Given $G$ connected linear algebraic group I want to find a $U \subset G$ maximal solvable subgroup which is not connected.
My attempt is to take $G=SO(n)$ for $n \geq 3$ and $U=D_{n}\cap SO(n)$: it'a trivial to see that $U \cong (\mathbb{Z}/_{2 \mathbb{Z}})^{n-1}$ so it's not connected and being abelian is solvable, but I cannot prove that it is not contained in any Borel subgroup of $G$.
If you can provide a different example your answer is accepted as well.
Thanks in advance.