A small farm in Illinois has 110 acres of land available on which to grow corn and soybeans. The following table shows the cultivation cost per acre, the labor cost per acre, and the expected profit per acre. The column on the right shows the amount of money available for each of these expenses. Find the number of acres of each crop that should be planted in order to maximize profit. Soybeans Corn Money Available Cultivation cost per acre $20 $20 $1,800 Labor cost per acre $60 $50 $4,800 Profit per acre $150 $250
To find the number of acres of each crop that should be planted in order to maximize profit on a small farm in Illinois, we can use linear programming.
Let's denote the number of acres of soybeans as 'x' and the number of acres of corn as 'y'.
We want to maximize the total profit, which is given by the equation:
Total Profit = (Profit per acre of soybeans * Number of acres of soybeans) + (Profit per acre of corn * Number of acres of corn)
So, the objective function can be formulated as:
f(x, y) = 150x + 250y
Now, let's consider the constraints:
1. The total number of acres available is 110:
x + y = 110
2. The amount of money available for cultivation and labor costs:
Cultivation cost per acre of soybeans * Number of acres of soybeans + Cultivation cost per acre of corn * Number of acres of corn <= Total money available for cultivation
(20x + 20y) <= 1,800
Labor cost per acre of soybeans * Number of acres of soybeans + Labor cost per acre of corn * Number of acres of corn <= Total money available for labor
(60x + 50y) <= 4,800
So, the system of equations and inequalities we need to solve is:
x + y = 110
20x + 20y <= 1,800
60x + 50y <= 4,800
To find the solution, we can use optimization techniques like the Simplex Method or graphical methods. Let me solve it for you.
To find the number of acres of each crop that should be planted to maximize profit, we can use linear programming.
Let's define:
x = number of acres of soybeans to be planted
y = number of acres of corn to be planted
We need to maximize the profit:
Profit = 150x + 250y
Subject to the following constraints:
Cultivation cost constraint: 20x + 20y ≤ 1,800
Labor cost constraint: 60x + 50y ≤ 4,800
Acreage constraint: x + y ≤ 110
We will solve this linear programming problem using the simplex method.
Step 1: Formulate the problem in standard form:
Maximize: Profit = 150x + 250y
Subject to:
20x + 20y + s1 = 1,800
60x + 50y + s2 = 4,800
x + y + s3 = 110
where s1, s2, s3 are slack variables.
Step 2: Convert the constraints to equations:
20x + 20y + s1 = 1,800
60x + 50y + s2 = 4,800
x + y + s3 = 110
Step 3: Convert the objective function to a minimization problem:
Minimize: -150x - 250y
Step 4: Set up the initial simplex tableau:
| x | y | s1 | s2 | s3 | RHS |
-----------------------------------------------------------
Row 0 | -150 | -250 | 0 | 0 | 0 | 0 |
Row 1 | 20 | 20 | 1 | 0 | 0 | 1,800 |
Row 2 | 60 | 50 | 0 | 1 | 0 | 4,800 |
Row 3 | 1 | 1 | 0 | 0 | 1 | 110 |
Step 5: Perform iterations until an optimal solution is reached:
First iteration:
Pivot on (1, 1) cell: Row 1 / 20
| x | y | s1 | s2 | s3 | RHS |
-------------------------------------------------------------------------
Row 0 | -150 | -250 | 0.00 | -7.50 | 0.00 | -45,000 |
Row 1 | 1 | 1 | 0.05 | 0.00 | 0.00 | 90 |
Row 2 | 40 | 40 | 2.00 | -0.40 | 0.00 | 1,200 |
Row 3 | 59 | 59 | 1.75 | -0.35 | 1.00 | 22,100 |
Next iterations:
Pivot on (1, 2) cell: Row 3 / 59
| x | y | s1 | s2 | s3 | RHS |
-------------------------------------------------------------------------
Row 0 | -150 | -250 | 0.00 | 6.78 | 0.00 | -30,036 |
Row 1 | 1 | 1 | -0.01 | 0.02 | -0.02| 90 |
Row 2 | 20 | 20 | 1.03 | -0.08 | -0.03| 960 |
Row 3 | 1 | 1 | 0.03 | -0.01 | 0.02 | 374 |
Optimal solution is reached when all coefficients in the top row are negative or zero.
Step 6: Interpret the results:
From the final tableau, we can see that the optimal solution is x = 20 acres of soybeans and y = 20 acres of corn, which will maximize the profit at $1,200.
Therefore, the farm should plant 20 acres of soybeans and 20 acres of corn to maximize profit.