A farmer wants to raise corn and soybean on a 2010 acre of crop land. Each acre of corn requires 2 hours of labor during growing season while each acre of soybean requires only 1 hour of labor. The total amount of hours of labor available during growing season is 340 hours.
The profit per acre for corn is $45 while the profit per acre for soybean is $35. Find how many acres of corn and soybean should be planted to maximize profit.
To solve this problem, we need to find the optimal allocation of acres for corn and soybean that maximizes the total profit.
Let's assume x represents the number of acres of corn planted, and y represents the number of acres of soybean planted.
We have two constraints to consider:
1. Total acres: x + y = 2010
2. Total available labor hours: 2x + y = 340
First, let's solve the system of equations:
Equation 1: x + y = 2010
Equation 2: 2x + y = 340
We can use the method of substitution to solve this system. From Equation 1, we can rewrite it as y = 2010 - x. Substituting this into Equation 2, we get:
2x + (2010 - x) = 340
2x + 2010 - x = 340
x + 2010 = 340
x = 340 - 2010
x = -1670
It appears that there may be a mistake in the problem statement since we have obtained a negative value for x. Please double-check the problem and provide the correct information to proceed with the calculations.