Question: For the cost function $C(x) = 4000 + 520x + 0.2x^2$ and the demand function $p(x) = 1560$, find the production level that will maximize profit. Round to the nearest integer as needed.
Production Level =
1. **State the problem:** We want to find the production level $x$ that maximizes profit given the cost function $$C(x) = 4000 + 520x + 0.2x^2$$ and the demand function $$p(x) = 1560$$.
2. **Recall the profit formula:** Profit $P(x)$ is revenue minus cost. Revenue $R(x)$ is price times quantity, so
$$R(x) = p(x) \cdot x = 1560x$$
Profit is
$$P(x) = R(x) - C(x) = 1560x - (4000 + 520x + 0.2x^2)$$
3. **Simplify the profit function:**
$$P(x) = 1560x - 4000 - 520x - 0.2x^2 = (1560x - 520x) - 4000 - 0.2x^2 = 1040x - 4000 - 0.2x^2$$
4. **Find the critical points by differentiating:**
$$P'(x) = \frac{d}{dx}(1040x - 4000 - 0.2x^2) = 1040 - 0.4x$$
5. **Set derivative equal to zero to find maximum:**
$$1040 - 0.4x = 0$$
$$0.4x = 1040$$
$$x = \frac{1040}{0.4}$$
6. **Simplify the fraction:**
$$x = \frac{1040}{\cancel{0.4}} \times \frac{\cancel{10}}{10} = \frac{10400}{4}$$
7. **Calculate the value:**
$$x = 2600$$
8. **Check the second derivative to confirm maximum:**
$$P''(x) = -0.4 < 0$$ which means $P(x)$ is concave down and $x=2600$ is a maximum.
**Final answer:** The production level that maximizes profit is approximately **2600** units.