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.