Problema Solution
An agriculture company has 80 tons of type I fertilizer and 120 tons of type II fertilizer. The company mixes these fertilizer into two products. Product X requires 2 parts of type I and 1 part of type II fertilizers. Product Y requires 1 part of type I and 3 parts of type II fertilizers. If each product sells for $2000, what is the maximum revenue and how many of each product should be made and sold to maximize revenue?
Answer provided by our tutors
Product X requires 2 parts of type I and 1 part of type II fertilizers:
2x tones of type I fertilizer
x tones of type II fertilizer
Product Y requires 1 part of type I and 3 parts of type II fertilizers:
y tones of type I fertilizer
3y tones of type II fertilizer
An agriculture company has 80 tons of type I fertilizer and 120 tons of type II fertilizer:
2x + y <= 80
x + 3y <= 120
are the constrains.
The objective function is:
F(x , y) = 2000(x + y)
Lets find the corner points
click here to see the graph of the system of inequalities:
2x + y <= 80
x + 3y <= 120
the corner points are:
(40, 0), (0, 40) and (24, 32)
(24, 32) is found by solving the system of equations:
2x + y=80,
x+3y=120
click here to see the step by step solution of the system of equations:
F(40 , 0) = 2000*40 = $80,000
F(0,40) = 2000*40 = $80,000
F(24, 32) = 2000(24 + 32) = $112,000
the maximum revenue is $112,000.
24 Product X and 32 Product Y should be made to maximize revenue.