SOLUTION: A farmer, in the business of growing fodder for livestock, has 90 acres available for planting alfalfa and corn. The cost of seed per acre is $32 for alfalfa and $48 for corn. The

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: A farmer, in the business of growing fodder for livestock, has 90 acres available for planting alfalfa and corn. The cost of seed per acre is $32 for alfalfa and $48 for corn. The       Log On


   

Question 867874: A farmer, in the business of growing fodder for livestock, has 90 acres available for planting alfalfa and corn. The cost of seed per acre is $32 for alfalfa and $48 for corn. The total cost of labor will amount to $60 per acre for alfalfa and $30 per acre for corn. The expected revenue (before costs are subtracted) is $500 per acre from alfalfa and $700 per acre from corn. If the farmer does not wish to spend more than $3888 for seed and $4200 for labor, how many acres of each crop should be planted to obtain the maximum profit?
alfalfa acres
corn acres

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x = number of acres of alfalfa
y = number of acres of corn.

costs for seed are equal to 32x + 48y <= 3885
costs for labor are equal to 60x + 30y <= 4200

revenue is equal to 500x + 700y

profit is equal to revenue minus cost.

this makes revenue equal to:

500x + 700y - 32x - 48y - 60x - 30y

combine like terms to get:

profit = 408x + 622y

you graph your constraint equations of:

32x + 48y <= 3885
60x + 30y <= 4200

solve for x in both of these equationt to get:

32x + 48y <= 3885 becomes:
y <= (3885-32x)/48

60x + 30y <= 4200 becomes:
y <= (4200-60x)/30

you will graph:

y = (3885-32x) / 48 and y = (4200-60x)/30 and you will shade the area under the lines of those equations and above the line of x = 0 and to the right of the line of y = 0

your graph will look like this:

graph%28600%2C600%2C-10%2C100%2C-10%2C100%2C%284200-60x%29%2F30%2C%283885-32x%29%2F48%29

your maximum profit will be at one of the following intersection points.

(0,80.94)
(44.3,51.41)
(70,0)

you analyze your profit equation at each of these points and you get:

(0,80.94) = 50344
(44.3,51.41) = 50051
(70,0) = 28560

the maximum profit equation you are using is 408x + 622y.

your maximum revenue is at (0,80.94)

your seed constraint equation is 32x + 48y
your labor constraint equation is 60x + 30y

at the maximum profit point of (0,80.94)...

your seed costs come out to be 3885.12 which is right at the max if you allow for some rounding error.

your labor costs come out to be 2428 which is well under the maximum labor costs.

if I did this right, corn is the winner.