\section*{Task 3 - 161213}

A school has the following tents available for a camping trip:

2 tents for 3 people each,

1 tent for 8 people,

2 tents for 10 people each, and

2 tents for 16 people each.

Each of these tents is either occupied by girls to exactly 50% of its maximum capacity or by boys to a maximum of 70% and a minimum of 50% of its capacity. In total, more girls than boys are to be considered for the camping trip.

a) How many people can participate in the camping trip under these conditions at most?

b) Provide a corresponding tent allocation for such a case.
}

The tents for 3 people are denoted by \(Z_{31}, Z_{32}\), the one for 8 people by \(Z_{8}\), the ones for 10 people by \(Z_{101}, Z_{102}\), and the ones for 16 people by \(Z_{161}, Z_{162}\).

First, according to the occupancy information, the tents \(Z_{31}, Z_{32}\) can only be occupied by 2 boys each. Furthermore, it follows that the number of people is at least \(34 (2 \cdot 2 + 1 \cdot 4 + 2 \cdot 5 + 2 \cdot 8)\).

For the individual tents, the following possible occupancies arise (where \(J\) denotes the number of boys and \(M\) the number of girls).

\begin{tabular}{lll} 
Tent & \(\mathrm{J}\) & \(\mathrm{M}\) \\
\hline\(Z_{31}\) & 2 & - \\
\(Z_{32}\) & 2 & - \\
\(Z_{8}\) & 4 or 5 & 4 \\
\(Z_{101}\) & 5 or 6 or 7 & 5 \\
\(Z_{102}\) & 5 or 6 or 7 & 5 \\
\(Z_{161}\) & 8 or 9 or 10 or 11 & 8 \\
\(Z_{162}\) & 8 or 9 or 10 or 11 & 8
\end{tabular}

Let \(n_{i}\) be the occupancy number for \(Z_{i} (i=31,32,9,101,102,161,162)\). If \(n_{161} \geq 9\) and \(n_{162} \geq 9\), then \(J \geq 2+2+9+9=22\) and since \(J+M \leq 44\), it follows that \(M \leq 22\), which contradicts \(M>J\). Without loss of generality, let \(n_{162}=8\).

Case distinction for \(n_{8}\):

Let \(n_{8}=5\) (i.e., this tent is occupied by 5 boys). If in addition \(n_{161} \geq 9\), then \(J \geq 2+2+2+5+9=18\) and thus \(M \leq 5+5+8=18\) (occupancy of the tents \(Z_{101}, Z_{102}\), and \(Z_{162}\)), which contradicts \(M>J\). Therefore, \(n_{161}=8\).

If \(n_{101}>5\) and \(n_{102}>5\), then \(J \geq 2+2+5+6+6=21\) and \(M \leq 8+8=16\), which contradicts \(M>J\).

For \(n_{101}=n_{102}=5\), we have \(J+M=2+2+5+5+5+8+8=35\). Since only the first two tents are more than 50% occupied, this occupancy can also be realized while considering \(M>J\).

If exactly one of the numbers \(n_{101}, n_{102}\) is equal to 5, then

\[
J+M \leq 2+2+5+5+7+8+8=37 \quad \text{where} \quad J \geq 2+2+5+6=15
\]
holds, so an occupancy with \(M>J\) is possible. In the case \(n_{8}=5\), a maximum of 37 students can be accommodated.

Let \(n_{8}=4\) (i.e., this tent can be occupied by either boys or girls). If at least one of the numbers \(n_{101}, n_{102}\) is greater than 5 and simultaneously \(n_{161} \geq 9\), then \(J \geq 2+2+6+9=19\) and, in contradiction to \(M>J\), \(M \leq 4+5+8=17\).

If \(n_{101}=n_{102}=5\), then \(J+M \leq 2+2+4+5+5+11+8=37\). Since \(J \geq 2+2+11=15\), such an occupancy is possible.

If exactly one of the numbers \(n_{101}, n_{102}\) is equal to 5 and thus \(n_{161}=8\), then

\[
J+M \leq 2+2+4+5+6+7+8+8=36
\]

The only remaining case \(n_{101}>5\) and \(n_{102}>5\) can be realized by occupying \(Z_{31}\) and \(Z_{32}\) with 2 boys each, \(Z_{8}\) with 4 girls, \(Z_{101}\) and \(Z_{102}\) with 7 boys each, and \(Z_{161}\) and \(Z_{162}\) with 8 girls each. This results in 18 boys and 20 girls, a total of 38 people. Therefore, the maximum occupancy number is 38.

\section*{Adapted from \([5]\)}