To find the marginal cost at a given production level, we need to calculate the derivative of the cost function $C(x)$ with respect to $x$, and then evaluate that derivative at $x = 100$.

The given cost function is:

$C(x) = 15,000 + 20x + \frac{1,000}{x}$

Step 1: Differentiate the cost function

We differentiate each term of the cost function $C(x)$:

1. The derivative of $15,000$ is $0$ (since it is a constant).
2. The derivative of $20x$ is $20$.
3. The derivative of $\frac{1,000}{x}$ is found using the power rule. Since $\frac{1,000}{x} = 1,000x^{-1}$, the derivative is: $\frac{d}{dx} \left( 1,000x^{-1} \right) = -1,000x^{-2} = -\frac{1,000}{x^2}$

Thus, the marginal cost function is:

$C'(x) = 20 - \frac{1,000}{x^2}$

Step 2: Evaluate the marginal cost at $x = 100$

Now, substitute $x = 100$ into the marginal cost function:

$C'(100) = 20 - \frac{1,000}{100^2}$ $C'(100) = 20 - \frac{1,000}{10,000} = 20 - 0.1 = 19.9$

Step 3: State the units

The cost function $C(x)$ is in dollars, so the marginal cost $C'(x)$ is in dollars per unit of production. Since $x$ is the number of units produced, the units of marginal cost are dollars per unit.

Final Answer:

The marginal cost at a production level of $x = 100$ is $19.90 per unit.

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Would you like more details on how we computed the derivative or other examples?

Here are some related questions to deepen your understanding:

1. What is the meaning of the marginal cost in economics?
2. How would the marginal cost change if the cost function included a quadratic term like $x^2$?
3. What is the difference between fixed and variable costs, and how do they relate to the cost function?
4. How do you interpret the cost function when $x$ approaches very large values?
5. What happens to the marginal cost if the number of units produced approaches zero?

Tip: Always check the units in your functions before differentiating. Consistency in units helps ensure that your final answer makes sense in the context of the problem.