In a new related question I give a conjecture for the unit square which agrees with the $k=5$ and $k=8$ solutions here.

If $s^2 \lt k \lt (s+1)^2$ then the optimal solution has $s$ or $s+1$ rows (depending on which square is closer) each with $s$ or $s+1$ rectangles. More specifically:

If $k=s^2+t$ with $0 \lt t \le s$ then the optimal solution has $s$ rows with $s-t$ rows having $s$ rectangles $b \times \frac1s $ and $t$ rows of $s+1$ rectangles $a \times \frac1{s+1}$ where $b^2+\frac{1}{s^2}=a^2+\frac1{(s+1)^2}$ and $(s-t)b+ta=1$

But if $k=s^2+t$ with $s \le t \lt 2s+1$ then the optimal solution has $s+1$ rows with $2s+1-t$ rows having $s$ rectangles $a \times \frac1s $ and $t-s$ rows of $s+1$ rectangles $\frac1{s+1} \times b$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(2s-t+1)a+(t-s)b=1$

Note that in case $k=s^2+s,$ either description gives all rectangles $\frac1s \times \frac1{s+1}.$

In a new related question I give a conjecture for the unit square which agrees with the $k=5$ and $k=8$ solutions here.

If $s^2 \lt k \lt (s+1)^2$ then the optimal solution has $s$ or $s+1$ rows (depending on which square is closer) each with $s$ or $s+1$ rectangles. More specifically:

If $k=s^2+t$ with $0 \lt t \le s$ then the optimal solution has $s$ rows with $s-t$ rows having $s$ rectangles $b \times \frac1s $ and $t$ rows of $s+1$ rectangles $a \times \frac1{s+1}$ where $b^2+\frac{1}{s^2}=a^2+\frac1{(s+1)^2}$ and $(s-t)b+ta=1$

But if $k=s^2+t$ with $s \le t \lt 2s+1$ then the optimal solution has $s+1$ rows with $2s+1-t$ rows having $s$ rectangles $a \times \frac1s $ and $t-s$ rows of $s+1$ rectangles $\frac1{s+1} \times b$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(2s-t+1)a+(t-s)b=1$

Note that in case $k=s^2+s,$ either description gives all rectangles $\frac1s \times \frac1{s+1}.$

In a new related question I give a conjecture for the unit square which agrees with the $n=5$ and $n=8$ solutions here.

If $s^2 \lt n \lt (s+1)^2$ then the optimal solution has $s$ or $s+1$ rows (depending on which square is closer) each with $s$ or $s+1$ rectangles. More specifically:

If $n=s^2+t$ with $0 \le t \lt s$ then the optimal solution has $s$ rows with $s-t$ rows having $s$ rectangles $a \times \frac1s $ and $t$ rows of $s+1$ rectangles $b \times \frac1{s+1}$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(s-t)a+tb=1$

But if $n=s^2+t$ with $s \le t \lt 2s+1$ then the optimal solution has $s+1$ rows with $2s+1-t$ rows having $s$ rectangles $a \times \frac1s $ and $t-s$ rows of $s+1$ rectangles $\frac1{s+1} \times b$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(2s-t+1)a+(t-s)b=1.$

In a new related question I give a conjecture for the unit square which agrees with the $k=5$ and $k=8$ solutions here.

If $s^2 \lt k \lt (s+1)^2$ then the optimal solution has $s$ or $s+1$ rows (depending on which square is closer) each with $s$ or $s+1$ rectangles. More specifically:

If $k=s^2+t$ with $0 \lt t \le s$ then the optimal solution has $s$ rows with $s-t$ rows having $s$ rectangles $b \times \frac1s $ and $t$ rows of $s+1$ rectangles $a \times \frac1{s+1}$ where $b^2+\frac{1}{s^2}=a^2+\frac1{(s+1)^2}$ and $(s-t)b+ta=1$

But if $k=s^2+t$ with $s \le t \lt 2s+1$ then the optimal solution has $s+1$ rows with $2s+1-t$ rows having $s$ rectangles $a \times \frac1s $ and $t-s$ rows of $s+1$ rectangles $\frac1{s+1} \times b$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(2s-t+1)a+(t-s)b=1$

Note that in case $k=s^2+s,$ either description gives all rectangles $\frac1s \times \frac1{s+1}.$