Example 2 As shown in Figure 2,  
there are soil piles $A$ and $B$, a depression $E$, and a pond $F$ at a construction site.  
The volumes of soil piles $A$ and $B$ are $781 \mathrm{~m}^{3}$ and $1584 \mathrm{~m}^{3}$, respectively. Depression $E$ requires $1025 \mathrm{~m}^{3}$ of soil to be filled, and pond $F$ can accommodate $1390 \mathrm{~m}^{3}$ of soil. Now, it is required to remove the two soil piles and use the soil to first fill the depression $E$, and then use the remaining soil to fill the pond $F$. How should the soil transportation plan be arranged to minimize labor?
Solution: Let " $\mathrm{m}^{3} \cdot \mathrm{m}$ " be the unit for the labor cost of transporting soil. Suppose the volume of soil transported from $A$ to $E$ is $x$, and the volume of soil transported to $F$ is $781-x$; the volume of soil transported from $B$ to $E$ is $1025-x$, and the volume of soil transported to $F$ is 1584 $(1025-x)=559+x$. Let the total $\mathrm{m}^{3} \cdot \mathrm{m}^{\prime}$ of labor cost be $W$. According to the problem, we have
$$
\begin{array}{l}
W=150(781-x)+50 x+30(1025-x)+ \\
=2140(559+x) \\
\text { Since }\left\{\begin{array}{l}
x \geqslant 0, \\
781-x \geqslant 0, \\
1025-x \geqslant 0, \\
599+x \geqslant 0,
\end{array}\right.
\end{array}
$$

Therefore, $0 \leqslant x \leqslant 781$.
Since $W$ is a linear function of $x$, and $k=-10<0$, $W$ decreases as $x$ increases.
Thus, when $x$ takes its maximum value 781, $W_{\text {min }}=207170$.
When $x=781$,
$$
781-x=0,1025-x=244 \text {. }
$$

Therefore, the most labor-saving soil transportation plan is: all $781 \mathrm{~m}^{3}$ of soil from pile $A$ is transported to depression $E$, and 244 $\mathrm{m}^{3}$ of soil from pile $B$ is transported to depression $E$, with the rest being transported to pond $F$.