\section*{Task 1 - 011231}

Two brickworks produce 6 million and 12 million bricks, respectively. They are to supply four construction sites with a demand of 5.2, 3.0, 5.7, and 4.1 million bricks, respectively.

The distances (in km) between the two brickworks and the four construction sites are shown in the following table:

\begin{tabular}{lcccc} 
Construction Site & 1 & 2 & 3 & 4 \\
Brickwork 1 & 28 & 30 & 37 & 21 \\
Brickwork 2 & 26 & 36 & 18 & 20
\end{tabular}

How many bricks must be transported from the 1st and 2nd brickworks to each of the individual construction sites to minimize the total transportation costs?

It is assumed that the transportation costs are proportional to the distance. Construction site 3 should only be supplied by brickwork 2.
}
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1735.jpg?height=436&width=1252&top_left_y=957&top_left_x=400)

To determine the optimal supply, the cost function must be expressed in terms of the controllable parameters. This function is the sum of the transportation costs to the individual construction sites, or (due to proportionality) the distance over which the bricks are transported.

If the deliveries from brickyard \(2\left(Z_{2}\right)\) - in million bricks - to construction sites 1, 2, and 4 (\(B_{1}\) etc.) are denoted by \(x, y\), and \(z\) (site \(B_{3}\) does not need to be considered), we get

\[
K(x, y, z)=28(5.2-x)+26x+30(3.0-y)+36y+21(4.1-z)+20z=321.7-2x+6y-z
\]

The side condition is \(x+y+z=6.3\), i.e., \(Z_{2}\) delivers 6.3 million bricks to \(B_{1}, B_{2}\), and \(B_{4}\); the remaining 5.7 million go to \(B_{3}\). The function \(K\) should be minimized.

This is the case when \(x\) and \(z\) take the largest possible values, and \(y\) takes a small value, as can be seen from the signs.

The side condition is incorporated by setting \(\tilde{K}(y, z)=K(6.3-y-z, y, z)=309.1+8y+z\).

It is also important to note \(0 \leq y \leq 3.0\) and \(0 \leq z \leq 4.1\) (only this range is reasonable for deliveries from \(Z_{2}\) to \(B_{2}\) and \(B_{4}\)), as well as \(1.1 \leq y+z \leq 6.3\) (\(B_{1}\) can receive a maximum of 5.2 million bricks from \(Z_{2}\)).

In \(\tilde{K}\), both \(y\) and \(z\) have positive signs, so both must be as small as possible. The smallest value for \(\tilde{K}(y, z)\) is obtained for \(y=0\) and \(z=1.1\), since \(y\) has the larger coefficient. It follows that \(x=5.2\).

In summary: \(Z_{1}\) delivers 3 million bricks each to \(B_{2}\) and \(B_{4}\), \(Z_{2}\) delivers 5.2 million to \(B_{1}\) and 1.1 million to \(B_{4}\).

Note: The method known from school for determining (local) minima using derivatives does not work here. The reason is that the function \(K\) (or \(\tilde{K}\)) is linear in its arguments. Extrema in this case only occur globally at the boundaries of the regions defined by the inequalities.