To find the value of $x$ that maximizes revenue, let's go through the steps. Revenue, $R(x)$, is the product of price $p$ and quantity $x$:

$R(x) = x \cdot p = x \cdot D(x)$

Given: $p = D(x) = 126 - 2\sqrt{x}$

Thus, the revenue function $R(x)$ becomes: $R(x) = x \cdot (126 - 2\sqrt{x})$ or $R(x) = 126x - 2x\sqrt{x}$

To maximize $R(x)$, take the derivative of $R(x)$ with respect to $x$ and set it to zero.

Step 1: Differentiate $R(x)$

1. Differentiate $126x$ with respect to $x$: $126$.
2. For $-2x\sqrt{x}$, rewrite $x\sqrt{x}$ as $x^{3/2}$, then differentiate: $\frac{d}{dx}(-2x^{3/2}) = -3x^{1/2}$

So: [ R'(x)