Problem 25 - Applied Calculus

Question

(a) Production of an item has fixed costs of \(\$ 10,000\) and variable costs of \(\$ 2\) per item. Express the cost, \(C\), of producing \(q\) items. (b) The relationship between price, \(p\), and quantity, \(q\), demanded is linear. Market research shows that 10,100 items are sold when the price is \(\$ 5\) and 12,872 items are sold when the price is \(\$ 4.50 .\) Express \(q\) as a function of price \(p\). (c) Express the profit earned as a function of \(q\). (d) How many items should the company produce to maximize profit? (Give your answer to the nearest integer.) What is the profit at that production level?

Step-by-Step Solution

Answer

Produce approximately 10,930 items to maximize profit, with a profit of about $42,786.

Step 1: Determine Cost Function

The total cost \( C \) of producing \( q \) items includes both fixed costs and variable costs. The fixed costs are \\(10,000 and the variable costs are \\)2 per item. So, the cost function is:\[C(q) = 10000 + 2q\]

Step 2: Establish Relationship Between Quantity and Price

To express \( q \) as a function of price \( p \), use the two data points given: (\( q = 10100, p = 5 \)) and (\( q = 12872, p = 4.5 \)). These points provide us with the slope of the line. Calculate the slope using:\[m = \frac{\Delta q}{\Delta p} = \frac{12872 - 10100}{4.5 - 5} = -5544\]Then use the point-slope form of the line equation:\[q - 10100 = -5544(p - 5)\]Simplifying this gives:\[q(p) = -5544p + 38120\]

Step 3: Define Profit Function

Profit \( P \) is the difference between total revenue \( R \) and total cost \( C \). Revenue is \( R = p \cdot q \). From Step 2, \( q = -5544p + 38120 \). Hence, \( R = p(-5544p + 38120) \). Substitute \( C \) from Step 1 to get the profit function:\[P(q) = R - C = p(-5544p + 38120) - (10000 + 2q)\]Since \( q = -5544p + 38120 \), you can replace in the final simplified form.

Step 4: Optimize Profit Function

To maximize profit, differentiate the profit function \( P \) with respect to \( q \), and set the derivative to zero to find the critical points. Solve the equation to find the value of \( q \) that maximizes \( P \). Then substitute the optimized quantity \( q \) back into the profit function to find the maximum profit. This calculation involves taking the derivative and critical point analysis.

Step 5: Calculate Maximum Profit

After calculating the critical point \( q \), ensure the second derivative test or confirmation of maxima is done by plugging back the values in the calculated profit function. This will give you the highest profit at that particular level of production, rounded to the nearest integer.

Key Concepts

Cost Function

The cost function is a fundamental concept in optimization problems, especially in calculus. It helps us calculate the total cost incurred by producing a specific number of items. In any production process, there are usually two main types of costs:

• Fixed Costs: These are costs that do not change with the level of output. For instance, rent or salaries might stay constant. In our example, the fixed cost is \(\\(10,000\).
• Variable Costs: These depend on the number of items produced. It might involve costs of materials or hourly wages. Here, it's \(\\)2\) per item.

To find the total cost of producing \(q\) items, we sum the fixed and variable costs, giving us the cost function: \[C(q) = 10000 + 2q\]
This linear function indicates that for every additional item produced, the total cost increases by \(\$2\). Understanding this function helps businesses manage expenses efficiently.

Revenue Function

Revenue function is crucial when analyzing a company's earnings from selling products. It tells us the total income from selling a certain number of items. Revenue is calculated as the product of price per item and the quantity sold.

• Price per Item (p): The amount of money customers pay for a single item.
• Quantity Sold (q): The number of items sold.

So, the revenue function can be expressed as: \[R(q) = p \times q\]
From the problem, we learn that demand affects revenue—the more the demand, the lower the price, but higher the quantity sold. Solving these relationships can help businesses set optimal pricing to maximize revenue.

Profit Function

Profit function is what businesses aim to maximize. It's the difference between total revenue and total cost. With our earlier functions, profit can be defined as:

• \(R(q)\) representing the income earned.
• \(C(q)\) representing the production expenses.

This leads to the profit function:\[P(q) = R(q) - C(q)\]
In the exercise, this becomes more detailed, using the inverse demand function to help understand how changing variables like price and quantity can impact profit. For maximizing profit, calculus methods such as taking derivatives are employed to find optimal production levels.

Linear Demand Curve

A linear demand curve is a simple representation of how price affects the quantity demanded. This curve slopes downward, indicating that as the price decreases, demand increases. This relationship is expressed linearly:\[q(p) = m \times p + b\]
In our exercise, we determined the slope (\(-5544\)), which shows the rate of change in quantity demanded per unit change in price. Using data points like price-quantity combinations, we derive the equation:\[q(p) = -5544p + 38120\]
This linear form helps predict quantity demanded for any given price, aiding in decisions like pricing strategies and inventory management. By understanding demand, companies can tailor their economic strategies for better profit margins.