Let $n=100, r=4$. Let $d=[2,7, 17, 22, 47]$. For each $d$, let $l= \frac{n}{r+d-1}$ and $t=\lfloor(\frac{r+d-2}{r})\rfloor$.
Here as the value of $d$ increases $l$ decreases and $t$ increases. And the product $l \times t$ is maximum when $d$ is the largest.
$$\begin{array}{|c|c|c|c|} d & l & t & l\times t \\ \hline 2 & 20 & 1 & 20 \\ 7 & 10 & 2 & 20 \\ 17 & 5 & 4 & 20 \\ 22 & 4 & 6 & 24 \\ 47 & 2 & 12 & 24 \\ \end{array}$$
In general is there a way to prove it that this product will be maximum when $d$ is large ?
Simplifying everything we can, $$l \times t =\left(\frac{100}{x + 3}\right)\left\lfloor\frac{x + 2}{4}\right\rfloor$$ and this is similar to a rational function. To maximize this value as much as we can, set some $x$ such that $(x + 2)/4$ is an integer. That is, set $x = -2 + 4c$ for $c \in \mathbb{Z}$. By rewriting the product, we have $$l \times t = \frac{100c}{4c + 1}.$$
Solving for the horizontal asymptote gives us $y = 25$ (assuming $x$-versus-$y$ plot) and since we are using values of $x$ greater than zero, therefore, $l \times t$ does not have a maximum regardless of how large $x$ is.
You can probably call $y = 25$ the supremum though.