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The average cost function helps us understand how much it costs to produce each unit, on average. It is essential for businesses to calculate their average costs, as it provides insights into pricing strategy and profitability.
To find the average cost function, we take the total cost function, denoted as \(C(x)\), and divide it by the number of units produced, \(x\). Thus, the average cost function, \(A(x)\), is given by:
\[ A(x) = \frac{C(x)}{x} = \frac{8x + 20 + \frac{x^3}{100}}{x} \] Simplifying, we get:
\[ A(x) = 8 + \frac{20}{x} + \frac{x^2}{100} \] This simplification reveals the contributions of different components to the average cost.

In calculus, a derivative represents the rate of change of a function with respect to a variable. For cost functions, the derivative gives us invaluable insights like marginal cost, which is the cost of producing one additional unit.
To find the derivative of the total cost function, \(C(x) = 8x + 20 + \frac{x^3}{100}\), we use basic derivative rules, such as the sum rule and the power rule:
\[ C'(x) = 8 + \frac{3x^2}{100} \] Similarly, for the average cost function \(A(x)\), we apply the quotient rule:
\[ A'(x) = 0 - \frac{20}{x^2} + \frac{2x}{100} = -\frac{20}{x^2} + \frac{x}{50} \]

Critical points are values of \(x\) where the derivative of a function is either zero or undefined. These points help us find maximum or minimum values of functions, which is crucial in cost analysis.
To locate critical points of the average cost function, we set its derivative equal to zero and solve for \(x\):
\[ -\frac{20}{x^2} + \frac{x}{50} = 0 \]
Simplify and solve:
\[ \frac{x}{50} = \frac{20}{x^2} \]
Multiplying both sides by \(x^2 \), we get: \[ x^3 = 1000 \]
Solving for \(x\), we find:
\[ x = 10 \]
Therefore, the critical point occurs at \(x = 10\).

Minimizing the cost function is crucial for businesses as it helps them achieve the lowest possible cost per unit, thereby maximizing profit. In this case, we focus on minimizing the average cost function.
To determine the minimum cost, we examine the critical point found from the average cost function's derivative. We know from previous calculations that:
\[ x_0 = 10 \]
Substitute \(x = 10\) into the total cost function's derivative:
\[ C'(10) = 8 + \frac{3(10)^2}{100} = 8 + 3 = 11 \]
Next, we calculate the average cost at \(x = 10\):
\[ A(10) = 8 + \frac{20}{10} + \frac{10^2}{100} = 8 + 2 + 1 = 11 \]
Here, the average cost \(A(10)\) equals the derivative of the total cost \(C'(10)\). This indicates that \[A(x_0) = C'(x_0) \], confirming the minimum average cost point.