SOLUTION: Define your variables and set up the following linear programming problem. You do not have to solve. A farmer has 120 acres of land available for planting crops. It costs $65 t

Question 1175633: Define your variables and set up the following linear programming problem. You do not have to solve.
A farmer has 120 acres of land available for planting crops. It costs $65 to prepare each acre for soybeans and it costs $45 to prepare each acre for wheat. He is not willing to spend more than $2,900 in preparation costs. Each acre of soybeans requires 4 workdays to maintains and each acre of what requires 5 workdays to maintain. The farmer can afford up to a total of 110 workdays. He expects to realize $230 profit per acre of soybean and $155 per acre profit for wheat. How many acres of each crop should he plant in order to maximize his profit? How many acres of soybean and wheat should he plant in order to maximize profit?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = number of acres of soybeans.
let y = number of acres of wheat.

your constraint inequalities are:

x + y <= 120 (acres of land available)
4x + 5y <= 110 (maintenance days)
65x + 45y <= 2900 (prep dollars)
x >= 0, y >= 0 (number of acres can't be negative)

your objective function is 240x + 155y
this is what you want to maximize.

using the desmos.com calculator, you would graph the opposite of the constraint inequalities.

those are:

x + y <= 120
4x + 5y <= 110
65x + 45y <= 2900
x >= 0, y >= 0

the area on the graph that is not shaded is your region of feasibility.
you would evaluate your objective function at each of the corner points of the feasible region to find your maximum profit.

the graph looks like this:

using the simplex method tool, i got the following results.

both tools lead to the answer being 27.5 acres of soybean planted will yield the maximum profit while still keeping within the constraints.

all the constraint inequalities are satisfied at the maximum profit point.

the point on the graph that has the maximum profit is (x,y) = (27.5,0).

65x + 45y = 1787 <= 2900
x + y = 27.5 <= 120
4x + 5y = 27.5 * 4 = 110 <= 110 *****

maximum profit is 27.5 * 230 = 6325.

it appears the maintenance days constraint is the limiting factor.

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