The largest number that cannot be made using a combination of $5$ and $11$?

This sort of question is usually framed in the form $x = ap + bq$, where $a, b > 0$ and $p, q$ are coprime. If you permit negative numbers then every number can be derived from the relation $1 = 1 \cdot 11 - 2 \cdot 5$, and hence $x = x \cdot 11 - 2x \cdot 5$, with transfers of $ 0 = 11 \cdot 5 - 5 \cdot 11$ to tidy things up.

The following supposes that $a, b$ are positive, in the standard case.

The largest number, that can not be made of the sum of a multiple of coprimes $p$ and $q$, is $pq-p-q$. Now, one simply sets $p=5, q=11$, and find that $pq-p-q = 39$.

Suppose $x = ap + bq$. Each multiple of $q$ leaves a different remainder when divided by $p$, which does not change when one adds multiplies of $p$ to it. This means that if $x$ leaves a remainder of $r$ when divided by $q$, there has been some $ap$ that leaves the same remainder. The last remainder to be taken is $q$ hapen with $p(q-1)$.

After this point, there is an $ap \le x$ that has the same remainder of $q$ as $x$ does.

The numbers not expressable in these terms is where one is looking at some $ap-bq$ or $bq-ao$. Since the highest number of this type is $p$ before $q(p-1)$, we find it to be $pq-p-q$.