Problem 104 - A Graphical Approach to Precalculus with Limits

Question

Profit from Farm Animals Farmer Jones raises only pigs and geese. She wants to raise no more than 16 animals, with no more than 12 geese. She spends \(\$ 50\) to raise a pig and \(\$ 20\) to raise a goose. She has \(\$ 500\) available for this purpose. Find the maximum profit she can make if she makes a profit of \(\$ 80\) per goose and \(\$ 40\) per pig. Indicate how many pigs and geese she should raise to achieve this maximum.

Step-by-Step Solution

Answer

Farmer Jones should raise 10 pigs and 12 geese for a maximum profit of $1,360.

Step 1: Understanding the Problem

We need to find out how many pigs and geese Farmer Jones should raise to maximize profit while considering constraints such as the number of animals, the cost, and her budget.

Step 2: Define Variables

Let \( x \) represent the number of pigs Farmer Jones raises and \( y \) represent the number of geese.

Step 3: Constraints Definition

There are three main constraints:1. \( x + y \leq 16 \) (no more than 16 animals)2. \( y \leq 12 \) (no more than 12 geese)3. \( 50x + 20y \leq 500 \) (total cost constraint)

Step 4: Profit Function

The objective is to maximize the profit function: \( P = 40x + 80y \).

Step 5: First Step: Identify Feasible Region

Using the constraints, plot inequalities on a graph to find a feasible region. The feasible region is bounded by the lines:\[ x + y = 16 \]\[ y = 12 \]\[ 50x + 20y = 500 \]

Step 6: Corner Points Calculation

Calculate the intersection points of the boundary lines to find the vertices of the feasible region. The points are:1. \((0, 12)\)2. \((10, 12)\)3. \((10, 6)\)4. \((0, 0)\)5. \((8, 8)\) (intersection of \(50x + 20y = 500\) and \( x + y = 16\))

Step 7: Maximize Profit at Corner Points

Calculate profit at each corner point:1. \( P(0, 12) = 40(0) + 80(12) = 960 \)2. \( P(10, 12) = 40(10) + 80(12) = 1,360 \)3. \( P(10, 6) = 40(10) + 80(6) = 920 \)4. \( P(0, 0) = 40(0) + 80(0) = 0 \)5. \( P(8, 8) = 40(8) + 80(8) = 960 \)

Step 8: Determine Maximum Profit

The maximum profit is at \((10, 12)\), which gives a profit of \$1,360. Thus, Farmer Jones should raise 10 pigs and 12 geese.

Key Concepts

Profit Maximization

Profit maximization is a key goal for businesses, including Farmer Jones's farm. The idea is to determine the optimal number of pigs and geese to raise so that the profits are as high as possible. For Farmer Jones, this means deciding on the best combination of 10 pigs and 12 geese, which yields a maximum profit of \( \\(1,360 \).
The profit formula used is \( P = 40x + 80y \), where \( x \) is the number of pigs, and \( y \) is the number of geese. Here, the profits per pig and goose are \( \\)40 \) and \( \$80 \) respectively.
By plugging in the different combinations tested — like \((10, 12)\) — we find the combination with the highest calculated profit.

Constraints in Linear Programming

In linear programming, constraints are limits or restrictions on the variables. These constraints guide the formulation of the problem and solutions, ensuring they remain realistic and feasible.
For Farmer Jones, the constraints are:

• The total number of pigs and geese should not exceed 16, represented as \( x + y \leq 16 \).
• The number of geese should not be more than 12, expressed as \( y \leq 12 \).
• The total cost for raising pigs and geese should not exceed her budget of \( \$500 \). This is modeled by the inequality \( 50x + 20y \leq 500 \).

These constraints help frame the problem within the real-world limitations of resources such as space and budget.

Feasible Region

The feasible region in a linear programming problem is the area on a graph where all constraints overlap, representing all possible solutions that satisfy these constraints.
Farmer Jones's constraints create a polygon on the coordinate graph. The feasible region is the space inside this polygon, bounded by the lines \(x + y = 16\), \(y = 12\), and \(50x + 20y = 500\).
The vertices of this region, where the solution is likely to lie, include points like \((0, 12)\), \((10, 12)\), and \((8, 8)\). By finding and calculating the profit for each of these points, Farmer Jones can identify where she can maximize her profits.

Objective Function

The objective function in linear programming defines what needs to be maximized or minimized. For Farmer Jones, the objective is profit maximization — thus the objective function is \( P = 40x + 80y \).
This function quantifies Farmer Jones's goal: to earn the most profit possible from raising pigs and geese. Each term in the equation reflects the profit contribution of pigs and geese.
By evaluating the objective function's value at each vertex of the feasible region, she determines the best combination of animals to maximize profits.