One, (20 points) A company spent 4.8 million yuan to purchase the production technology of a certain product, and then invested another 15.2 million yuan to buy production equipment to process and produce the product. It is known that the cost of producing each piece of this product is 40 yuan. Market research has found that: the reasonable range for the unit selling price of the product is between 100 yuan and 300 yuan. When the unit selling price is set at 100 yuan, the annual sales volume is 200,000 pieces; when the unit selling price exceeds 100 yuan but does not exceed 200 yuan, for every 10 yuan increase in the unit selling price, the annual sales volume will decrease by 0.8 ten thousand pieces; when the unit selling price exceeds 200 yuan but does not exceed 300 yuan, for every 10 yuan increase in the unit selling price, the annual sales volume will decrease by 1 ten thousand pieces. Let the unit selling price be \( x \) (yuan), the annual sales volume be \( y \) (ten thousand pieces), and the annual profit be \( w \) (ten thousand yuan).
(1) Directly write the function relationship between \( y \) and \( x \).
(2) Find the function relationship between the first year's annual profit \( w \) and \( x \), and explain whether the company is profitable or at a loss in the first year of investment. If profitable, what is the maximum profit? If at a loss, what is the minimum loss?
(3) The company hopes that by the end of the second year, the total profit over two years will be no less than 18.42 million yuan. Please determine the range of the unit selling price at this time. Under this condition, to maximize the product sales volume, what should the unit selling price be set to?
$-、(1) y=\left\{\begin{array}{ll}-\frac{2}{25} x+28, & 100 \leqslant x \leqslant 200 \\ -\frac{1}{10} x+32, & 200<x \leqslant 300\end{array}\right.$
(2) When $100 \leqslant x \leqslant 200$, $w=x y-40 y-(1520+480)$.
Substituting $y=-\frac{2}{25} x+28$ into equation (1) gives $w=x\left(-\frac{2}{25} x+28\right)-40\left(-\frac{2}{25} x+28\right)-2000$.
Simplifying, we get $w=-\frac{2}{25}(x-195)^{2}-78$.
When $200<x \leqslant 300$, similarly, we get $w=-\frac{1}{10}(x-180)^{2}-40$.
Thus, $w=\left\{\begin{array}{ll}-\frac{2}{25}(x-195)^{2}-78, & 100 \leqslant x \leqslant 200 ; \\ -\frac{1}{10}(x-180)^{2}-40, & 200<x \leqslant 300 .\end{array}\right.$
If $100 \leqslant x \leqslant 200$, when $x=195$, $w_{\max }=-78$;
If $200<x \leqslant 300$, $w_{\max }<-80$.
Therefore, the company is in a loss in the first year of investment, with the minimum loss being 780,000 yuan.
(3) According to the problem, the function relationship between $w$ and $x$ in the second year is
$$
w=\left\{\begin{array}{ll}
\left(-\frac{2}{25} x+28\right)(x-40), & 100 \leqslant x \leqslant 200 ; \\
\left(-\frac{1}{10} x+32\right)(x-40), & 200<x \leqslant 300 .
\end{array}\right.
$$

When the total profit over two years is exactly 18,420,000 yuan, according to the problem, we have
$$
\left\{\begin{array}{l}
\left(-\frac{2}{25} x+28\right)(x-40)-78=1842 \\
100 \leqslant x \leqslant 200
\end{array}\right.
$$

or $\left\{\begin{array}{l}\left(-\frac{1}{10} x+32\right)(x-40)-78=1842, \\ 200<x \leqslant 300 .\end{array}\right.$
Solving, we get $x_{1}=190, x_{2}=200$.
Therefore, when $190 \leqslant x \leqslant 200$, the total profit is no less than 18,420,000 yuan.

From $y=-\frac{2}{25} x+28(100 \leqslant x \leqslant 200)$, we know that when the selling price is set to 190 yuan, the sales volume is maximized.