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
3 answers
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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
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
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
max
180x+60y
x+y</=60
20x+10y<=700
x>=0
y>=0
x+y>=1 phony but fills boxes