\section*{Task 1 - V11131}

In the field of national economic planning, mathematical methods are also applied. The following is a highly simplified example from our socialist construction industry:

In a city, apartments are to be built in 1962, specifically of Type A (brick construction) and Type B (prefabricated construction). The requirements per housing unit are as follows:

\begin{tabular}{cc|c} 
& Cement & Wall components \\
\hline Type A & \(5.23 \mathrm{t}\) & - \\
Type B & \(4.19 \mathrm{t}\) & \(22.1 \mathrm{t}\)
\end{tabular}

In total, 8000 t of cement and 24000 t of wall components are available. If \(x\) represents the number of Type A apartments and \(y\) represents the number of Type B apartments to be built, the following inequalities must be satisfied:

\[
5.23 x + 4.19 y \leq 8000 \quad ; \quad 22 y \leq 24000
\]

The total number of apartments \((x + y)\) should be as large as possible.

What are the numbers \(x\) of Type A apartments and \(y\) of Type B apartments?
}

In terms of material consumption, it is most cost-effective to first use the building materials for apartments of type B. It is

\[
y \leq \frac{24000}{22} \approx 1090.9
\]

and thus 1090 apartments of type B will be built. Therefore,

\[
\begin{aligned}
5.23 x + 4.19 y & \leq 8000 \\
5.23 x & \leq 3432.9 \\
x & \leq \frac{3432.9}{5.23} \approx 656.386
\end{aligned}
\]

so 656 apartments of type A will be built. Thus, \(x=656\) and \(y=1090\) must hold.

If one wanted to build only apartments of type A, \(x<1600\) would be the case, and thus the most cost-effective scenario described above maximizes the total number.