For $1 \le i \le 20$ and $1 \le j \le 11$ we define $a_{i,j}$ as follows; $a_{i,j} = 1$ if the $i$-th mathematics problem and $j$-th physics problem are chosen by some student, $a_{i,j} = 0$ otherwise. Now we can reformulate the problem: Find the maximal possible value of the expression
$$A = \sum_{i=1}^{20} \sum_{j=1}^{11} a_{i,j} \text{ under the following two conditions:}$$
$$\bullet\ a_{i,j} = 0 \text{ or } 1$$
* if $a_{k,l} = 1$ for some $k$ and $l$, then at least one of the sums $\sum_{j=1}^{11} a_{k,j}$ and $\sum_{i=1}^{20} a_{i,l}$ does not exceed $2$.

First of all, let us show that $A \le 54$. Suppose that $a_{k,l} = 1$. We say that $k$ is *1-good*, if
$$\sum_{j=1}^{11} a_{k,j} \le 2; \text{ we say that } l \text{ is 2-good if } \sum_{i=1}^{20} a_{i,l} \le 2.$$
If the total number of 1-good values of $k$ is $20$, then $A \le 2 \cdot 20 = 40$.
If the total number of 2-good values of $l$ is $11$, then $A \le 2 \cdot 11 = 22$.
If the total number of 1-good values of $k$ is $19$, then $A \le 2 \cdot 19 + 11 = 49$.
If the total number of 2-good values of $l$ is $10$, then $A \le 2 \cdot 11 + 20 = 32$.
Finally, if the total number of 1-good values of $k$ is less than or equal to $18$ and the total number of 2-good values of $l$ is less than or equal to $9$, then the total number of good values is at most $27$ and readily $A \le 2 \cdot 27 = 54$, since the number of nonzero terms of $A$ is less than or equal to twice the number of good values. Thus, $A \le 54$.

Now we give an example for $A = 54$. Let $a_{i,j} = 1$ only for
$$(i,j) \in \{(i,j) : i \in \{1,20\} \text{ or } j \in \{1,11\}\} \setminus \{(1,1), (20,1), (1,11), (20,11)\}.$$
The conditions are readily satisfied and we are done.