Waterbrook Farm includes 280 acres of cropland. The farm owner wishes to plant this acreage in both corn and soybeans. The profit per acre in corn production is $\$ 340$ and in soybeans is $\$ 255$. A total of $720 \mathrm{hr}$ of labor is availat Each acre of corn requires $3 \mathrm{hr}$ of labor, whereas each acre of soybean requires $2 \mathrm{hr}$ of labor. How should the land b divided between corn and soybeans in order to yield the maximum profit? What is the maximum profit?

The farm owner should plant $\square$ acres of corn and $\square$ acres of soybeans to yield a maximum profit of $\$ \square$.

Step 1 ：Let's denote the number of acres of corn as \(x\) and the number of acres of soybeans as \(y\).

Step 2 ：From the problem, we know that the total acreage is 280, so we have the equation: \(x + y = 280\)

Step 3 ：We also know that the total labor hours is 720, and each acre of corn requires 3 hours of labor, while each acre of soybeans requires 2 hours of labor. This gives us another equation: \(3x + 2y = 720\)

Step 4 ：We want to maximize the profit, which is $340 per acre for corn and $255 per acre for soybeans. So, the profit \(P\) can be represented as: \(P = 340x + 255y\)

Step 5 ：First, let's solve the system of the first two equations. We can multiply the first equation by 2 and subtract it from the second equation: \(3x + 2y - 2x - 2y = 720 - 2*280\) which simplifies to \(x = 720 - 560 = 160\)

Step 6 ：Substitute \(x = 160\) into the first equation: \(160 + y = 280\) which simplifies to \(y = 280 - 160 = 120\)

Step 7 ：So, the farm owner should plant 160 acres of corn and 120 acres of soybeans.

Step 8 ：Now, let's calculate the maximum profit. Substitute \(x = 160\) and \(y = 120\) into the profit equation: \(P = 340*160 + 255*120 = 54400 + 30600 = $85000\)

Step 9 ：So, the farm owner should plant 160 acres of corn and 120 acres of soybeans to yield a maximum profit of \(\boxed{85000}\) dollars.