item on average. The marginal profit of 20 indicates that the company would make an additional profit of $20 if they sell one more item when 1000 items have already been sold. This information can be helpful for the company in determining if it is beneficial to sell more items or adjust their production and pricing strategies.

Step-by-step solution

Find the profit function P(x)

Based upon the profit function definition \(P(x)=x p(x)-C(x)\), our goal is to plug in the given cost and price functions into the equation. \(P(x)=x(200-0.1x)-(-0.04x^2+100x+800)\).

Simplifying P(x)

Let's simplify P(x) by expanding the expression and combining like terms. \(P(x)=(200x-0.1x^2)-(0.04x^2-100x+800)\) \(P(x)=-0.14x^2+300x-800\)

Find the average profit function P(x)/x

Divide P(x) by x to obtain the average profit function. \(\frac{P(x)}{x}=\frac{-0.14x^2+300x-800}{x}\) \(\frac{P(x)}{x}=-0.14x+300-\frac{800}{x}\)

Find the marginal profit function dP/dx

Differentiate the profit function P(x) with respect to x using the power rule. \(\frac{dP}{dx}=-0.28x+300\)

Find the average profit and marginal profit if x=a

We are given \(\bar{a}=1000\). Plug in the value of \(a\) into the average profit function and marginal profit function. \(\frac{P(\bar{a})}{\bar{a}}=-0.14(1000)+300-\frac{800}{1000}=-0.14(1000)+300-0.8=-140+300-0.8=160-0.8=159.2\) \(\frac{dP}{dx}|_{x=\bar{a}}=-0.28(1000)+300=-280+300=20\)

Interpret the results

The average profit when 1000 units are sold is \(159.2. This means that on average, the company makes a profit of \)159.2 per item when selling 1000 items. The marginal profit when selling 1000 units is \(20. This means that the company would make an additional profit of \)20 if they sell one more item when 1000 items have already been sold.

Key concepts

Profit Function

Understanding the profit function is crucial for any business. It calculates the earnings after covering all the expenses and is given by the formula:
\[ P(x) = x \times p(x) - C(x) \]
Here, \( x \) is the quantity of items sold, \( p(x) \) is the price per item, and \( C(x) \) represents the total production costs for \( x \) items. Simplifying this equation helps to visualize how profit changes with the level of production and sales.

Average Profit

Average profit is a metric to analyze profitability per unit. It is calculated by dividing the total profit, \( P(x) \), by the number of items, \( x \):
\[ \text{Average profit} = \frac{P(x)}{x} \]
By expressing it in this way, you determine the profit attributed to each item sold, which aids in pricing strategies and helps assess if the production scale is efficient.

Marginal Profit

Marginal profit is the extra profit from selling one additional unit. It is the derivative of the profit function with respect to \( x \):
\[ \text{Marginal profit} = \frac{dP}{dx} \]
This concept is pivotal for decision-making, as it can indicate

• Optimal production levels
• How profits are affected by changes in sales volume
• When to stop increasing production to avoid decreasing profitability

Cost Functions

A cost function, \( C(x) \), is an equation that represents the total cost of producing \( x \) items. It often includes both fixed costs (independent of \( x \)) and variable costs (which change with \( x \)). Analyzing the behavior of cost functions helps businesses to manage and optimize their expenses.

Price Functions

Dynamic Pricing Strategies

The price function, \( p(x) \), reflects how the selling price per item can vary with the number of items, \( x \). This function allows businesses to incorporate:

• Discounts for bulk purchases
• Price increases due to scarcity
• Dynamic pricing strategies

It's a tool for maximizing revenue by adapting to market conditions and consumer behavior.

Differentiating Profit Function

Differentiation is a mathematical tool used to find the rate at which a function is changing. When applied to the profit function, it reveals the rate of change of profit with respect to the number of items sold, otherwise known as the marginal profit. Calculating the derivative, \( \frac{dP}{dx} \), provides insight into the profitability of selling additional units and can help in finding the maximum profit point.