Question

A farmer has 70 acres of land available for planting either soybeans or wheat. The cost of preparing the soil, the workdays required, and the expected profit per acre planted for each type of crop are given in the following table. The farmer cannot spend more than \(\$ 1800\) in preparation costs and cannot use a total of more than 120 workdays. How many acres of each crop should be planted to maximize the profit? What is the maximum profit? What is the maximum profit if the farmer is willing to spend no more than \(\$ 2400\) on preparation?

Short answer

Plant 60 acres of soybeans and 10 acres of wheat for a maximum profit of \(2700. New budget: maximum profit is \)3600.

Step-by-step solution

Define variables

Let \( x \) be the number of acres planted with soybeans and \( y \) be the number of acres planted with wheat.

Formulate constraints

Based on the given information, set up the constraints:1. Total land constraint: \( x + y \ \leq 70 \)2. Preparation cost constraint: \( 10x + 20y \ \leq 1800 \)3. Workdays constraint: \( 2x + 4y \ \leq 120 \)

Formulate the objective function

The objective is to maximize the profit: \( \text{Profit} = 40x + 30y \)

Solve the linear programming problem

Solve the system of inequalities to find the feasible region and then evaluate the objective function at each vertex of the feasible region.

Evaluate the new maximum profit with the increased budget

Repeat steps for the new preparation cost of \( 2400 \) and find the new solution.

Key concepts

Constraints

Constraints are the conditions that must be met for a solution to be feasible. In our problem, we have three main constraints:

1. **Total Land Constraint:** The farmer has a total of 70 acres for planting. This gives us the equation: \( x + y \leq 70 \), where 𝑥 is the number of acres of soybeans and 𝑦 is the number of acres of wheat.

2. **Preparation Cost Constraint:** The farmer cannot exceed a preparation cost of \$1800\. The cost for planting soybeans and wheat per acre are \$10\ and \$20\ respectively. Therefore, we get the constraint: \( 10x + 20y \leq 1800 \).

3. **Workdays Constraint:** The total available workdays are limited to 120. Soybeans require 2 workdays per acre while wheat needs 4 workdays per acre. This provides the equation: \( 2x + 4y \leq 120 \).

These constraints help define the feasible region, which is the set of all possible solutions that satisfy all the constraints. Understanding these conditions accurately is essential for setting up the linear programming problem.

Objective Function

The objective function represents what we are trying to optimize—in this case, profit. For the farmer, the formula to calculate profit is given by \( \text{Profit} = 40x + 30y \), where \( x \) and \( y \) are the acres of soybeans and wheat planted, respectively.

The coefficients 40 and 30 represent the profits per acre for soybeans and wheat, respectively. Our goal is to maximize this function while adhering to the constraints. By substituting different values of 𝑥 and 𝑦 within the feasible region into the objective function, we can determine the maximum profit.

Feasible Region

The feasible region is the area on a graph that contains all the possible solutions that meet the problem's constraints. To find this region, we graph all constraints on the same coordinate plane:

- The line \( x + y \leq 70 \) represents the total land constraint.
- The line \( 10x + 20y \leq 1800 \) represents the preparation cost constraint.
- The line \( 2x + 4y \leq 120 \) represents the workdays constraint.

By identifying the area where these inequalities overlap, we define the feasible region. Any point in this region satisfies all constraints. The vertices (corner points) of this region are particularly important because, in linear programming, the maximum and minimum values of the objective function are found at these vertices.

Optimization

Optimization in linear programming involves finding the maximum or minimum value of the objective function within the feasible region. For this problem, we want to maximize profit.

The process includes:

- Graphing the constraints to establish the feasible region.
- Identifying the vertices of the feasible region.
- Evaluating the objective function at each vertex.

Let's find our vertices from the constraints defined earlier. By substituting these points into the objective function \(40x + 30y\), we determine the profits. The vertex yielding the highest value provides the optimal solution.

If the budget increases to \$2400\, we adjust the constraint \( 10x + 20y \leq 2400 \), redefine the feasible region, find new vertices, and re-evaluate the objective function at those vertices to find the new maximum profit.