Problem 18 - 5 Steps to A 5: AP Calculus AB 2021

Question

Given the cost function \(\mathrm{C}(x)=2500+0.02 x+0.004 x^{2},\) find the product level such that the average cost per unit is a minimum.

Step-by-Step Solution

Answer

Answer: The production level at which the minimum average cost per unit is achieved is 2500 units, and the minimum average cost is 3.

Step 1: Find the Average Cost function

The average cost function can be found by dividing the cost function C(x) by the production level x: $$ \text{AC} (x) = \frac{\text{C}(x)}{x} = \frac{2500+0.02x+0.004x^2}{x} $$

Step 2: Calculate the first derivative

The first derivative of the AC function can be found by using the quotient rule which states: $$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} $$ where u = 2500+0.02x+0.004x^2 and v = x Firstly, find the derivatives \(u'\) and \(v'\): $$ u' = \frac{d}{dx} (2500+0.02x+0.004x^2) = 0.02+0.008x $$ $$ v' = \frac{d}{dx} (x) = 1 $$ Now we apply the quotient rule and find the first derivative of AC: $$ \frac{d\text{AC}}{dx} = \frac{x(0.02+0.008x) - (2500+0.02x+0.004x^2)}{x^2} $$ Simplify the expression: $$ \frac{d\text{AC}}{dx} = \frac{0.02x + 0.008x^2 - 2500 - 0.02x - 0.004x^2}{x^2} = \frac{0.004x^2 - 2500}{x^2} $$

Step 3: Solve for the critical point

Set the first derivative equal to zero and solve for x: $$ \frac{0.004x^2 - 2500}{x^2} = 0 \Rightarrow 0.004x^2 = 2500 $$ Now we can solve for x: $$ x^2 = \frac{2500}{0.004} \Rightarrow x^2 = 625000 \Rightarrow x = 2500 $$ Thus, the critical point occurs at x = 2500.

Step 4: Calculate the minimum average cost per unit

Plug the critical point x = 2500 into the original AC function: $$ \text{AC}(2500) = \frac{2500+0.02(2500)+0.004(2500)^2}{2500} $$ After evaluating the expression, we get: $$ \text{AC}(2500) = 3 $$ So, the minimum average cost per unit is 3 when the production level is 2500 units.

Key Concepts

Cost Function

A cost function is a mathematical equation used to represent the total cost a company incurs in producing a specific number of goods. It is often expressed in terms of production output, denoted by \( x \). For instance, in this particular exercise, the cost function is given by \( C(x) = 2500 + 0.02x + 0.004x^2 \). This expression shows how total costs increase with the number of units produced.

The cost function usually includes:

• A fixed cost, which does not change with production level (here, 2500).
• A linear term that indicates variable costs per unit (here, 0.02x).
• A quadratic term that may represent increasing or diminishing returns to scale (here, 0.004x²).

By understanding the components of a cost function, we can better analyze how costs behave with changes in production levels. This is crucial for minimizing costs and optimizing production strategies.

Quotient Rule

The Quotient Rule is an essential calculus tool when dealing with derivatives of functions that are ratios of two differentiable functions. In simple terms, it helps us find the derivative of a fraction where both the numerator and denominator are functions of \( x \).

For a function \( \frac{u(x)}{v(x)} \), the Quotient Rule is expressed as: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} \]

Here's what this means:

• \( u \) is the numerator function.
• \( v \) is the denominator function.
• \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

In our example, the average cost (AC) was calculated as a ratio of the cost function and production level \( x \). The application of the Quotient Rule allowed us to find the derivative of this average cost function, crucial for determining where the average cost is minimized.

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. They provide important information about the behavior of a function, including potential minima or maxima.

To locate critical points, we:

• Take the derivative of the function.
• Equal this derivative to zero.
• Solve for the values of \( x \).

In the given problem, after applying the Quotient Rule and simplifying the derivative of the average cost function, we set it to zero to find \( x \), resulting in the equation \( 0.004x^2 - 2500 = 0 \). Solving these, we found \( x = 2500 \), identifying the critical point for this function. Critical points are a stepping stone to further analyze if these points are minima or maxima using second derivative tests or other methods.

Calculus Optimization

Calculus optimization involves using derivatives to find maximum or minimum values of a function within a given domain. This process is essential in determining optimal solutions in various fields, such as economics and engineering.

Optimization steps include:

• Finding the function to optimize (here, average cost).
• Calculating its derivative.
• Identifying critical points by setting the derivative to zero.
• Evaluating these points to decide whether they represent minima or maxima.

In our exercise, optimization was performed on the average cost function. After determining critical points, we verified that \( x = 2500 \) indeed led to a minimized average cost by evaluating it in the original average cost function. This optimization ensures that the production level set at 2500 units results in the lowest possible cost per unit, enhancing efficiency.