Given a complex simple Lie algebra $\mathfrak{g}$ of rank $n\in\mathbb{N}$ with $n$ sufficiently large (say $n\ge10$), is there a way to determine whether $\mathfrak{g}$ contains a simple subalgebra of a prescribed type with rank "close" to $n$? For example, if $\mathfrak{g}$ is of type $B_n$ with $n\ge10$, does $\mathfrak{g}$ contain a subalgebra of type $C_{n2}$?

3$\begingroup$ Some more tags are natural here, such as rt.representationtheory. Many of us rely on the tags to identify questions which may be interesting. (Also, it's useful to highlight the basic question being asked with > followed by space.) $\endgroup$– Jim HumphreysSep 11 '18 at 0:04
The answer to the question in your example is no in general: $B_n$ does not contain $C_{n2}$ for large $n$. To see this, observe that $B_n$ has an irreducible orthogonal representation $V$ of dimension $2n+1$. The Weyl dimension formula shows that for any simple Lie algebra, the dimension of an irreducible representation is the smallest for fundamental representations.
[Edited]
For large $n$, the smallest dimensional fundamental representation of $C_{n2}$ is the standard one, of dimension $2n4$. Therefore the restriction of $V$ to $C_{n2}$ can only be some copies of the trivial representation and ONE copy of the standard representation $W$ of dimension $2n4$. This means that the group of type $C_{n2}$ must preserve both a symplectic form and a quadratic form on $W$, which is impossible by Schur's lemma. Hence the restriction of $V$ to $C_{n2}$ can only be a sum of copies of the trivial representation which is impossible.
I had previously given a wrong reason (assuming that the rep of $B_n$ had dimension $n$; thanks to @BS for pointing this out.

2$\begingroup$ I thought $B_n$ was $so(2n+1)$. What is its $n$dim irrep ? $\endgroup$– BS.Sep 16 '18 at 8:16

1$\begingroup$ Thank you. You are right (I had a brain fade). I have given the correct reason now, I hope. $\endgroup$ Sep 16 '18 at 8:45

2$\begingroup$ Schur's lemma usually refers to the statement that the space of equivariant endomorphisms of an irreducible representation is onedimensional. How does that implies that $C_{n2}$ cannot preserve two binary forms? $\endgroup$ Sep 16 '18 at 9:06

1$\begingroup$ The existence of an invariant Bilinear form on W gives an equivariant map between W and it’s dual. By Schur’s lemme, there is only one such up to scalars $\endgroup$ Sep 16 '18 at 10:00
I am not sure if this is exactly what you are looking for, but there have been some classic works, developing general methods for such topics:
 In Dynkin, Semisimple subalgebras of semisimple Lie algebras, Mat. Sb. (N.S.), 1952, Volume 30(72), Number 2, p. 349–462, the author among others classifies (up to linear equivalence) the simple Lie subalgebras of the Lie algebras of exceptional type. The original paper is in Russian (and is free) but you can find a translation in the translation series of AMS here (not free). Some of these results have been refined and generalized in a 2006 paper by Minchenko (but i do not have available the exact reference right now),
 In M. Lorente, B. Gruber, Classification of Semisimple Subalgebras of Simple Lie Algebras, Journal of Mathematical Physics 13, 1639 (1972), the authors extend Dynkin's methods and classify the simple subalgebras of Lie algebras of classical type of rank $\leq 6$, up to linear equivalence,
 In W. A.de Graaf, Constructing semisimple subalgebras of semisimple Lie algebras, Journal of Algebra, v. 325, 1, 2011, p.416430, general methods are obtained for classifying the semisimple subalgebras of a given semisimple Lie algebra, up to linear equivalence. Some of these methods are then applied to obtain a classification of the semisimple subalgebras of the simple Lie algebras of rank $\leq 8$.
Simple algebras of rank >8 are classical, so you are asking is there a representation (linear, orthogonal or symplectic) of a given dimension of a prescribed Lie algebra. This amounts to the question what is the minimal dimension of a nontrivial linear, orthogonal or symplectic representation of the Lie algebra. A table in Bourbaki answers this.

3$\begingroup$ Can you be more specific about the location of the "table in Bourbaki"? $\endgroup$ Sep 10 '18 at 23:59

2$\begingroup$ Tables I and II in "Groupes et algèbres de Lie, Chapitres 7 et 8" (I use Russian translation but I hope the tables are in the original text as well and not added by a translator). $\endgroup$ Sep 11 '18 at 12:08

$\begingroup$ Since my copy is the first printing of the French edition, I can confirm that these tables occur at the end of the text of Chapter 8 before the exercises. (There is an obvious misprint $F_6$ for $E_6$ in Table II.) Note however that the tables claim only to provide data for the fundamental representations in each Lie type. $\endgroup$ Sep 11 '18 at 13:24

2$\begingroup$ The table in Bourbaki yields the fundamental reps, their dimension, and say if they are orthogonal, symplectic, or none. If $V$ is any rep of dimension $d$, then $V\oplus V$ can be made both orthogonal and symplectic. So how do we conclude? the issue is that possibly, if there's a $d$dimensional rep, are there nonfundamental irreducibles of dimensions $<2d$? is there a way to list them, and know if they're orthogonal/symplectic? $\endgroup$– YCorSep 11 '18 at 16:10

1$\begingroup$ For the last part of the question, Proposition 12 in Chapter 8 of Bourbaki gives a nice criterion to be orthogonal or symplectic in terms of the coordinates of the highest weight of a representation. The dimension, as mentioned above, can be computed via the Weyl dimension formula. But I don't know an explicit procedure how to list all representations of dimension less than a given bound. $\endgroup$ Sep 11 '18 at 18:09