Problem 70 - Precalculus

Question

After Hurricane Charley devastated central Florida unexpectedly, Orlando residents prepared for Hurricane Frances by boarding up windows and filling up their cars with gas. It took 5 hours of standing in line to get plywood, and lines for gas were just as time-consuming. A student at Seminole Community College decided to do a spoof of the "Got Milk" ads and created two T-shirts: "Got Plywood" showing a line of people in a home improvement store, and "Got Gas" showing a street lined with cars waiting to pump gasoline. The "Got Plywood" shirts cost \(\$ 8\) to make, and she sold them for \(\$ 13 .\) The "Got Gas" shirts cost \(\$ 6\) to make, and she sold them for \(\$ 10 .\) She decided to limit her costs to \(\$ 1,400 .\) She estimated that demand for these T-shirts would not exceed 200 T-shirts. Find the number of each type of T-shirt she should have made to yield maximum profit.

Step-by-Step Solution

Answer

Make 100 'Got Plywood' and 100 'Got Gas' shirts to maximize profit.

Step 1: Understand the Problem

The student created two types of T-shirts. The profit for 'Got Plywood' shirts is \(13 - 8 = 5 \) dollars per shirt, and for 'Got Gas,' it's \(10 - 6 = 4 \) dollars per shirt. The production cost is limited to \(1,400 \) dollars, and the total number of T-shirts should not exceed \(200 \).

Step 2: Define Variables

Let \( x \) represent the number of 'Got Plywood' shirts, and \( y \) represent the number of 'Got Gas' shirts. Our goal is to maximize the profit \( P \), given by the equation:\[ P = 5x + 4y \]

Step 3: Set Up Constraints

We have two main constraints:1. Costume budget: \[8x + 6y \leq 1,400 \]2. Total shirts: \[ x + y \leq 200 \]Additionally, since the number of shirts cannot be negative: \(x \geq 0 \) and \(y \geq 0 \).

Step 4: Graph the Constraints

Graph the inequalities in the \( xy \)-plane:- For \( 8x + 6y \leq 1,400 \): Rewrite as \( y \leq \frac{1400 - 8x}{6} \).- For \( x + y \leq 200 \): Rewrite as \( y \leq 200 - x \).These form the feasible region on the graph.

Step 5: Find Corner Points

The feasibility region is bounded by the lines from our constraints. Identify potential corner points of this region by solving equations:1. \((0, 0)\)2. Intersection of \( 8x + 6y = 1,400 \) with \( x + y = 200 \).3. Intersections with the axes (e.g., \(y = 0\) and \(x=0\)).

Step 6: Calculate Profit at Each Corner Point

Evaluate the profit \( 5x + 4y \) at each corner point:1. At \((0, 0): P = 5(0) + 4(0) = 0\)2. At the intersection found from constraints solving.3. Check axes intersection points.

Step 7: Determine Maximum Profit

Choose the point with the highest profit from calculation in Step 6. The intersection of \( 8x + 6y = 1,400 \) and \( x + y = 200 \) depends on solving: \[ \begin{cases} 8x + 6y = 1,400 \ x + y = 200 \ \end{cases} \] Solve to find \( x \) and \( y \).

Step 8: Solve the Equations for Intersection

First solve for \( y \) in the second equation: \( y = 200 - x \). Substitute in the first:\[ 8x + 6(200 - x) = 1,400 \]Solve to find \( x \): \[ 8x + 1200 - 6x = 1,400 \]\[ 2x = 200 \implies x = 100 \]\[ y = 200 - 100 = 100 \]

Step 9: Verify and Maximize the Profit

Substitute \( x = 100 \) and \( y = 100 \) into the profit equation: \[ P = 5(100) + 4(100) = 500 + 400 = 900 \]Thus, making 100 of each shirt maximizes the profit.

Key Concepts

Profit Maximization

Profit maximization is a core objective in many business scenarios. This concept involves determining the number of units to produce to achieve the highest possible profit. In the given problem, the student aims to maximize profits from selling two types of T-shirts: 'Got Plywood' and 'Got Gas.'

To find the maximum profit, calculate the profit from each T-shirt style using their selling and production costs. The profit per 'Got Plywood' T-shirt is calculated as:

• Profit per 'Got Plywood': \(13 - 8 = 5 \) per shirt
• Profit per 'Got Gas': \(10 - 6 = 4 \) per shirt

The total profit, represented by \( P\), is expressed as a linear function:\[ P = 5x + 4y \] where \( x \) is the quantity of 'Got Plywood' shirts, and \( y \) is the 'Got Gas' shirts. The task is to find the best combination of \( x \) and \( y \) that maximizes \( P \).

Understanding and solving for profit maximization requires knowledge of the constraints that limit production choices, which in turn guides businesses to optimum decisions.

Constraints

Constraints are factors that limit the available solutions to a problem. These constraints are crucial in linear programming, as they define the boundaries within which profit maximization can be pursued. In this scenario, several constraints affect the production of T-shirts.

First, the budget constraint limits the production cost to $1,400, expressed by the inequality:

• \( 8x + 6y \leq 1,400 \)

Here, \( x \) and \( y \) must adhere to the cost per shirt for 'Got Plywood' and 'Got Gas,' respectively.

The second constraint is on the total number of T-shirts, which cannot exceed 200. This is represented by:

• \( x + y \leq 200 \)

Additionally, another essential condition is that the number of shirts cannot be negative:

• \( x \geq 0 \)
• \( y \geq 0 \)

These constraints need to be carefully analyzed to find a solution within the possible range of values, guiding decisions that conform to all restrictions.

Feasible Region

The feasible region is critical in understanding which combinations of variables are possible given the constraints. Graphically, it is the area where the solutions to the inequalities make sense.

To find this region in the problem, plot the constraints on the xy-plane. The budget constraint \( 8x + 6y \leq 1,400 \) and the total shirts constraint \( x + y \leq 200 \) are boundaries of this region.

Transform the equations to solve for \( y \):

• \( y \leq \frac{1400 - 8x}{6} \)
• \( y \leq 200 - x \)

The shaded area between the lines, where all inequalities overlap, represents the feasible region. Points within this area are potential solutions.

Inside the feasible region, select points that could yield the highest profit based on the profit function. Calculated intersections and critical boundary points provide the maximum or optimal solutions.

Inequalities

Inequalities are mathematical expressions showcasing relationships between two expressions that use symbols like \( \leq \) and \( \geq \). They define the limits of a system and are foundational to linear programming.

In this exercise, inequalities set the constraints within which the optimal production plan must fit. They are used to express conditions like budget limits and production capacities:

• \( 8x + 6y \leq 1,400 \) - This implies that the combined cost of producing both T-shirt types must not exceed $1,400.
• \( x + y \leq 200 \) - This shows the production should not go beyond 200 shirts in total.

Understanding and transforming these inequalities helps in drawing boundaries on a graph, labeling the feasible region. They ensure that calculated solutions for profit maximization adhere to all systemic requirements and restrictions.

Getting comfortable with working through inequalities can aid in finding effective solutions to optimize results and assess feasibility.