Two variables $x$ and $y$ are related by the equation $8y = \left(\frac{x}{2} - 1\right)^4$. Given that both $x$ and $y$ vary with time, find the value of $x$ at the instant when the rate of change of $y$ is twice the rate of change of $x$.

1. Find the derivative of $y$ with respect to $x$.

$y = \frac{1}{8}\left(\frac{x}{2} - 1\right)^4$
$\frac{dy}{dx} = \frac{1}{8} \times 4\left(\frac{x}{2} - 1\right)^3 \times \frac{1}{2}$
$\frac{dy}{dx} = \frac{1}{4}\left(\frac{x}{2} - 1\right)^3$

2. Use the given relationship between the rates of change to find the value of $\frac{dy}{dx}$.

Given $\frac{dy}{dt} = 2\frac{dx}{dt}$.
By the chain rule for connected rates of change:
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$2\frac{dx}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dx} = 2$

3. Equate the derivative to 2 and solve for $x$.

$\frac{1}{4}\left(\frac{x}{2} - 1\right)^3 = 2$
$\left(\frac{x}{2} - 1\right)^3 = 8$
$\frac{x}{2} - 1 = \sqrt[3]{8}$
$\frac{x}{2} - 1 = 2$
$\frac{x}{2} = 3$
$x = 6$