Three. (Full marks 25 points) A steel plant uses 2 tons of raw material $A$ and 5 tons of raw material $B$ to produce 1 ton of type 甲 steel pipe, and 3 tons of raw material $A$ and 1 ton of raw material $B$ to produce 1 ton of type 乙 steel pipe. The unit profit (unit: ten thousand yuan/ton) from selling these two types of steel pipes in Beijing, Shanghai, and Guangzhou is shown in the table below:

Now, based on market supply and demand information: the weekly supply of raw materials $A$ and $B$ is 10 tons and 12 tons, respectively; the weekly production of type 甲 steel pipe cannot exceed 2.5 tons, and the weekly production of type 乙 steel pipe cannot exceed 1.5 tons, and all steel pipes can only be sold in one place. How many tons of each type of steel pipe should be produced, and where should they be sold, to maximize the total profit for the week? What is the maximum total profit?

| Sales Location | Type 甲 (ten thousand yuan/ton) | Type 乙 (ten thousand yuan/ton) |
|----------------|--------------------------------|--------------------------------|
| Beijing        | 10                            | 8                             |
| Shanghai       | 12                            | 10                            |
| Guangzhou      | 14                            | 12                            |
Three, let the weekly production of two types of steel pipes, A and B, be $x$ tons and $y$ tons, respectively. According to the problem, we have:
$$
\left\{\begin{array}{l}
2 x+5 y \leqslant 10, \\
4 x+3 y \leqslant 12, \\
0 \leqslant x \leqslant 2.5, \\
0 \leqslant y \leqslant 1.5 .
\end{array}\right.
$$

The weekly total profit for selling all to Beijing is $f_{1}=2 x+6 y$, to Shanghai is $f_{2}=3 x+4 y$, and to Guangzhou is $f_{3}=4 x+2 y$ (unit: ten thousand yuan).

As shown in Figure 5, the shaded area is the feasible region of this linear programming problem,
$$
\begin{array}{l}
A\left(\frac{5}{4}, \frac{3}{2}\right), \\
B\left(\frac{15}{7}, \frac{8}{7}\right) \text {. } \\
C\left(\frac{5}{2}, \frac{2}{3}\right) .
\end{array}
$$

By comparing the slopes of the lines $f_{1}=2 x+6 y$, $f_{2}=3 x+4 y$, $f_{3}=4 x+2 y$ with $4 x+3 y=12$ and $2 x+3 y=10$, we have
$$
-2<-\frac{4}{3}<-\frac{3}{4}<-\frac{2}{5}<-\frac{1}{3} \text {. }
$$
$\therefore$ In the parallel line system $f_{1}=2 x+6 y$, when passing through point $A$,
$$
f_{1 \text { max }}=2 \times \frac{5}{4}+6 \times \frac{3}{2}=11 \frac{1}{2} \text {; }
$$

In the parallel line system $f_{2}=3 x+4 y$, when passing through point $B$,
$$
f_{2_{\max }}=3 \times \frac{15}{7}+4 \times \frac{8}{7}=11 \text {; }
$$

In the parallel line system $f_{3}=4 x+2 y$, when passing through point $C$,
$$
f_{\text {3nax }}=4 \times \frac{5}{2}+2 \times \frac{2}{3}=11 \frac{1}{3} \text {. }
$$

Among these, $11 \frac{1}{2}$ is the largest.
Therefore, producing $\frac{5}{4}$ tons of steel pipe A and $\frac{3}{2}$ tons of steel pipe B and selling all to Beijing can maximize the total weekly profit, with the maximum total profit being 11.5 ten thousand yuan.
（Fei Zhenxian, Yongfeng Middle School, Chengqu District, Yancheng City, Jiangsu Province, 224054）