SOLUTION: A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $30 per acre. Seed for crop B costs $15 per acre. The farmer can spend at most $1200 on seed.

Question 1184214: A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $30 per acre. Seed for crop B costs $15 per acre. The farmer can spend at most $1200 on seed.
If crop B brings in a profit of $80 per acre, and crop A brings in a profit of $120 per acre, how many acres of each crop should the farmer plant to maximize her profit?
acres of crop A
acres of crop B
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = number of acres for crop A.
y = number of acres for crop B.

your objective function is equal to 120x + 80y.
this is the profit function that you want to maximize the value of.

your constraint functions are:
x + y <= 60 (acres)
30x + 15y <= 1200 (money spent on seeds)
x >= 0, y >= 0 (must not be negative)

using the desmos.com calculator, you would graph the opposite of these inequalities.

specifically, you would graph:
x + y >= 60
30x + 15y >= 1200
x <= 0, y <= 0

the area on the graph that is not shaded is your region of feasibility.

the corner points of this region are where the maximum profit lies.

all the constraints must be met at the corner point where the maximum profit lies.

the graph is shown below:

the maximum profit is at the point (x,y) = (20,40).

at this point, the profit is 120x + 80y = 5600.
the number of acres is x + y = 60.
the amount spent on seeds is 30x + 15y = 1200.
all values of x and y are greater than or equal to 0.

your solution is that the farmer should plant 20 acres of crop A and 40 acres of crop B to maximize profits.

here's a picture of what the graph shows.

the graphing calculator can be found at https://www.desmos.com/calculator

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