Problem 15 - Algebra and Trigonometry with Analytic Geometry

Question

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 $$\$ 3840$$ for seed and $$\$ 4200$$ for labor, how many acres of each crop should be planted to obtain the maximum profit?

Step-by-Step Solution

Answer

The farmer should plant 30 acres of alfalfa and 60 acres of corn to maximize profit.

Step 1: Assign Variables

Let \( x \) represent the number of acres planted with alfalfa and \( y \) represent the number of acres planted with corn. We know the total available acres is 90, so we can write the equation \( x + y = 90 \).

Step 2: Cost Constraints

Determine the cost constraints based on seed and labor costs:- Seed cost constraint: \( 32x + 48y \leq 3840 \)- Labor cost constraint: \( 60x + 30y \leq 4200 \)

Step 3: Profit Function

The profit function is calculated as the revenue minus costs:- Revenue from alfalfa: \( 500x \)- Revenue from corn: \( 700y \)The profit function, \( P \), is \( P = 500x + 700y - (32x + 48y) - (60x + 30y) \), which simplifies to \( P = 408x + 622y \).

Step 4: Simplify Constraints

Simplify the constraints: - Seed cost: \( 32x + 48y \leq 3840 \) simplifies to \( 2x + 3y \leq 240 \)- Labor cost: \( 60x + 30y \leq 4200 \) simplifies to \( 2x + y \leq 140 \).

Step 5: Graph Constraints

Graph the linear inequalities derived from the constraints: - \( x + y = 90 \)- \( 2x + 3y \leq 240 \)- \( 2x + y \leq 140 \)These lines will form a feasible region on a graph, and it's this region we need to further analyze.

Step 6: Identify Corner Points

Find the vertices of the feasible region by solving the equations:1. Intersection of \( x + y = 90 \) and \( 2x + 3y = 240 \)2. Intersection of \( x + y = 90 \) and \( 2x + y = 140 \)3. Intersection of \( 2x + 3y = 240 \) and \( 2x + y = 140 \)4. Check intersection with x and y-axis where equations hit or cross, adhering to constraints.

Step 7: Calculate Profit at Corner Points

Calculate the profit for each corner point using \( P = 408x + 622y \). Choose the corner point that gives the maximum profit.

Step 8: Verify Feasibility

Ensure that the chosen corner point satisfies all original constraints. This includes both the maximum budget for seed and labor costs.

Key Concepts

Optimization Problem

When we talk about an optimization problem in linear programming, it involves finding the best solution from a set of feasible solutions. Here, the goal is to determine the optimal number of acres for planting alfalfa and corn that would maximize the farmer's profit.

Optimizing means we're either maximizing or minimizing a specific objective. In this case, the objective is profit, which is maximized when costs are minimized and revenues are increased. When defining an optimization problem:

• Objective Function: This represents the goal to optimize, such as maximizing profit or minimizing costs.
• Decision Variables: Variables that decision makers will decide upon which impact the outcome. For this farmer, variables are the number of acres allocated to each crop.
• Constraints: These are limitations or requirements, like budget restrictions and the amount of available land.

The main idea is to find values for the variables that maximize the objective while still satisfying every constraint.

Constraints Analysis

Constraints are conditions that must be met for a solution to be considered feasible in an optimization problem. In this scenario, the farmer has several constraints:

• Land Constraint: The farmer has a limited amount of land, 90 acres, which can be used for both alfalfa and corn. This is expressed as the equation \( x + y = 90 \) where \( x \) and \( y \) represent the acres for alfalfa and corn, respectively.
• Seed Cost Constraint: The budget for seeds cannot exceed \(3840. The cost for alfalfa seed per acre is \)32 and for corn is \(48, leading to the inequality \( 32x + 48y \leq 3840 \).
• Labor Cost Constraint: Labor expenses should not surpass \)4200, with \(60 per acre for alfalfa and \)30 per acre for corn. This is represented by \( 60x + 30y \leq 4200 \).

Constraints must be carefully analyzed to ensure any proposed solution is achievable given the real-world restrictions the farmer faces.

Profit Maximization

Profit maximization involves adjusting the decision variables to get the highest possible profit from the available resources. For our farmer, profit maximization means determining the number of acres to plant with each crop to maximize financial returns.

The profit function in this problem is calculated by subtracting associated costs from the revenue. Revenue from alfalfa and corn are \( 500x \) and \( 700y \) respectively, while costs for seed and labor for the two crops are calculated and subtracted from the revenue.

• Profit Function: This is given by the equation \( P = 408x + 622y \). Here, \( 408 \) is the profit contribution per acre from alfalfa and \( 622 \) from corn after costs.

By evaluating this profit function at various feasible corner points determined from the constraints, the optimal planting strategy is found.

Graphical Method

The graphical method is a visual way to solve linear programming problems involving two variables. It involves sketching the constraints as lines on a graph and identifying the feasible region where all constraints overlap. For this problem, the graph includes:- The land constraint \( x + y = 90 \),- The simplified seed constraint \( 2x + 3y \leq 240 \), and- The labor constraint \( 2x + y \leq 140 \).To solve using the graphical method:

• Plot the Constraints: Each constraint is plotted as a line on the graph, with shading indicating regions where the constraint conditions are satisfied.
• Identify Feasible Region: The overlapping area satisfies all constraints simultaneously and represents all possible solutions.
• Corner Points Analysis: The corners of this region are analyzed as potential solutions where the objective function can achieve its optimal value.

By checking these points, one can quickly identify where the maximum profit occurs, offering a feasible and optimal solution to the problem.