Maximal points on an $n$-dimensional ellipsoid For an $n$-dimensional ellipsoid in $\mathbb{R}^n$ centered at $\mathbf{0}_n$, defined by a set of vectors $\mathbf{v}_i$ giving the directions of the principle axes and $\lambda_i$ representing the corresponding magnitudes of each axis (for $i=1,\ldots, n$), is there a straightforward way to work out the maximal $x$-coordinate, for example, on the ellipsoid?
$\textbf{2D case:}$

For the 2D case let $\mathbf{v}_1=\left(x_1,\;y_1\right)^T$ with $\lambda_1=a$ represent the major axis and $\mathbf{v}_2=\left(x_2,\;y_2\right)^T$ with $\lambda_2=b$ represent the minor axis.
Here the geometry is not too complicated.  We can define a parametric equation for the x-coordinate of the curve in terms of an angle $t$ as
$$x=a\cos(t)\cos(c)-b\sin(t)\sin(c)$$
which with some basic calc and trig can be shown to be maximized at
$$ x_\text{max} = \frac{a^2\cos(c)^2+b^2-b^2\cos(c)^2}{a\cos(c)\sqrt{\frac{a^2\cos(c)^2+b^2-b^2\cos(c)^2}{a^2\cos(c)^2}}} $$
Similarly a maximal point for $y$ can also be obtained.
$\textbf{General case:}$
For the general case, however, I am stumped. I imagine an analytic result is pretty difficult, so any advice on a numerical approach for finding such a solution in general would also be appreciated.
 A: First, rewrite the equation of your ellipsoid using matrices:
$$\sum_{i=1}^n \frac{(\vec{x} \cdot \vec{v}_i )^2}{\lambda_i^2} = 1
\quad\iff\quad
\vec{x}^T \Lambda\,\vec{x} = 1
\quad\text{ where }\quad \Lambda = \sum_{i=1}^n\frac{\vec{v}_i \otimes \vec{v}_i}{\lambda_i^2}
$$
where $\otimes$ stands for outer product between two column vectors.
The unit normal vector $\vec{n}$ at a point $\vec{x}$ on the ellipsoid will be in the direction of the gradient:
$$\vec{n}\;\propto\; \vec{\nabla}(\vec{x}^T \Lambda\,\vec{x}) = 2 \Lambda\vec{x}$$
Let $\mu \in \mathbb{R}$ be the number such that
$$\Lambda \vec{x} = \mu \vec{n} \quad\iff\quad \vec{x} = \mu \Lambda^{-1} \vec{n}$$
We have
$$\mu^2 \vec{n}^T \Lambda^{-1} \vec{n} = \vec{x}^T\Lambda\vec{x} = 1
\quad\implies\quad
\mu = \frac{1}{\sqrt{\vec{n}^T\Lambda^{-1}\vec{n}}}
\quad\implies\quad
\vec{x} = \frac{\Lambda^{-1}\vec{n}}{{\sqrt{\vec{n}^T\Lambda^{-1}\vec{n}}}}
$$
Given any direction in the form of a unit vector $\vec{e}$, at the point $\vec{x}$ on the ellipsoid which maximizes $\vec{x}\cdot\vec{e}$, the corresponding normal vector $\vec{n}$ lies in the direction of $\vec{e}$. This means the extend of the ellipsoid
along direction $\vec{e}$ will be given by
$$\vec{e}\cdot\vec{x} = \vec{n}\cdot\vec{x} =
\sqrt{ \vec{n}\Lambda^{-1}\vec{n} } = \sqrt{ \vec{e}\Lambda^{-1}\vec{e} }$$
For example, to obtain the extend of the ellipsoid along the $x$-direction, one just need to compute the inverse matrix $\Lambda^{-1}$ and take the square root of its first diagonal element!
Update
Let us use the $2$-dim example in question as an example. 
Instead of using $c$ as the angle between the major- and $x$-axis, 
we will use $\theta$ instead.
Redefine $c$ now as $\cos\theta$ and let $s = \sin\theta$. we have
$$
\lambda_1 = a,\;\lambda_2 = b,\;
\vec{v}_1 = \begin{pmatrix} c \\ s\end{pmatrix},\;
\vec{v}_2 = \begin{pmatrix}-s \\ c\end{pmatrix}
$$
The matrix $\Lambda$ is now given by
$$\frac{1}{a^2} 
\begin{pmatrix}c \\ s\end{pmatrix} \otimes 
\begin{pmatrix}c \\ s\end{pmatrix} +
\frac{1}{b^2} 
\begin{pmatrix}-s \\ c\end{pmatrix} \otimes
\begin{pmatrix}-s \\ c\end{pmatrix}
= \frac{1}{a^2} 
\begin{pmatrix} c^2 & sc \\ sc & s^2\end{pmatrix} + 
\frac{1}{b^2}
\begin{pmatrix} s^2 & -sc\\-sc & c^2\end{pmatrix}
$$
Expand everything out, we find
$$\Lambda = \begin{pmatrix}
\frac{c^2}{a^2} + \frac{s^2}{b^2} & 
\frac{sc}{a^2} - \frac{sc}{b^2}\\
\frac{sc}{a^2} - \frac{sc}{b^2} &
\frac{s^2}{a^2} + \frac{c^2}{b^2}
\end{pmatrix}
\quad\implies\quad \Lambda^{-1} = \frac{1}{\Delta} 
\begin{pmatrix}
\frac{s^2}{a^2} + \frac{c^2}{b^2}  & 
\frac{sc}{b^2} - \frac{sc}{a^2}\\
\frac{sc}{b^2} - \frac{sc}{a^2} &
\frac{c^2}{a^2} + \frac{s^2}{b^2}
\end{pmatrix}
$$
where $\Delta = \det\Lambda = \frac{1}{a^2b^2}$.
From this, we can read off the maximal $x$-extend of the ellipse as
$$
x_{max} 
= \sqrt{\begin{pmatrix}1 \\ 0\end{pmatrix}^T \Lambda^{-1}\begin{pmatrix}1 \\ 0\end{pmatrix}}
= ab \sqrt{\frac{s^2}{a^2} + \frac{c^2}{b^2}} 
= \sqrt{a^2\cos^2\theta + b^2\sin^2\theta}
$$
This is a simplified version of the expression of $x_{max}$ in question.
A: The problem is to maximize $\langle x,e_1\rangle$ subject to $\sum_i \lambda_i^2\langle x,v_i\rangle^2 = 1$.  (Well, $\le 1$, but by convexity the extreme value is attained on the boundary.)  Since the $v_i$ are an orthonormal basis,
$$ \langle x,e_1\rangle = \sum_i \langle x,v_i\rangle \langle e_1,v_i\rangle
= \sum_i \lambda_i \langle x,v_i\rangle \cdot \frac{\langle e_1,v_i\rangle}{\lambda_i} $$
Now use Cauchy-Schwarz (the equality case of which will tell you enough to get a formula for $x = \sum_i \langle x,v_i\rangle v_i$).
