II. (25 points) Sustainable development has become a common concern for human development, and China has made it a basic national policy. To protect the environment and maintain ecological balance, a policy of zoned grazing has been implemented in the northern pastoral areas of China. Mr. Li, a herdsman, has a grassland that can be used for raising cattle and sheep. However, once the grass on the grassland is eaten up, the grassland will lose its ability to regenerate and will no longer be able to support cattle and sheep. If the grassland is used to raise 200 cattle and 200 sheep, the grass will be eaten up in 10 days; if it is used to raise 100 cattle and 200 sheep, the grass will be eaten up in 20 days; if it is used to raise 50 cattle and 300 sheep, the grass will be eaten up in 30 days. Cattle and sheep need to be raised for a full year before they can be sold (otherwise, the herdsman can only produce and sell them himself), and when sold, each cow is worth 2500 RMB, and each sheep is worth 300 RMB. If Mr. Li wants to maximize his income from raising cattle and sheep in one year, please design a feasible breeding plan for Mr. Li.
(Note: The original amount of grass on the grassland should not decrease after one year.)
Let the amount of grass consumed by each cow and each sheep per day be $x$ and $y$, respectively, the amount of grass that grows on the pasture each day be $z$, and the original amount of grass on the pasture be $A$. According to the problem, we have
$$
\left\{\begin{array}{l}
200 \times 10 x + 200 \times 10 y = 10 z + A, \\
100 \times 20 x + 200 \times 20 y = 20 z + A, \\
50 \times 30 x + 300 \times 30 y = 30 z + A.
\end{array}\right.
$$

Solving these equations, we get $x = 6y$, $z = 200y$, and $A = 12000y$.
To maximize the profit after one year of breeding, it is necessary to ensure that the total amount of grass consumed by the cows and sheep each day does not exceed the amount of grass that the pasture can regenerate. Otherwise, the pasture will lose its ability to regenerate, and the cows and sheep will have no grass to eat. To maximize the profit, let Mr. Li raise $m$ cows and $n$ sheep, then $mx + ny \leq z$. At this point, Mr. Li's income is $w = 2500m + 300n$.
Therefore, $6my + ny \leq 200y \Rightarrow 6m + n \leq 200$.
Clearly, since $x = 6y$, the amount of grass consumed by one cow per day is equivalent to the amount consumed by 6 sheep per day, but the value of one cow is 2500 RMB, while 6 sheep are worth $300 \times 6 = 1800$ RMB. Therefore, to maximize the profit after one year, Mr. Li should raise as many cows as possible. Since $6m + n \leq 200$, when $m = 33$ and $n = 2$, Mr. Li's profit is maximized, at which point the profit is
$$
w = 2500 \times 33 + 300 \times 2 = 83100 \text{ RMB.}
$$

Therefore, to maximize his profit after one year, Mr. Li can raise 33 cows and 2 sheep, at which point the maximum profit is 83100 RMB.