Problem 88 - A Graphical Approach to College Algebra

Question

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

Raise 6 pigs and 10 geese for a maximum profit of $920.

Step 1: Understanding the Variables

Let's define our variables: \( x \) will represent the number of pigs, and \( y \) will represent the number of geese. We know: Farmer Jones raises only pigs and geese.

Step 2: Establishing Constraints

We identify the constraints based on the problem statement: 1. Total number of animals: \[ x + y \leq 16 \]2. Maximum number of geese: \[ y \leq 12 \]3. Budget:\[ 50x + 20y \leq 500 \]

Step 3: Objective Function for Profit

The profit function based on the profits per animal is: \[ P = 40x + 80y \] where \( P \) is the total profit.

Step 4: Setting Up a Linear Programming Problem

The objective is to maximize the profit \( P = 40x + 80y \) subject to the constraints:- \( x + y \leq 16 \)- \( y \leq 12 \)- \( 50x + 20y \leq 500 \)Also, \( x, y \geq 0 \).

Step 5: Solving for Maximum Profit

To find the maximum profit, evaluate the objective function at critical points, typically where the constraints intersect with units:1. Solve for \( x \) and \( y \) at the intersections: - Solving \( x + y = 16 \) and \( 50x + 20y = 500 \): - Simplify and solve: \( 5x + 2y = 50 \to y = 16 - x \to 5x + 2(16-x) = 50 \) - \( 5x + 32 - 2x = 50 \to 3x = 18 \to x = 6 \) - Substitute \( x = 6 \) into \( x + y = 16 \): \( y = 10 \)2. Other vertex calculations: - Intersection of \( y = 12 \) with budget constraint: - Substitute \( y = 12 \) into \( 50x + 20 \times 12 = 500 \): - \( 50x + 240 = 500 \to 50x = 260 \to x = 5.2 \) - not integer - Intersection of budget line \( 50x + 20y = 500 \) with \( x = 0 \): - \( 20y = 500 \to y = 25 \) - exceeds max geese. - With \( y = 0 \): - \( 50x = 500 \to x = 10 \) (This gives \( P = 400 \))3. Evaluate \( P \) at integer solutions from constraints: - \( (x, y) = (6, 10) \Rightarrow P = 40\times6 + 80\times10 = 920 \) - Above solution satisfies all constraints and provides a maximum profit.

Step 6: Conclusion

The farmer should raise 6 pigs and 10 geese to achieve the maximum profit of 920 dollars.

Key Concepts

Objective Function

In linear programming, the objective function is a mathematical expression you want to optimize, either by maximizing or minimizing its value. In this problem, the goal is to maximize profits through the ideal combination of pigs and geese.
The profit per pig is given as \(40, and the profit per goose is \)80. These figures are assembled into the objective function, allowing us to calculate the total profit based on the number of animals Farmer Jones raises. This can be expressed as: \[ P = 40x + 80y \]where:\

• \( x \) is the number of pigs.
• \( y \) is the number of geese.

By plugging in different values of \( x \) and \( y \), we can find different profit scenarios and select the best possible outcome.

Constraints

Constraints in linear programming are conditions that restrict the possible solutions to the problem. For Farmer Jones, there are several constraints she has to work within:

• She cannot have more than 16 animals in total, represented by the inequality \( x + y \leq 16 \).
• She is limited to a maximum of 12 geese, represented by the inequality \( y \leq 12 \).
• Her budget for raising the animals cannot exceed \(500. Each pig costs \)50, and each goose costs $20 to raise, leading to the budget constraint \( 50x + 20y \leq 500 \).

These constraints form the boundaries of the solution, and the challenge is to find a combination of \( x \) and \( y \) that meets all these conditions while optimizing the profit.

Profit Maximization

Profit maximization involves finding the point where our objective function reaches its highest value within the given constraints.
In this exercise, we look for integer solutions because Farmer Jones can only raise whole numbers of animals. By solving the various linear equations formed by our constraints and intersecting them, we can find possible solutions to examine.
The intersections of constraints often show the optimal points to test. Calculating the profit at these points helps identify the maximum possible profit. For Farmer Jones, raising 6 pigs and 10 geese provides a maximum achievable profit of $920, satisfying all the constraints and maximizing her earnings.

Integer Solutions

In many linear programming problems, including this one, solutions must be integers. This is because partial units of pigs or geese don't make sense in real-life scenarios.
While the general solution to a system of linear inequalities may give rise to fractional numbers, it's important to round them to the nearest integer or find alternative integer points that satisfy all constraints and are still optimal.
For this problem, one calculated solution was \( x = 6 \) and \( y = 10 \), which not only satisfies the constraints but also makes logical sense because Farmer Jones cannot have half a pig or goose. Exploring integer solutions ensures solutions are feasible and occur in real-world situations.