Problem 24 - Calculus Single Variable

Question

A cost function \(c\) and \(a\) form of the demand equation are given. Calculate the sales level \(x\) that maximizes profits. \(C(x)=7000+x, D(p)=2402-2 p\)

Step-by-Step Solution

Answer

The sales level that maximizes profits is 1200 units.

Step 1: Understanding the Problem

To maximize profits, we need to find the sales level \(x\) where profit is maximum. Profit \(P(x)\) is calculated as revenue minus cost. The revenue \(R(x)\) comes from selling goods at price \(p\), i.e., \(R(x) = p \times x\). The cost function is given as \(C(x) = 7000 + x\), and the demand function is \(D(p) = 2402 - 2p\). We need to express \(x\) in terms of \(p\) using the demand function first.

Step 2: Relate Sales Level and Price

Using the demand function \(D(p) = 2402 - 2p\), which implies that \(x = D(p)\), we express \(x = 2402 - 2p\). This relation allows us to express \(x\) in terms of the price \(p\).

Step 3: Express Revenue in Terms of Price

Revenue \(R(x)\) can be written in terms of \(p\) by substituting for \(x\): \(R(p) = p \times (2402 - 2p) = 2402p - 2p^2\). This gives us the revenue as a function of price \(p\).

Step 4: Formulate the Profit Function

Profit \(P(x)\) is given by revenue minus cost: \(P(p) = 2402p - 2p^2 - (7000 + (2402 - 2p)) = 2402p - 2p^2 - 7000 - 2402 + 2p\), simplifying to \(P(p) = 2404p - 2p^2 - 9402\).

Step 5: Find the Derivative of Profit Function

To maximize the profit, find the derivative \(P'(p)\) with respect to \(p\): \(P'(p) = 2404 - 4p\).

Step 6: Solve for Critical Points

Set \(P'(p) = 0\) to find the critical points: \(2404 - 4p = 0\). Solving, we get \(p = \frac{2404}{4} = 601\).

Step 7: Determine the Sales Level

Substitute \(p = 601\) back into \(x = 2402 - 2p\): \(x = 2402 - 2 \times 601 = 2402 - 1202 = 1200\). Thus, the sales level that maximizes profits is \(x = 1200\).

Key Concepts

Cost Function

A cost function is a crucial component in economics and business modeling. It represents the total cost incurred by a company to produce a certain quantity of goods. In our exercise, the cost function is given as \(C(x) = 7000 + x\). This formula is structured to include:

• A fixed cost, which is \(7000\), representing costs that do not change with the level of output, such as rent or salaries.
• A variable cost, denoted by \(x\), which varies directly with the level of production or sales. This could include costs for materials or labor.

Understanding the cost function helps in determining how costs change with different levels of production. It allows businesses to plan their budget and pricing strategies effectively.

Demand Function

The demand function is key in understanding how the quantity demanded by consumers is influenced by the price of the goods. In this problem, our demand function is \(D(p) = 2402 - 2p\). This function tells us that:

• As the price \(p\) increases, the quantity demanded \(x\), represented by \(2402 - 2p\), decreases. This is a typical behavior, where higher prices can deter consumers.
• The coefficient \(-2\) in front of \(p\) illustrates how sensitive the quantity demanded is to changes in price. A steeper slope implies greater sensitivity.

The insight from the demand function allows businesses to understand market demand and adjust pricing to optimize sales and visibility.

Profit Function

The profit function is derived from calculating the difference between revenue and cost. It provides a clear picture of the financial gain or loss from business operations. Given by the formula \(P(p) = 2404p - 2p^2 - 9402\), this function originates from:

• Revenue, modeled as \(2404p - 2p^2\), which represents the total earnings from sales.
• Cost, modeled as \(7000 + (2402 - 2p)\).

Combining these parts shows us how profit changes with the price \(p\). By maximizing this function, businesses can determine the optimal selling price to achieve maximum profitability.

Derivative for Critical Points

The derivative is a mathematical tool used to find critical points where a function's value could be at its maximum or minimum. For profit maximization, we use the derivative of the profit function \(P(p)\).

• The derivative \(P'(p) = 2404 - 4p\) indicates the rate of change of profit with respect to price.
• Setting \(P'(p) = 0\) helps us find the point where the profit doesn't change, indicating a possible maximum or minimum.

Solving \(2404 - 4p = 0\), we find \(p = 601\). Substituting back, we determine the sales level \(x\) that maximizes profit. This shows how derivatives are vital in practical business strategies for finding optimal conditions.