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.