14. (15 points) As shown in Figure 4,

there are two mounds $A$ and
$B$, a depression $E$,
and a pond $F$. The volumes of the two mounds are 781 cubic meters and
1584 cubic meters, respectively. The depression $E$ needs
to be filled with 1025 cubic meters of soil, and the pond $F$
can be filled with 1390 cubic meters of soil. Now it is required to dig out the two mounds 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?
$\vdots 1$. "cubic meter・meter" as the unit of labor cost for transporting soil, let the amount of soil transported from $A$ to $E$ be $x_{1}$, and to $F$ be $y_{1}$; the amount of soil transported from $B$ to $E$ be $x_{2}$, and to $F$ be $y_{2}$. The total "cubic meter・meter" of soil transported is $W$. According to the problem, we have
$$
\begin{array}{l}
x_{1}+y_{1}=781, \\
x_{2}+y_{2}=1584, \\
x_{1}+x_{2}=1025, \\
y_{1}+y_2=1340, \\
W=50 . x_{1}+150 y_{1}+30 x_{2}+120 y_{2} .
\end{array}
$$

where $0 \leqslant x_{1} \leqslant 781,0 \leqslant x_{2} \leqslant 1584$.
From equation (1), we get $y_{1}=781-x_{1}$.
From equation (3), we get $x_{2}=1025-x_{1}$.
Thus, from equation (4), we get
$$
\begin{aligned}
y_{2} & =1340-y_{1}=1340-\left(781-x_{1}\right) \\
& =559+x_{1} .
\end{aligned}
$$

Then $W=50 x_{1}+150\left(781-x_{1}\right)+30(1025-$
$$
\begin{aligned}
& \left.x_{1}\right)+120\left(559+x_{1}\right) \\
= & 214980-10 x_{1} .
\end{aligned}
$$

Therefore, when $x_{1}$ takes the maximum value of 781, $W_{\text {group }}=207170$.
Thus, the soil transportation plan that minimizes labor is:
All 781 cubic meters of soil from mound $A$ are transported to depression $E$, 244 cubic meters of soil from mound $B$ are transported to depression $E$, and the remaining soil from mound $B$ is all transported to pond $F$.
(Provided by Nan Xiuquan, Huanggang City Teaching and Research Office, Hubei Province)