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The problem provided is asking to determine the quantity of products that maximizes the storage cost function, which is given by:

\[
C(x) = 1000x - \frac{x^2}{2}
\]

Where \( x \) represents the quantity of products in stock.

### Step-by-step solution:

1. **Identify the cost function**:
   \[
   C(x) = 1000x - \frac{x^2}{2}
   \]

2. **Find the derivative of the cost function**:
   To maximize or minimize the cost, we take the derivative of \( C(x) \) with respect to \( x \) and set it to zero:
   \[
   C'(x) = 1000 - x
   \]
   
3. **Set the derivative equal to zero** to find the critical points:
   \[
   1000 - x = 0
   \]
   Solving for \( x \), we get:
   \[
   x = 1000
   \]

4. **Second derivative test**:
   To confirm that this critical point is a maximum, we take the second derivative of \( C(x) \):
   \[
   C''(x) = -1
   \]
   Since \( C''(x) = -1 \) is negative, the function is concave down at \( x = 1000 \), meaning \( x = 1000 \) is a maximum.

Thus, the quantity of products that maximizes the cost is **1000 units**.

### Answer:
- 1000 unidades

Would you like further details or have any questions?

### 5 related questions:
1. What does the second derivative test tell us about the behavior of the function?
2. How would the cost function change if we added a linear term?
3. What is the significance of maximizing the cost in inventory management?
4. How can we minimize the cost instead of maximizing it?
5. What happens to the cost function as \( x \) approaches large values?

### Tip:
Always use the second derivative to verify whether a critical point is a maximum or minimum when working with optimization problems.

Uma empresa fez um levantamento sobre o seu sistema de estocagem e obteve um modelo matemático para o custo de armazenamento em função da quantidade de peças no estoque, considerando que o modelo do custo seja dado por, C(x) = 1000x - (x^2)/2. Onde x indica a quantidade de produtos no estoque. Determine a quantidade de produtos que maximiza o custo.

In this problem, you will learn how to maximize a storage cost function using calculus. We explore the derivative of the function C(x) = 1000x - (x^2)/2 and find that the maximum occurs at x = 1000. Suitable for advanced high school students.

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