Problem 37 - Algebra 2

Question

FARMING For Exercises \(34-37\) , use the following information. Dean Stadler has 20 days in which to plant corn and soybeans. The corn can be planted at a rate of 250 acres per day and the soybeans at a rate of 200 acres per day. He has 4500 acres available for planting these two crops. How much of each should Mr. Stadler plant if the profit on corn is \(\$ 29\) per acre and the profit on soybeans is \(\$ 24\) per acre? What is the maximum profit?

Step-by-Step Solution

Answer

Mr. Stadler should plant 4500 acres of corn, zero acres of soybeans for a maximum profit of $130500.

Step 1: Define Variables

Let \( x \) be the number of acres planted with corn, and \( y \) be the number of acres planted with soybeans. Our objective is to determine the values of \( x \) and \( y \) that will maximize profit.

Step 2: Set Up Constraints

There are three constraints to consider:- The total number of acres must not exceed 4500: \( x + y \leq 4500 \).- The total planting days must not exceed 20: \( \frac{x}{250} + \frac{y}{200} \leq 20 \). Simplifying this gives \( x + 1.25y \leq 5000 \).- Both \( x \) and \( y \) must be non-negative: \( x \geq 0 \) and \( y \geq 0 \).

Step 3: Write the Objective Function

The profit function can be expressed as: \[ P = 29x + 24y \]Our goal is to maximize this profit function subject to the given constraints.

Step 4: Graph the Constraints

Plot the lines \( x + y = 4500 \) and \( x + 1.25y = 5000 \) on a coordinate plane. Shade the region that satisfies both constraints and includes the feasible region for \( x, y \geq 0 \).

Step 5: Find the Corner Points

Determine the points of intersection for the boundary lines and the axes, and check the corner points of the feasible region. The potential corner points include: \((0, 0)\), \((0, 3600)\), \((2500, 2000)\), and \((4500, 0)\).

Step 6: Evaluate the Objective Function at Each Corner Point

Calculate the profit \( P \) for each corner point:- At \((0, 3600)\): \( P = 29(0) + 24(3600) = 86400 \)- At \((2500, 2000)\): \( P = 29(2500) + 24(2000) = 122500 \)- At \((4500, 0)\): \( P = 29(4500) + 24(0) = 130500 \)

Step 7: Identify the Maximum Profit

Among calculated profits, the highest is \(130500\) at the point \((4500, 0)\). This indicates that planting all acres with corn yields the highest profit.

Key Concepts

Objective Function

In linear programming, an objective function is a mathematical expression that represents a goal we wish to achieve. Often, this goal is maximizing or minimizing some quantity. In the exercise provided, Dean Stadler wants to maximize profit from planting corn and soybeans. The objective function here is the profit function, represented as: \[ P = 29x + 24y \] where \( x \) is the number of corn acres planted, and \( y \) is the number of soybean acres planted. Each acre of corn contributes \\(29 to the profit, while each soybean acre contributes \\)24. We want to find the values of \( x \) and \( y \) that make \( P \) as large as possible.

Constraints

Constraints in linear programming are conditions that the solution to a problem must satisfy. For Dean Stadler, these constraints are:

• The sum of the acres planted, \( x + y \), must not exceed the total land available, which is 4500 acres.
• The time constraint: the planting days for corn and soybeans combined must not exceed the 20 days available. Corn takes \( 1/250 \) days per acre and soybeans take \( 1/200 \) days per acre, simplifying to the constraint \( x + 1.25y \leq 5000 \).
• Both \( x \) and \( y \) must be non-negative, as they represent real, physical quantities (acres) that can't be negative.

These constraints create a system of inequalities that define the limits of our solution space.

Feasible Region

The feasible region is the area on a graph where all the constraints are satisfied simultaneously. In our problem, this is the area where both constraints:

• \( x + y \leq 4500 \)
• \( x + 1.25y \leq 5000 \)

intersect, along with the non-negativity conditions \( x \geq 0 \) and \( y \geq 0 \).By shading this intersected area on a graph, we visually identify the feasible region. Any point within this area can be a potential solution, but only specific corner points will possibly maximize or minimize the objective function.

Profit Maximization

Profit maximization in this context involves identifying the point within the feasible region that offers the highest profit according to the objective function. By evaluating the objective function at each corner point of the feasible region, we find possible solutions:

• \((0, 3600)\) gives a profit of \(86400\).
• \((2500, 2000)\) yields \(122500\).
• \((4500, 0)\) achieves the maximum profit of \(130500\).

Thus, Strategically planning 4500 acres of corn yields the maximum profit, illustrating successful profit maximization within the boundaries set by constraints.