Farmer McDonald has 150 acres of land to put into crops and 8000 dollars to spend planting and raising them. He can plant two crops, which we'll call crop A and crop B. Crop A costs $40/acre to raise and crop B costs $60/acre. He also has time costs to consider: Crop A requires 25 hours of labor per acre to raise and crop B takes 20 hours of labor. Altogether, he has 3500 hours of labor that he can spend on raising the crops. If he can expect to make $150/acre for crop A and $200/acre for crop B at market, what is the maximum revenue he can realize at harvest time?
Answer: P = ?
To find the maximum revenue Farmer McDonald can realize at harvest time, we need to determine the optimal combination of crop A and crop B that maximizes his revenue while taking into account the land and budget constraints.
Let's denote the number of acres of crop A as x, and the number of acres of crop B as y.
The objective is to maximize the revenue function P(x, y), which is given by:
P(x, y) = 150 * (150/40) x + 200 * (150/60) y
P(x, y) = 3.75 x + 5 y
Subject to the following constraints:
1. Budget constraint: 40x + 60y ≤ 8000, since crop A costs $40/acre and crop B costs $60/acre.
2. Land constraint: x + y ≤ 150, since Farmer McDonald has 150 acres of land.
3. Labor constraint: 25x + 20y ≤ 3500, since crop A requires 25 hours of labor per acre and crop B requires 20 hours of labor, and he has a total of 3500 hours of labor available.
Now we can solve this optimization problem to find the maximum revenue.
Step 1: Convert the budget constraint to the same units as the revenue function:
40x + 60y ≤ 8000
Divide both sides by 20:
2x + 3y ≤ 400
Step 2: Plot the feasible region:
Plot the region determined by the three constraints on a graph to find the feasible region where the constraints are satisfied.
Step 3: Find the vertices of the feasible region:
The vertices of the feasible region are the points where the boundary lines intersect. There are typically 3-4 vertices.
Step 4: Calculate the revenue at each vertex:
Plug the values of x and y from each vertex into the revenue function P(x, y) = 3.75x + 5y to calculate the revenue at each vertex.
Step 5: Compare the revenues and find the maximum:
Compare the revenues calculated at each vertex and identify the vertex with the highest revenue. This will give us the maximum revenue Farmer McDonald can realize at harvest time.
Therefore, P = Maximum revenue at harvest time.
To find the maximum revenue Farmer McDonald can realize at harvest time, we need to determine the number of acres he should plant for each crop in order to maximize his profit.
Let's assume the number of acres planted for crop A is x, and the number of acres planted for crop B is y.
Given that Farmer McDonald has 150 acres of land available, we have the constraint:
x + y ≤ 150 (Equation 1)
Next, we need to consider the cost of planting and raising the crops. Crop A costs $40/acre, and Crop B costs $60/acre. Since Farmer McDonald has $8000 to spend, we have the following equation:
40x + 60y ≤ 8000 (Equation 2)
Now, let's consider the time constraint. Crop A requires 25 hours of labor per acre, and Crop B requires 20 hours of labor per acre. Since Farmer McDonald has 3500 hours of labor available, we have the equation:
25x + 20y ≤ 3500 (Equation 3)
Finally, we need to maximize the revenue. Crop A can be sold for $150/acre, and Crop B can be sold for $200/acre. Thus, the total revenue can be calculated as:
Revenue = 150x + 200y
Now, to find the maximum revenue, we can set up the following linear programming problem:
Maximize: Revenue = 150x + 200y
Subject to:
x + y ≤ 150 (Equation 1)
40x + 60y ≤ 8000 (Equation 2)
25x + 20y ≤ 3500 (Equation 3)
Solving this linear programming problem will give us the values of x and y that maximize the revenue. This can be done using a graphical method or by using linear programming software.
Once we have the values of x and y, we can substitute them into the revenue equation (Revenue = 150x + 200y) to find the maximum revenue Farmer McDonald can realize at harvest time.