Question

Average and marginal profit 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 the marginal profit function. c. Find the average profit and the 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

In summary, given the cost function \(C(x)=-0.02 x^{2}+50 x+100\) and the price function \(p(x)=100\), we found that the profit function is \(P(x)=0.02x^2+50x-100\). The average profit function is \(\frac{P(x)}{x}\) and the marginal profit function is \(\frac{dP(x)}{dx}\). When \(x=a=500\) units are sold, the average profit is $48 per item and the marginal profit is $70 for the additional item sold. This means that the company makes a profit of $48 per item on average when selling 500 items, and if they sell one more item, their profit will increase by approximately $70 for that additional item.

Step-by-step solution

Recall that the profit function \(P(x)\) is given by the formula: \(P(x) = x p(x) - C(x)\). Here, \(C(x)=-0.02 x^{2}+50 x+100\) and \(p(x)=100\). Substitute \(C(x)\) and \(p(x)\) in the formula to get: $$P(x) = x (100) - (-0.02 x^{2}+50 x+100)$$ Simplify the expression: $$P(x)=100x+0.02x^2-50x-100$$ Combining like terms, we have: $$P(x)=0.02x^2+50x-100$$ #Step 2: Find the average profit function and the marginal profit function#

The average profit function is given by \(\frac{P(x)}{x}\). Using the profit function from step 1, this will be: $$\text{Average Profit}=\frac{0.02x^2+50x-100}{x}$$ The marginal profit function is the derivative of the profit function with respect to \(x\), so we have: $$\text{Marginal Profit}=\frac{dP(x)}{dx}$$ Using the profit function from step 1, differentiate with respect to \(x\): $$\frac{d}{dx}(0.02x^2+50x-100)=0.04x+50$$ #Step 3: Find the average profit and the marginal profit when x=a units are sold#

We are given that \(a=500\). To find the average and marginal profit when \(x=a\), we need to substitute \(x=500\) into both the average profit function and the marginal profit function that we found in step 2. Average profit when \(x=500\): $$\text{Average Profit} = \frac{0.02(500)^2 + 50(500) -100}{500}$$ Calculate the value: $$\text{Average Profit}=48$$ Marginal profit when \(x=500\): $$\text{Marginal Profit} = 0.04(500) + 50$$ Calculate the value: $$\text{Marginal Profit}=70$$ #Step 4: Interpret the meaning of the values obtained in part (c)#

The average profit of 48 when \(x=500\) means that, on average, the company makes a profit of $48 per item when selling 500 items. The marginal profit of 70 when \(x=500\) means that if the company sells one more item (a total of 501 items), their profit will increase by approximately $70 for that additional item.

Key concepts

Average Profit

The concept of average profit is fundamental in understanding a business's performance. It represents the profit per unit of product sold and is calculated by dividing the total profit by the number of items sold, \( \frac{P(x)}{x} \). Imagine a lemonade stand; if after a day's sales you determine your total profit and divide it by the number of cups sold, you'd get your average profit per cup.

In our exercise, we discovered that calculating the average profit for a particular number of units sold is relatively simple. When the company sells 500 items, for example, we found the average profit to be \(48. This means that for every item sold, out of the 500, the company effectively earns \)48 in profit on average. This is a crucial metric for businesses to understand how profitable their products are on the scale of unit sales.

Marginal Profit

Moving on to marginal profit, which is a concept from calculus invaluable when we consider changes in production levels. The marginal profit is the derivative of the profit function, denoted as \( \frac{dP(x)}{dx} \). It approximates how much additional profit a business will earn by selling one more unit. In the world of manufacturing, this is akin to asking, 'If I sell one more gadget, how much extra money will I make?'

From our example, we've deduced that when 500 units are sold, the marginal profit is \(70. This signals that after selling 500 gadgets, if one more gadget is sold, the business can expect approximately \)70 increase in profit just for that single additional unit. It reveals the incremental benefit of boosting production, which is essential when a company considers expanding or optimizing production quantities.

Cost Function

The cost function, often denoted as \( C(x) \), plays a pivotal role in a company's profitability analysis. It details the expenses involved in producing a certain number of units. In our textbook exercise, it was represented by the quadratic equation \( C(x) = -0.02x^2 + 50x + 100 \), where \( x \) is the number of items produced. This function is crucial because it incorporates both fixed costs (such as rent and salaries) and variable costs that increase with production, like materials and labor.

Understanding the shape and behavior of the cost function is key to making efficient production and pricing decisions. As it's a quadratic function in the exercise, the costs don't just increase linearly but accelerate after a certain point of production, emphasizing the importance of finding an optimal output level to maximize profits.

Derivative

Lastly, the concept of a derivative is a cornerstone of calculus and is central to understanding many areas of economics, including profit maximization. In essence, the derivative represents the rate of change of a function at any given point and is a measure of sensitivity to change. When applied to the profit function, it measures how much profit changes with each additional unit sold, which we refer to as marginal profit.

A good analogy here is driving a car: the derivative of your distance traveled with respect to time gives you your speed at any given moment. In our exercise, we calculated the marginal profit by finding the derivative of the profit function, which was a simple process of applying differentiation rules to the quadratic function to find \( 0.04x + 50 \). This step is essential in understanding how profit is affected as production quantity changes.