14. The quality of a product can be divided into 6 different levels. If the working hours remain unchanged, 40 pieces of the lowest level product can be produced daily; if the level is increased by one, the profit per piece can increase by 1 yuan, but 2 fewer pieces of the product will be produced daily.
(1) If the profit per piece of the lowest level product is 16 yuan, which level of product will yield the maximum profit?
(2) If the profit per piece of the lowest level product is 22 yuan, which level of product will yield the maximum profit?
(3) Due to market price fluctuations, the profit per piece of the lowest level product can vary from 8 yuan to 24 yuan. In this case, which level of product will yield the maximum profit?
14. (1) Let the profit be maximized when producing the $x$-th grade product. Clearly, $1 \leqslant x \leqslant 6$, and $x$ is a natural number. Let the daily profit be $y$ yuan. Then
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$

Thus, when $x=3$, $y$ is maximized.
Therefore, producing the third-grade product yields the highest profit.
(2) Similarly to (1), we get
$$
\begin{array}{l}
y=[40-2(x-1)][22+(x-1)] \\
=-2 x^{2}+882 .
\end{array}
$$

Since $1 \leqslant x \leqslant 6$, and $x$ is a natural number, $y$ is maximized when $x=1$. Therefore, producing the first-grade product yields the highest profit.
(3) Let the profit per unit of the lowest-grade product be $a$ yuan, and $8 \leqslant a \leqslant 24$. Then
$$
\begin{array}{l}
y=[40-2(x-1)][a+(x-1)] \\
=(42-2 x)[(a-1)+x] \\
=-2\left(x-\frac{22-a}{2}\right)^{2}+\frac{a^{2}+40 a+400}{2} .
\end{array}
$$

Through analysis and discussion, the following conclusions can be drawn:
$$
x=\left\{\begin{array}{ll}
6, & 8 \leqslant a \leqslant 10 ; \\
\frac{22-a}{2} \text { (rounded to the nearest integer) }, & 10<a<20 ; \\
1, & 20 \leqslant a \leqslant 24 .
\end{array}\right.
$$