Problem 21 - Calculus Single Variable

Question

In each of Exercises \(21-24,\) a cost function \(c\) and \(a\) form of the demand equation are given. Calculate the sales level \(x\) that maximizes profits. \(C(x)=1000+2 x, D^{-1}(x)=1602-8 x\)

Step-by-Step Solution

Answer

The sales level that maximizes profits is 100 units.

Step 1: Define the Revenue Function

The demand equation, given by the inverse demand function \(D^{-1}(x)\), provides the price \(p\) as a function of \(x\). Thus, \(p = 1602 - 8x\). The revenue function \(R(x)\) is the product of quantity \(x\) and price \(p\), so \(R(x) = x \times (1602 - 8x) = 1602x - 8x^2\).

Step 2: Define the Profit Function

The profit function \(\Pi(x)\) is given by the difference between the revenue function \(R(x)\) and the cost function \(C(x)\). Thus, \(\Pi(x) = R(x) - C(x) = (1602x - 8x^2) - (1000 + 2x) = 1600x - 8x^2 - 1000\).

Step 3: Determine the Critical Points

The maximum profit occurs at the critical points of \(\Pi(x)\). To find these, take the derivative of \(\Pi(x)\) with respect to \(x\), and set it equal to zero. The derivative is \(\frac{d\Pi}{dx} = 1600 - 16x\). Set \(1600 - 16x = 0\) and solve for \(x\).

Step 4: Solve for Maximum Sales Level x

Solving the equation \(1600 - 16x = 0\) gives us \(16x = 1600\). Dividing both sides by 16, we find \(x = 100\). Therefore, the sales level \(x\) that maximizes profit is 100.

Key Concepts

Cost Function

The cost function is an expression that describes the total cost of producing a certain quantity of goods. It is crucial for understanding how much it costs to produce at different levels of output. In the given exercise, the cost function is represented as \(C(x) = 1000 + 2x\). Here, the cost function has two components:

• A fixed cost of \(1000\), which remains constant regardless of how many units are produced.
• A variable cost of \(2x\), which depends on the quantity \(x\). The term \(2x\) implies that each unit produced adds \(2\) to the total cost.

Understanding the cost function helps businesses determine the break-even point and analyze how costs will change with different levels of production. Remember, knowing your cost structure is crucial for effective decision-making, as it directly impacts pricing and profit margins.

Revenue Function

A revenue function describes how much money a company earns from selling its products at different levels of output. In this exercise, the demand equation has been given in an inverse form, \(D^{-1}(x) = 1602 - 8x\), which defines price \(p\) in terms of quantity \(x\). To derive the revenue function \(R(x)\), we multiply the price by the quantity sold:

• Price is \(p = 1602 - 8x\)
• Revenue is \(R(x) = x imes (1602 - 8x)\)

This simplifies to \(R(x) = 1602x - 8x^2\). This quadratic equation shows that revenue initially increases with sales, but after a certain point, the negative term \(-8x^2\) causes it to decrease, reflecting the diminishing returns of selling too many units at a lower price.

Demand Equation

The demand equation is a crucial concept, as it links consumer demand for a product with its price. In mathematical problems, the inverse demand equation is often provided, like here, to directly determine price based on quantity. The equation is \( D^{-1}(x) = 1602 - 8x \).

• This is an inverse demand equation that shows how price decreases as quantity \(x\) increases.
• When \(x = 0\), the price is \(p = 1602\), which is the highest price a consumer is willing to pay if no units are sold.
• As \(x\) increases, the price consumers are willing to pay decreases by \(8\) for each additional unit.

This relationship not only helps determine optimal pricing but also indicates customer behavior and market capacity. Identifying how changes in price might affect demand can lead to more strategic pricing decisions.

Critical Points

Critical points are essential in finding maximum or minimum values for a function, which in business analysis translates to finding optimal sales levels or maximum profit. In this exercise, critical points are found by taking the derivative of the profit function with respect to \(x\). The profit function \(\Pi(x)\) is given by:
\[\Pi(x) = (1602x - 8x^2) - (1000 + 2x) = 1600x - 8x^2 - 1000\]To find the critical points, calculate the derivative \(\frac{d\Pi}{dx} = 1600 - 16x\). Setting this derivative to zero solves for \(x\) values where the slope of the profit function is zero (peak or valley points):

• Set \(1600 - 16x = 0\)
• Solve for \(x\), which gives \(x = 100\)

This means that at a sales level of \(100\), the profit is maximized. These calculations help businesses adjust their production and sales strategies to maximize profitability efficiently.