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?
acres of crop A
acres of crop B
a+b<=60
20 a + 10b </= 700
a and b >=0
maximize 180 a + 60 b
intersections at
0,60
35,0
10,50
at a = 0, b = 60
180 (0) + 60 *60 = 3600
at a = 10, b = 50
180(10) + 60(50) = 4800
at a = 35, b = 0
180(35) = 6300
so plant 35 acres of a and zero of b.
Use the whole $700 for seed a
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue
note, to make the program format happy I put in
max
180x+60y
x+y</=60
20x+10y<=700
x>=0
y>=0
x+y>=1 phony but fills boxes
To find out how many acres of each crop the farmer should plant to maximize her profit, we can use linear programming.
Let's assign variables:
Let x be the number of acres of crop A.
Let y be the number of acres of crop B.
We have the following constraints:
1) x + y = 60 (Since the total number of acres is 60)
2) 20x + 10y ≤ 700 (Since the farmer can spend at most $700 on seed)
We want to maximize the profit, which is given by:
Profit = 180x + 60y
Now, let's solve the problem step by step:
1) Solve for y in the first constraint:
y = 60 - x
2) Substitute y in the second constraint:
20x + 10(60 - x) ≤ 700
20x + 600 - 10x ≤ 700
10x ≤ 100
x ≤ 10
3) The third constraint shows that the maximum number of acres of crop A the farmer can plant is 10.
4) Now, substitute y = 60 - x in the profit equation:
Profit = 180x + 60(60 - x)
Profit = 180x + 3600 - 60x
Profit = 120x + 3600
5) Find the value of x that maximizes the profit:
To find the maximum profit, we need to find the value of x that results in the highest value for the profit equation. Since x ≤ 10 (from constraint 3), we can evaluate the profit equation for x = 10 and x = 0.
For x = 10, Profit = 120(10) + 3600 = 1200 + 3600 = 4800
For x = 0, Profit = 120(0) + 3600 = 3600
The maximum profit is achieved when x = 10, which means the farmer should plant 10 acres of crop A.
To find the number of acres for crop B, we can substitute the value of x in the first constraint:
x + y = 60
10 + y = 60
y = 60 - 10
y = 50
Therefore, the farmer should plant 10 acres of crop A and 50 acres of crop B to maximize her profit.