TITLE: Decomposing square of side length $n$ into $n$ squares in a certain "maximal" way
QUESTION [10 upvotes]: I was wondering if anything is known about this problem. We are given a square of side length $n$ and we wish to embed $n$ smaller (integer) squares inside it such that the sum of the side-lengths of the squares inside is maximised. Let $f(n)$ be the maximum sum of side-lengths one can produce in this way. I have computed some values of $f(n)$, and have displayed them below. Apologies for the handwritten drawing, but I think the picture does help explain what I'm trying to get at.

The inequalities I have highlighted in red are written so because I am unsure if this is the maximum that can be achieved. In particular, for $n=8$ we don't even use $8$ squares, but I cannot figure out a configuration of $8$ squares that beats $20$. (I have reason to believe that $f(8)\leq 21$, and it would be nice if this could be attained with equality.) The values I have written as equality I am more sure about, based on exhaustive computer simulation. Is anything known about this function $f$? Even basic upper and lower bounds would be really helpful. It certainly seems to be roughly $n^{3/2}$.
Update. I made my simulations a bit more efficient, and they seem to indicate that $f(8) = 20$ and $f(10)=30$ are indeed equalities. We also have $f(11)\geq 35$ (and this is likely equality, since my program is taking a while trying to find an example with sum $36$.).
Edit. I appreciate all of the comments! Since the constraints of original question may be a bit too difficult (and possibly artificially so), I would like to weaken the requirements slightly. In this new version of the problem, we have a rectangle of integer side lengths $a$ and $b$ and wish to fill it with $c$
squares in order to maximise the sum of side lengths of the inner squares. Thus the original problem was when $a = b = c = n$. The weaker assumptions may make it easier to form an inductive argument. I am still working on this, but would appreciate any help!

REPLY [8 votes]: $f(8)=21$:

The other reported values up through $n=11$ are optimal.  You can solve the problem via integer linear programming as follows.  For square with top left corner $(i,j)$ and side length $s$, let binary decision variable $x_{i,j,s}$ indicate whether that square is used.  The problem is to maximize $\sum_{i,j,s} s x_{i,j,s}$ subject to
\begin{align}
\sum_{i,j,s} x_{i,j,s} &= n \tag1 \\
\sum_{\substack{i,j,s:\\ i \le i_0 \le i+s-1 \\ j \le j_0 \le j+s-1}} x_{i,j,s} &\le 1 &&\text{for $(i_0,j_0)\in \{1,\dots,n\}\times \{1,\dots,n\}$} \tag2 \\
\end{align}
Constraint $(1)$ forces $n$ squares to be used.
Constraint $(2)$ prevents covering cell $(i_0,j_0)$ more than once.
Here are results for $n \le 50$, where $u_1(n)$ is the one-dimensional integer knapsack upper bound and $u_2(n)$ is the one-dimensional continuous knapsack upper bound:
\begin{matrix}
n  &  \lfloor \sqrt{n}\rfloor^3  &  f(n)  & u_1(n)  &  \lfloor u_2(n) \rfloor   &  u_2(n) &  \lfloor n^{3/2} \rfloor \\
\hline
1  &  1  &  1  &  1  &  1  &  1  &  1  \\
2  &  1  &  2  &  2  &  2  &  2.666666667 &  2  \\
3  &  1  &  4  &  5  &  5  &  5  &  5  \\
4  &  8  &  8  &  8  &  8  &  8  &  8  \\
5  &  8  &  10 &  11 &  11 &  11 &  11 \\
6  &  8  &  14 &  14 &  14 &  14.4  &  14 \\
7  &  8  &  17 &  18 &  18 &  18.2  &  18 \\
8  &  8  &  21 &  22 &  22 &  22.4  &  22 \\
9  &  27 &  27 &  27 &  27 &  27 &  27 \\
10 &  27 &  30 &  31 &  31 &  31.42857143 &  31 \\
11 &  27 &  35 &  36 &  36 &  36.14285714 &  36 \\
12 &  27 &  40 &  41 &  41 &  41.14285714 &  41 \\
13 &  27 &  45 &  46 &  46 &  46.42857143 &  46 \\
14 &  27 &  51 &  52 &  52 &  52 &  52 \\
15 &  27 &  56 &  57 &  57 &  57.85714286 &  58 \\
16 &  64 &  64 &  64 &  64 &  64 &  64 \\
17 &  64 &  68 &  69 &  69 &  69.88888889 &  70 \\
18 &  64 &  75 &  76 &  76 &  76 &  76 \\
19 &  64 &  80 &  82 &  82 &  82.33333333 &  82 \\
20 &  64 &  88 &  88 &  88 &  88.88888889 &  89 \\
21 &  64 &  94 &  95 &  95 &  95.66666667 &  96 \\
22 &  64 &  100   &  102   &  102   &  102.6666667 &  103   \\
23 &  64 &  108   &  109   &  109   &  109.8888889 &  110   \\
24 &  64 &  115   &  117   &  117   &  117.3333333 &  117   \\
25 &  125   &  125   &  125   &  125   &  125   &  125   \\
26 &  125   &  130   &  132   &  132   &  132.3636364 &  132   \\
27 &  125   &  138   &  139   &  139   &  139.9090909 &  140   \\
28 &  125   &  145   &  147   &  147   &  147.6363636 &  148   \\
29 &  125   &  154   &  155   &  155   &  155.5454546 &  156   \\
30 &  125   &  162   &  163   &  163   &  163.6363636 &  164   \\
31 &  125   &  170   &  171   &  171   &  171.9090909 &  172   \\
32 &  125   &  178   &  180   &  180   &  180.3636364 &  181   \\
33 &  125   &  186   &  189   &  189   &  189   &  189   \\
34 &  125   &  195   &  197   &  197   &  197.8181818 &  198   \\
35 &  125   &  204   &  206   &  206   &  206.8181818 &  207   \\
36 &  216   &  216   &  216   &  216   &  216   &  216   \\
37 &  216   &  222   &  224   &  224   &  224.8461539 &  225   \\
38 &  216   &  232   &  233   &  233   &  233.8461539 &  234   \\
39 &  216   &  240   &  243   &  243   &  243   &  243   \\
40 &  216   &  250   &  252   &  252   &  252.3076923 &  252   \\
41 &  216   &  259   &  261   &  261   &  261.7692308 &  262   \\
42 &  216   &  270   &  271   &  271   &  271.3846154 &  272   \\
43 &  216   &  278   &  281   &  281   &  281.1538462 &  281   \\
44 &  216   &  288   &  291   &  291   &  291.0769231 &  291   \\
45 &  216   &  299   &  301   &  301   &  301.1538462 &  301   \\
46 &  216   &  308   &  311   &  311   &  311.3846154 &  311   \\
47 &  216   &  319   &  321   &  321   &  321.7692308 &  322   \\
48 &  216   &  329   &  332   &  332   &  332.3076923 &  332   \\
49 &  343   &  343   &  343   &  343   &  343   &  343   \\
50 &  343   &  350   &  353   &  353   &  353.3333333 &  353   \\
\end{matrix}