Example 1: Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, A, B, and C, according to the signed contracts. They are to supply 45 tons to A, 75 tons to B, and 40 tons to C. Base A can arrange 60 tons, and Base B can arrange 100 tons. The distances in kilometers between A and B and markets A, B, and C are shown in Table 1. Assuming the transportation cost is 1 yuan/(km·t). How should the supply be arranged to minimize the total transportation cost? Find the minimum total transportation cost.

Table 1
\begin{tabular}{|c|c|c|c|}
\hline & A & B & C \\
\hline A & 10 & 5 & 6 \\
\hline B & 4 & 8 & 15 \\
\hline
\end{tabular}
Solution: Let the base B supply $x \mathrm{t}$ to $A$, $y \mathrm{t}$ to $B$, and $[100-(x+y)] \mathrm{t}$ to $C$, then base A supplies $(45-x) \mathrm{t}$ to $A$, $(75-y) \mathrm{t}$ to $B$, and $[40-(100-x-y)] \mathrm{t}=(x+y-60) \mathrm{t}$ to $C$.
Let the total transportation cost be $W$ yuan. According to the problem, we have
$$
\begin{array}{l} 
W= 10(45-x)+5(75-y)+ \\
6(x+y-60)+4 x+8 y+ \\
15[100-(x+y)] \\
= 1965-15 x-6 y \\
= 1965-6(x+y)-9 x . \\
\text { Because }\left\{\begin{array}{l}
x \geqslant 0, \\
y \geqslant 0, \\
100-(x+y) \geqslant 0, \\
45-x \geqslant 0, \\
75-y \geqslant 0, \\
x+y-60 \geqslant 0,
\end{array}\right.
\end{array}
$$

Then, $60 \leqslant x+y \leqslant 100,0 \leqslant x \leqslant 45$.
Therefore, when and only when $x+y=100, x=45$, $W$ has the minimum value, and
$$
W_{\text {min }}=1965-6(x+y)-9 x=960 .
$$

When $x+y=100, x=45$, $y=55$.
Thus, the optimal distribution plan is:
Base B supplies $45 \mathrm{t}$ to $A$, $55 \mathrm{t}$ to $B$, and does not supply to $C$; Base A does not supply to $A$, supplies 20 $\mathrm{t}$ to $B$, and 40 $\mathrm{t}$ to $C$.