The problem gives the total cost function $C(x) = 0.5x^2 + 15x + 5000$, where $x$ represents the number of units produced. We are tasked with finding the level of production where the average cost per unit is minimized. The average cost per unit is given by $\text{AC}(x) = \frac{C(x)}{x}$.

Steps to solve:

1. Calculate the Average Cost Function: $\text{AC}(x) = \frac{C(x)}{x} = \frac{0.5x^2 + 15x + 5000}{x}$ Simplifying: $\text{AC}(x) = 0.5x + 15 + \frac{5000}{x}$

2. Minimize the Average Cost: To minimize the average cost function, we need to take its derivative with respect to $x$, set it equal to zero, and solve for $x$.

   The derivative of $\text{AC}(x)$ is: $\frac{d}{dx} \left( 0.5x + 15 + \frac{5000}{x} \right) = 0.5 - \frac{5000}{x^2}$

3. Set the Derivative Equal to Zero: $0.5 - \frac{5000}{x^2} = 0$ Solving for $x$: $\frac{5000}{x^2} = 0.5$ $x^2 = \frac{5000}{0.5} = 10000$ $x = \sqrt{10000} = 100$

Thus, the level of production that minimizes the average cost per unit is $x = 100$ units.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:

1. How would changes in fixed costs (e.g., 5000) affect the minimum average cost?
2. Can we find the minimum value of the average cost at $x = 100$?
3. What if the quadratic term had a negative coefficient (i.e., if costs decreased as production increased)?
4. How do changes in the coefficient of $x^2$ affect the shape of the cost curve?
5. How would this method change if the cost function included logarithmic or exponential terms?

Tip: When minimizing functions, always remember to check if your solution corresponds to a minimum by using the second derivative test.