A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $20 per acre. Seed for crop B costs $10 per acre. The farmer can spend at most $700 on seed. If crop B brings in a profit of $60 per acre, and crop A brings in a profit of $180 per acre, how many acres of each crop should the farmer plant to maximize her profit?

Just graph each boundary line for the constraints, then evaluate the function at each vertex.

To find out how many acres of each crop the farmer should plant to maximize her profit, we can set up a system of equations.

Let's say the farmer plants x acres of crop A and y acres of crop B.

According to the given information, the total area to be planted is 60 acres. So, we have the equation:
x + y = 60 ----- Equation 1

The cost of seed for crop A is $20 per acre, and the cost of seed for crop B is $10 per acre. The farmer can spend at most $700 on seed. Therefore, the cost equation is:
20x + 10y ≤ 700 ----- Equation 2

The profit per acre for crop A is $180, and the profit per acre for crop B is $60. So, the profit equation is:
180x + 60y ----- Equation 3

Now, we can solve this system of equations to find the values of x and y.

First, let's solve Equations 1 and 2. Rearrange Equation 1 to solve for x:
x = 60 - y

Substitute this value of x in Equation 2:
20(60 - y) + 10y ≤ 700

Simplify the equation:
1200 - 20y + 10y ≤ 700
-10y ≤ 700 - 1200
-10y ≤ -500
Divide both sides by -10 (since we are dividing by a negative number, the inequality symbol will flip):
y ≥ -500 / -10
y ≥ 50

So, from Equation 1, we have x = 60 - y:
x = 60 - 50
x = 10

Therefore, the farmer should plant 10 acres of crop A (x = 10) and 50 acres of crop B (y = 50) to maximize her profit.