Given a vector $\vec{c}$ and a radius $r$, solve the problem:
\begin{equation*} \begin{aligned} & \underset{x}{\text{maximise}} & & \vec{c} \cdotp \vec{x}=a \\ & \text{subject to} & & ||x^2||=r^2 \end{aligned} \end{equation*}
Given a vector $\vec{c}$ and a radius $r$, solve the problem:
\begin{equation*} \begin{aligned} & \underset{x}{\text{maximise}} & & \vec{c} \cdotp \vec{x}=a \\ & \text{subject to} & & ||x^2||=r^2 \end{aligned} \end{equation*}
Let $\theta$ be the angle between $\vec{x}$ and $\vec{c}$. Then assuming the constraint is $\|\vec{x}\|^2 = r^2$, we have
$$ \vec{c} \cdot \vec{x} = \|\vec{c}\| \|\vec{x}\| \cos\theta = \|\vec{c}\| |r| \cos\theta $$
which is maximised when $\theta = 0$.
The geometric interpretation is that for $\vec{x}$ of fixed length $|r|$, the dot product $\vec{c} \cdot \vec{x}$ is maximised by taking $\vec{x}$ pointing in the same direction as $\vec{c}$,
$$ \vec{x} = |r| \frac{\vec{c}}{\|\vec{c}\|} .$$