# Let x &#x2208;<!-- ∈ --> <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="doub

Let $x\in {\mathbb{R}}^{n}$ and let $P$ be a $n×n$ positive definite symmetrix matrix. It is known that the maximum of
$\begin{array}{ll}\text{maximize}& {x}^{T}P\phantom{\rule{thinmathspace}{0ex}}x\\ \text{subject to}& {x}^{T}x\le 1\end{array}$
is ${\lambda }_{\text{max}}\left(P\right)$, the largest eigenvalue of $P$. Now consider the following problem
$\begin{array}{ll}\text{maximize}& {x}^{T}P\phantom{\rule{thinmathspace}{0ex}}x\\ \text{subject to}& \left(x-a{\right)}^{T}\left(x-a\right)\le 1\end{array}$
where $a\in {\mathbb{R}}^{n}$ is known. What is the analytical solution?
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hospitaliapbury
This is (a variant of) the trust-region problem, and the solution can be computed rather easily, although not an analytical solution
I'm reparameterizing your problem slightly, you can easily change your model to be $max{x}^{T}Qx+{c}^{T}x+b$ subject to ${x}^{T}x\le 1$. At optimality, you will be at the border of feasibility, and the constraint gradient will be pointing in the same direction as the objective gradient (draw this geometrically!), i.e., you have $2Qx+c=\lambda 2x$ for some unknown $\lambda$. Solve for $x$ to obtain $x=-\left(2Q-\lambda I{\right)}^{-1}c$. To find lambda, solve ${x}^{T}x=1$ which is an equation in $\lambda$ (possibly tricky, numerical methods required, line-search, bisection etc)