To find the value of \( x \) that minimizes the function \( a(x) = 2700x - 2x^2 \), we need to use **calculus**—specifically by finding the derivative of \( a(x) \) and setting it equal to zero.

### Steps:

1. **Differentiate the function** \( a(x) \) with respect to \( x \):
   \[
   a'(x) = \frac{d}{dx}(2700x - 2x^2)
   \]
   Applying the derivative rules:
   \[
   a'(x) = 2700 - 4x
   \]

2. **Set the derivative equal to zero** to find the critical point:
   \[
   2700 - 4x = 0
   \]

3. **Solve for \( x \)**:
   \[
   4x = 2700
   \]
   \[
   x = \frac{2700}{4} = 675
   \]

So, \( x = 675 \) is the critical point. Since this is a quadratic equation with a negative coefficient for \( x^2 \), this point corresponds to a **maximum** rather than a minimum. Therefore, this is the **largest** value of \( x \) where the function reaches its maximum.

Since you are asking for the "small" \( x \), the relevant context would need clarification—perhaps you are asking for another value or a minimum if one exists within a domain. Could you clarify?

### Questions for further exploration:
1. Would you like me to explain how I differentiated the function?
2. Would you like me to verify if there are any restrictions on \( x \)?
3. Do you want me to explore the behavior of the function for smaller values of \( x \)?
4. Should I plot the function to show how \( a(x) \) behaves visually?
5. Would you like to calculate the value of the function at \( x = 675 \)?

**Tip**: For maximizing or minimizing functions, always check the sign of the second derivative for confirmation.

Discover how to find the critical point of the function a(x) = 2700x - 2x^2 using derivatives. This problem involves calculus concepts like differentiation and optimization. Ideal for high school students in Grades 10-12, it includes step-by-step instructions and explores the behavior of the quadratic function.

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