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 \((c)\). $$C(x)=-0.02 x^{2}+50 x+100, p(x)=100, a=500$$

Short answer

Answer: When 500 units are sold, the average profit is $140, which means that the company makes a profit of $140 per item sold. The marginal profit at 500 units is $130, which indicates that if the company sells one more unit after reaching 500, the additional profit obtained from selling that single unit is approximately $130. Since the marginal profit is lower than the average profit, the profit per item is decreasing as more items are sold.

Step-by-step solution

The profit function, \(P(x)\), is given by the revenue (\(x\cdot p(x)\)) minus the cost of production (\(C(x)\)). In this case, we have: $$ P(x) = x\cdot p(x) - C(x) $$ Using the given cost function \(C(x) = -0.02x^2 + 50x + 100\) and price function \(p(x) = 100\), we can find the profit function by substituting these expressions into the equation for \(P(x)\). $$ P(x) = x\cdot 100 - (-0.02x^2 + 50x +100) $$ Calculating, we obtain the profit function: $$ P(x) = -0.02x^2 + 50x + (100x - 100) $$ $$ P(x) = -0.02x^2 + 150x - 100 $$ #b. Find the average profit function and marginal profit function.#

To find the average profit function, we need to divide the profit function by \(x\). Thus, we have: $$ \text{Average Profit }(A(x)) = \frac{P(x)}{x} $$ Using the profit function we just derived, we have: $$ A(x) = \frac{-0.02x^2 + 150x - 100}{x} $$ The marginal profit function, \(MP(x)\), is the derivative of the profit function with respect to \(x\). So, taking the derivative of \(P(x)\), we get: $$ MP(x) = \frac{dP(x)}{dx} = -0.04x + 150 $$ #c. Find the average profit and marginal profit if \(x=a\) units are sold.#

Given that \(a = 500\) units are sold, we want to find the average profit, \(A(500)\), and the marginal profit, \(MP(500)\). Plugging the value of \(a\) into our previously derived formulas, we have: $$ A(a) = A(500) = \frac{-0.02(500)^2 + 150(500) - 100}{500} $$ $$ A(500) = \frac{-5000 + 75000 - 100}{500} $$ $$ A(500) = \frac{70000}{500} $$ $$ A(500) = 140 $$ Now let's find the marginal profit at \(x = a = 500\): $$ MP(a) = MP(500) = -0.04(500) + 150 $$ $$ MP(500) = -20 + 150 $$ $$ MP(500) = 130 $$ #d. Interpret the meaning of the values obtained in part (c).#

The average profit found for selling \(a = 500\) units is \(140. That means that the company makes an average profit of \)140 per item sold when 500 units are sold. The marginal profit found for selling \(a = 500\) units is \(130. This means that if the company sells one more unit after reaching 500 sold units, the additional profit obtained from selling that single unit is approximately \)130. Note that the marginal profit is lower than the average profit, which implies that the profit per item is decreasing as more items are sold.

Key concepts

Cost Function

In economics, a cost function shows the total cost of producing a certain quantity of goods. It essentially summarizes how production costs behave with varying levels of output. The cost function is usually expressed as a mathematical equation, with variables representing the number of items produced and specific cost parameters like fixed costs and variable costs. In our example, the cost function is given by: \( C(x) = -0.02x^2 + 50x + 100 \).

• The term \(-0.02x^2\) represents the change in cost due to economies or diseconomies of scale. Here, it's quadratic, indicating that the cost could increase or decrease at an increasing rate.
• The linear term \(50x\) reflects variable costs, which are directly proportional to the number of items produced.
• \(100\) is a fixed cost, which remains constant regardless of the quantity produced.

Understanding the cost function helps businesses manage production expenses and make informed pricing decisions.

Average Profit

Average profit provides a measure of the profit earned on each item sold. It is a useful indicator for understanding how well a company is performing at a micro level per unit sold. The formula for average profit is simply the total profit divided by the number of items sold, given by: \(A(x) = \frac{P(x)}{x}\)
For our specific scenario, substituting the profit function \( P(x) = -0.02x^2 + 150x - 100 \) into the formula, we get: \(A(x) = \frac{-0.02x^2 + 150x - 100}{x}\).
This simplifies to: \(A(x) = -0.02x + 150 - \frac{100}{x}\).
When evaluating average profit at \( x = 500 \), we find \( A(500) = 140 \). It shows that on average, \( $140 \) profit per item is made when 500 items are sold. This measure enables firms to assess their pricing strategy and operational efficiency per product sold.

Marginal Profit

Marginal profit provides insight into the additional profit gained by selling one more unit of a product. It is critical for making decisions about whether to increase production. The marginal profit is derived by taking the derivative of the profit function with respect to the number of items sold. For our case: \(P(x) = -0.02x^2 + 150x - 100\),
its derivative is: \(MP(x) = \frac{dP}{dx} = -0.04x + 150\).
Calculating this at \( x = 500 \) gives \( MP(500) = 130 \). This implies that the addition of selling the 501st item generates approximately \( $130 \) more in profit. Understanding marginal profit is essential for strategic planning in terms of sales and production, as it indicates the profitability of scaling up production.

Derivative in Economics

In economics, derivatives are powerful tools used to assess rates of change, helping economists understand how small changes in one variable can affect another. Derivatives are particularly useful for calculating marginal values, such as marginal cost, marginal revenue, and in our case, marginal profit. These marginal analyses utilize calculus to provide immediate insight into the effects of changing conditions on a firm's economic decisions.
For the problem at hand, we calculated the derivative of the profit function to find the marginal profit. Generally, the derivative of a function \( f(x) \) with respect to \( x \) gives the slope of the tangent line to the function at any given point \( x \). In business, this translates into practical terms, allowing firms to judge whether an additional unit of production will increase profits favorably. Mastery of derivatives in economics allows managers to make informed decisions quickly and adaptively.