Problema Solution
A farmer has a 300-acre farm on which she plants two crops: corn and soybeans, for each acre of corn planted, her expenses are $50 and for each acre of soybeans planted, her expenses are $100. each acre of corn requires 160 bushels of storage and yields a profit of $60; each acre of soybeans requires 60 bushels of storage and yields a profit of $90. If the total amount of storage space available is 24,000 bushels and the farmer has only $20,000 on hand, how many acres of each crop should she plant in order to maximize her profit? what will her profit be if she follows this strategy?
Answer provided by our tutors
let
x = the number of acre of corn planted, x>=0
y = the number of acre of soybeans planted, y>=0
for each acre of corn planted, her expenses are $50 and for each acre of soybeans planted, her expenses are $100 and the farmer has only $20,000 on hand:
50x + 100y <= 20000 divide both sides by 50
x + 2y <= 400
each acre of corn requires 160 bushels of storage, each acre of soybeans requires 60 bushels of storage and the total amount of storage space available is 24,000 bushels:
160x + 60y <= 24000 divide both sides by 20
8x + 3y <= 1200
each acre of corn yields a profit of $60, each acre of soybeans yields a profit of $90 thus the total profit is the objective function:
F(x, y) = 60x + 90y
lets graph the system of inequalities (the constrains)
x >=0
y >= 0
x + 2y <= 400
8x + 3y <= 1200
click here to see the graph
the corner points are: (0, 200), (150, 0) and (1200/13, 2000/13) = (92.31, 153.85)
F(x, y) = 60x + 90y
F(0, 200) = 60*0 + 90*200 = $18,000
(150, 0) = 60*150 + 90*0 = $9,000
F(1200/13, 2000/13) = 60*(1200/13) + 90*(2000/13) = $19,384.62 is the maximum of the objective function
In order to maximize her profit she should plant 92.31 acres corn and 153.85 acres soybeans. In that case her profit will be $19,384.62.