Problem 18 - Precalculus: Functions and Graphs

Question

Maximizing profit A man plans to operate a stand at a one-day fair at which he will sell bags of peanuts and bags of candy. He has $$ 2000\( available to purchase his stock, which will cost $$ 2.00 per bag of peanuts and $$ 4.00\) per bag, of candy. He intends to sell the peanuts at $$ 3.00$ and the candy at 55.50 per bag. His stand can accommodate up to 500 bags of peanuts and 400 bags of candy. From past experience he knows that he will sell no more than a total of 700 bags. Find the number of bags of each that he should have available in order to maximize his profit. What is the maximum profit?

Step-by-Step Solution

Answer

He should sell 300 bags of peanuts and 400 bags of candy for a maximum profit of 900.

Step 1: Define the Variables

Let \( x \) be the number of bags of peanuts and \( y \) be the number of bags of candy. We need to determine values of \( x \) and \( y \) that maximize profit, subject to constraints.

Step 2: Construct the Objective Function

The profit for peanuts is \( 3x - 2x = x \) and the profit for candy is \( 5.5y - 4y = 1.5y \). Therefore, the objective function to maximize the total profit is given by \( P = x + 1.5y \).

Step 3: Set Up the Constraints

The constraints are: 1. \( 2x + 4y \leq 2000 \) (Budget constraint)2. \( x \leq 500 \) (Peanut storage limit)3. \( y \leq 400 \) (Candy storage limit)4. \( x + y \leq 700 \) (Sales limit)5. \( x, y \geq 0 \) (Non-negativity constraint)

Step 4: Graph the Constraints

Graph the constraints on a coordinate plane to identify the feasible region. The lines are: - \( 2x + 4y = 2000 \) - \( x = 500 \) - \( y = 400 \) - \( x + y = 700 \)The feasible region is bounded by these constraints.

Step 5: Identify the Vertices of the Feasible Region

The vertices (corners) of the feasible region, which are potential candidates for maximizing the profit, are determined by solving the intersections of the constraints. They are: - (0, 400) - (500, 0) - (300, 400) - (500, 200)

Step 6: Evaluate the Objective Function at Each Vertex

Calculate the profit \( P = x + 1.5y \) for each vertex:1. At \((0,400)\): \(P = 0 + 1.5(400) = 600\)2. At \((500,0)\): \(P = 500 + 1.5(0) = 500\)3. At \((300,400)\): \(P = 300 + 1.5(400) = 900\)4. At \((500,200)\): \(P = 500 + 1.5(200) = 800\)

Step 7: Select the Maximum Profit

The maximum profit occurs at the vertex (300, 400), where the profit is 900.

Key Concepts

Objective Function

In linear programming, the objective function is a crucial element that acts as the foundation for what you are trying to achieve. In this exercise, our man wants to maximize his profit from selling bags of peanuts and candy.
The objective function is a mathematical representation of this goal.
It tells us how the profit depends on the number of bags of peanuts, represented as \( x \), and candy, represented as \( y \). For the peanuts, the profit is \(3x - 2x = x\), while for the candy, it is \(5.5y - 4y = 1.5y\). Collectively, the equation we need to maximize is given by:

• \( P = x + 1.5y \)

The objective function guides all subsequent calculations to find the optimal numbers of each type of bag to be sold for maximum profit.

Constraints

Constraints are the limitations or restrictions imposed on the decision variables, \(x\) and \(y\), in linear programming. Here, you must consider financial, physical, and logical limits the man faces at his stand.
Constraints can be likened to rules that must be respected while trying to achieve the objective function.
In our problem, we have the following constraints:

• Budget Limit: \( 2x + 4y \leq 2000 \)
• Peanut Storage Constraint: \( x \leq 500 \)
• Candy Storage Constraint: \( y \leq 400 \)
• Total Sales Constraint: \( x + y \leq 700 \)

These rules ensure that you do not overspend, overextend storage, or exceed the maximum sales based on past experiences. Remember, \( x, y \geq 0 \) due to non-negativity constraints, ensuring no negative number of bags.

Feasible Region

The feasible region is a critical concept in linear programming. It represents all possible combinations of the variables \(x\) and \(y\) that satisfy the constraints. Visualizing the feasible region often involves plotting it on a graph based on the constraints.
By graphing the constraints, you create a map that shows you what combinations of peanuts and candy can be sold simultaneously.
The feasible region is the intersection area on the graph where all these constraints overlap and the solutions remain valid.Within this region, the boundaries are shaped by:

• \( 2x + 4y = 2000 \)
• \( x = 500 \)
• \( y = 400 \)
• \( x + y = 700 \)

It's within this area that we find what mix of \(x\) and \(y\) offers the maximum profit, making the feasible region a target for finding the optimal solution.

Profit Maximization

Profit maximization is the ultimate goal in this exercise, aimed at determining the best number of bags of peanuts \(x\) and candy \(y\) to sell. Once the feasible region is identified, the next step is evaluating the objective function at several critical points, called vertices, within this region.
These vertices represent potential solutions that might optimize profit.
At each vertex, we calculate the expected profit by substituting the coordinates back into the objective function \( P = x + 1.5y \).For instance:

• At Point (0, 400): Profit is \(0 + 1.5 \, \times \, 400 = 600\).
• At Point (500, 0): Profit is \(500 + 1.5 \, \times \, 0 = 500\).
• At Point (300, 400): Profit is \(300 + 1.5 \, \times \, 400 = 900\).
• At Point (500, 200): Profit is \(500 + 1.5 \, \times \, 200 = 800\).

The maximum profit is precisely where it's the highest among these calculations, which is at point (300, 400) with a profit of 900. Thus, the man should prepare 300 bags of peanuts and 400 bags of candy to maximize his profit at the fair.