Question: Suppose $C(x) = 0.03x^2 + 2x + 3500$ is the total cost of a company to produce $x$ units of a certain product. Find the production level $x$ that minimizes the average cost $A(x) = \frac{C(x)}{x}$. (Round the production level to two decimal places.)
x = __________ units
1. **State the problem:** We want to find the production level $x$ that minimizes the average cost function $$A(x) = \frac{C(x)}{x}$$ where $$C(x) = 0.03x^2 + 2x + 3500$$.
2. **Write the average cost function:**
$$A(x) = \frac{0.03x^2 + 2x + 3500}{x} = \frac{0.03x^2}{x} + \frac{2x}{x} + \frac{3500}{x} = 0.03x + 2 + \frac{3500}{x}$$
3. **Find the derivative of $A(x)$ to locate critical points:**
$$A'(x) = \frac{d}{dx}\left(0.03x + 2 + \frac{3500}{x}\right) = 0.03 - \frac{3500}{x^2}$$
4. **Set the derivative equal to zero to find critical points:**
$$0.03 - \frac{3500}{x^2} = 0$$
5. **Solve for $x^2$:**
$$0.03 = \frac{3500}{x^2} \implies 0.03x^2 = 3500$$
6. **Divide both sides by 0.03:**
$$\cancel{0.03}x^2 = \frac{3500}{\cancel{0.03}} \implies x^2 = \frac{3500}{0.03}$$
7. **Calculate $x^2$:**
$$x^2 = 116666.67$$
8. **Take the positive square root (since production level cannot be negative):**
$$x = \sqrt{116666.67} \approx 341.56$$
9. **Verify that this critical point is a minimum by checking the second derivative:**
$$A''(x) = \frac{d}{dx}\left(0.03 - \frac{3500}{x^2}\right) = \frac{7000}{x^3}$$
Since $x > 0$, $A''(x) > 0$, confirming a minimum.
**Final answer:** The production level that minimizes the average cost is approximately **$341.56$ units**.