Here is the optimal solution for an arbitrary capacity $c$, amount $n$ and distance $x$:
First note that the path of the camel can be divided into segments determined by all the turning points. Then the number of times each same segment is crossed is even except for the first and last segment. Also the camel never needs to cross a former segment once it has crossed a later segment because otherwise it could have traversed the former segment first as there would have been a sufficient supply already brought to that point earlier. Thus there is one optimal solution such that the camel brings a certain amount in one or more trips from one point to the next and never returns to an earlier point. Also, there should not be any amount left at any point because otherwise that amount $r$ can be used to move the amount that is transported $t$ in $2 \lceil \frac{t}{c} \rceil - 1$ trips by the distance $\frac{r}{ 2 \lceil \frac{t}{c} \rceil - 1 }$, and it would also be optimal.
Now, let $f(c,n,x)$ be the maximum amount that can be brought a distance of $x$ where an amount $n$ and the camel with capacity $c$ are at the starting point. Then we have:
$f(c,n,x) = 0$ for all $n < x$
$f(c,n,0) = n$ for all $n \geq 0$
$f(c,n,x) = \max( \{ f( n - ( 2 \lceil \frac{n}{c} \rceil - 1 ) y , x - y ) : 0 < y \leq x \} )$ for all x > 0
The value of $f(c,n,x)$ is clear from the equivalent geometric interpretation:
Given point $(n,x)$, f(c,n,x) \geq f(c,n',x') for every point (n',x') where the line segment has gradient $\frac{ 1 }{ 2 \lceil \frac{n}{c} \rceil - 1 }$ and $0 \leq x' \leq x$
Since any solution corresponds to a series of such segments connected end to end, ending at $(m,0)$, and $f(m,0)$ is strictly increasing in $m$, and the gradient of the segments are strictly increasing, an optimal solution clearly must start a new segment when it is at each line $n=i$ where $i \in \mathbb{Z}^+$
Thus $ f(c,n,x) = \begin{cases} 0 & if & n < x \\\ n & if & n \geq x = 0 \\\ n - x ( 2 \lceil \frac{n}{c} \rceil - 1 ) & if & 2 \lceil \frac{n}{c} \rceil - 1 \leq r \\\ f( c , n - r , x - \frac{ r }{ 2 \lceil \frac{n}{c} \rceil - 1 } ) & if & otherwise \end{cases} $ where $r = n + c - c \lceil \frac{n}{c} \rceil$
For this example,
$f(1000,3000,1000)$
$ = f(1000,2000,\frac{4}{5}(1000))$
$ = f(1000,1000,\frac{7}{15}(1000))$
$ = \frac{8}{15}(1000) = 533\frac{1}{3}$
Interestingly, the maximum distance that the camel can go with a starting heap of $cn$ bananas is $c ( \frac{1}{2} ln(n) + O(1) )$, which means that an exponential number of bananas is needed with respect to the distance from the starting point!