Problem 26. A company has 2 workshops and 3 warehouses. It is necessary to determine the most profitable organization of transportation.

| Warehouses | No. 1 | No. 2 | No. 3 | Produced |
| :---: | :---: | :---: | :---: | :---: |
| I | 3 | 3 | 2 | 10000 |
| II | 6 | 5 | 1 | 5000 |
| Required to deliver | 4000 | 8000 | 3000 |  |
## S o l u t i o n.

Let $x$ units be transported from workshop I to warehouse 1, and $y$ units to warehouse 2. Then, from workshop I to warehouse 3, $(10000-x-y)$ units will be transported, and from workshop II to warehouses 1, 2, and 3, $4000-x$, $8000-y$, and $x+y-7000$ units will be transported, respectively (Table 13).

|  | № 1 | № 2 | № 3 |  |
| :---: | :---: | :---: | :---: | :---: |
| I | 个 | 3 | 2 | 10000 |
|  | $x$ | $y$ | $\|10000-x-y\|$ |  |
| II | 6 | $\mid 5$ | 1 | 5000 |
|  | $4000-x$ | $8000-y$ | $x+y-7000$ |  |
|  | 4000 | 8000 | 3000 |  |

Table 13

The total transportation cost $F$ under this assumption will be:

$$
\begin{gathered}
F=3 x+3 y+2 \cdot(10000-x-y)+6 \cdot(4000-x)+ \\
+5 \cdot(8000-y)+1 \cdot(x+y-7000)= \\
=-x-3(x+y)+77000
\end{gathered}
$$

Since $x+y=10000$, then $F=x-30000+77000=-x+47000$. We see that reducing $F$ is possible by increasing $x$ (where $x$ is an integer from $[0 ; 4000]$). Therefore, when $x=4000$, $F$ reaches its minimum value. The remaining 6000 units in workshop I can be transported to warehouse 2, which has a capacity of 8000 units. This means that when $x=4000$, the variable $y$ takes its maximum value, which is 6000. Thus, the function $F$ reaches its minimum value when $x=4000$ and $y=6000$.

The optimal transportation plan is shown in Table 14.

![](https://cdn.mathpix.com/cropped/2024_05_21_07b5a7f6f561b0ff634dg-33.jpg?height=531&width=788&top_left_y=1910&top_left_x=637)

32

The minimum transportation cost for all 15000 units will be 43000 monetary units:

$$
F_{\text {min }}=43000
$$