The constraints on the small $x\times y$ rectangles are that $10=a\le x\le b=20$, $a\le y\le b$ and $xy\le c=250$. Observe that $ab\le c\le b^2$.
The quality achieved by covering $A\times B$ with $n$ rectangles can be measured by the "loss" $n-\frac{AB}c$. A covering with loss $<1$ is clearly optimal.
The following method describes how to find a covering that is optimal in many cases and guaranteed to use at most two rectangles more than an optimal solution.
Let us see what $A\times B$ rectangles can be "losslessly" reduced to smaller rectangles.
Case 1: If $B\ge A\ge 33\frac13=\frac{bc}{b^2-c}$, then $\frac{b A}c- \frac Ab=A\cdot \frac{b^2-c}{bc}\ge1$, hence there exists an integer $n$ such that
$$\tag1\frac {aA}c\le\frac Ab\le n\le \frac{b A}c\le\frac Aa,$$
where the outer inequalities follow from $ab\le c$.
From $(1)$ we conclude $a\le\frac An\le b $ and $a\le \frac{nc}A\le b$.
Therefore, $x:=\frac An$ and $y:=\frac {nc}A$ are valid side lengths for a small rectangle and because of $xy=c$, also the area constraint is obeyed.
Therefore we can use $n$ rectangles $x\times y$ to reduce the $A\times B$ rectangle losslessly to a $A\times (B-y)$ rectangle. Note that $B-y\ge A-b$.
Case 1a: If $A\ge B\ge 33\frac13$, we may apply case 1 symmetrically, thus reducing $A\times B$ to $(A-y)\times B$ with $A-y\ge B-b$.
Case 2: If $33\frac13>A\ge\frac{2c}b=25$ and $B\ge \frac{2c}A$ (note that $ \frac{2c}A\le b$), then $\frac Ab<2$ and $\frac{bA}c\ge 2$. Hence $(2)$ holds woith the choice $n=2$ and we can continue as in case 1, thus reducing $A\times B$ losslessly to $A\times (B-y)$ with $B-y\ge 0$.
Case 2a: If $33\frac13> B\ge 25$ and $A\ge\frac{2c}B$, by symmetry with case 2, we can achieve losslessly a reduction of $A\times B$ to $(A-y)\times B$ with $A-y\ge 0$.
Case 3: If $b\ge A\ge \frac cb = 12\frac12$ and $B\ge \frac cA$, then $(2)$ holds with $n=1$, thus we can reduce $A\times B$ to $A\times (B-y)$ with $B-y\ge0$.
Case 3a: If $b\ge B\ge 12\frac12$ and $A\ge \frac cB$, by symmetry we reduce to $(A-y)\times B$ with $A\ge 0$.
If we start with $A\ge 25$ and $B\ge 25$ and apply the above cases repeatedly while applicable, we will achieve a lossless reduction until reaching a rectangle $A'\times B'$.
Depending on what was the last case applied, we have
- if it was case 1, $A'\ge 33\frac13$ and $B'\ge A'-b\ge 13\frac13$; moreover $B'<33\frac13$ as otherwise case 1 or 1a would apply; but then also $B'<25$ as otherwise case 2a would apply; moreover, $B'>20$ as otherwise case 3a would apply; we conclude $A'\le B'+b<45$. Since $B'>20$, we need at least two rectangles vertically; since $B'<25$ we may choose (allowing overlap) $y=12\frac12$ and $x=20$. If $33\frac13\le A'\le40$, this produces a covering with 4 rectangles and loss $<4-\frac{33\frac13\cdot20}{250}=1\frac13$, hence at most one rectangle from optimum. If $40<A'\le 45$, this produces 6 rectangles and loss $<6-\frac{40\cdot20}{250}=2\frac45$, hence at most two rectangles from optimum.
- if it was case 1a, the situation is symmetric to above.
- if it was case 2, $25\le A'<33\frac13$ and $B'\ge 0$; moreover $B'<\frac{2c}{A'}\le b$ as otherwise case 2 would apply; but then also $B'<12\frac12$ as otherwise case 3a would apply. If $B'=0$, we are done with no loss. Otherwise, we may choose $y=12\frac12>B'$ and $x=20$. If $A'\le 20$, this produces a covering by one rectangle with loss $<1$, hence optimal. If $A'>20$, we need two rectangles and the loss is $< 2$, that is possibly one from optimum (but optimum if $A'B'>c$).
- if it was case 2a, the situation is symmetric to above.
- if it was case 3, $20\ge A'\ge 12\frac12$ and $B'<\frac c{A'}\le 20$; if $B'=0$, we are done (without any loss); otherwise we can complete the task with a single $A'\times \frac c{A'}$ rectangle; the loss $1-\frac{A'B'}\c$ is less than $1$ and hence the covering is optimal.
- if it was case 3a, the situation is symmetric to above.
In summary, the above method produces a covering that uses at most two rectangles more than an optimal covering as given by the area constraints. It does so for all $A\times B$ rectangles with $A\ge25$, $B\ge 25$. Personally I conjecture that even the coverings with loss $\ge1$ are truely optimal in many cases.
The method above neglects "long strips", i.e. where the shortest side is less than $25. For these, the "obvious" method should be optimal: If $A\le12\frac12$, stack $12\frac12\times 20$ rectangles. If $12\frac12