Problem 23 - Precalculus: Functions and Graphs

Question

Planning sterage units A contractor has a large building that she wishes to convert into a series of rental storage spaces. She will construct basic \(8 \mathrm{ft} \times 10 \mathrm{ft}\) units and deluxe \(12 \mathrm{ft} \times 10 \mathrm{ft}\) units that contain extra shelves and a clothes closet. Market considerations dictate that there be at least twice as many basic units as deluxe units and that the basic units rent for $$ 75\( per month and the deluxe units for $$ 120 per month. At most \)7200 \mathrm{ft}^{2}\( is available for the storage spaces, and no more than \)\$ 80,000 can be spent on construction. If each basic unit will cost $$ 800$ to make and each deluxe unit will cost $$ 1600, how many units of each type should be constructed to maximize monthly revenue?

Step-by-Step Solution

Answer

Construct 40 basic units and 20 deluxe units to maximize revenue.

Step 1: Define Variables

Let \( x \) represent the number of basic units and \( y \) represent the number of deluxe units. We aim to determine the values of \( x \) and \( y \) that maximize the monthly revenue while satisfying the given constraints.

Step 2: Establish Revenue Function

The revenue from basic units is \( 75x \) and from deluxe units is \( 120y \). Thus, the revenue function to maximize is: \[ R = 75x + 120y \]

Step 3: Formulate Constraints

There are several constraints to consider:- Space constraint: \[ 80x + 120y \leq 7200 \] (in square feet)- Cost constraint: \[ 800x + 1600y \leq 80000 \] (in dollars)- Market constraint: \[ x \geq 2y \]- Non-negativity constraints: \[ x \geq 0, \; y \geq 0 \]

Step 4: Simplify Constraints

Simplify the inequalities:- Space: \[ 8x + 12y \leq 720 \]- Cost: \[ x + 2y \leq 100 \]

Step 5: Determine Feasible Region

Plot the constraints on a graph to identify the feasible (allowable) region. The feasible region is the area where all constraints overlap after considering only non-negative values for both \( x \) and \( y \).

Step 6: Identify Corner Points

The potential solutions lie at the intersection points (corner points) of the feasible region (lines). Determine these intersection points by solving the system of equations derived from each pair of constraint lines.

Step 7: Evaluate Revenue at Each Corner Point

Substitute the coordinates of each corner point into the revenue function \( R = 75x + 120y \) to find which combination yields the highest revenue. Compare these values to determine the optimal solution.

Key Concepts

Constraints

In linear programming, constraints are the conditions that limit the possibilities of obtaining a solution. They are essentially the rules we must follow to make our planned scenario feasible and credible. In the given exercise, the constraints are the boundaries that dictate how many basic and deluxe storage units can be constructed based on available resources and market conditions.

Here are the crucial types of constraints in the scenario:

• Space constraint: This represents the physical limit of 7200 square feet of available space. The total area occupied by the basic and deluxe units must not exceed this limit.
• Cost constraint: The available budget for construction is $80,000. Total spending on creating these units cannot surpass this amount.
• Market constraint: Market conditions demand at least twice as many basic units as deluxe units. This ensures the contractor meets probable demand.
• Non-negativity constraints: Negative amounts of units are impractical, which means the numbers of both unit types must be zero or more.

These constraints help transform a real-world problem into a mathematical model ready to be solved by optimization techniques.

Feasible Region

The feasible region in linear programming is the set of all possible points that satisfy the given constraints. It represents all potential combinations of basic and deluxe units that the contractor can build given her limitations. Visualizing a problem often helps simplify an abstract concept, and this is where plotting constraints on a graph becomes handy.

Key points of the feasible region include:

• Graphing boundaries: Each constraint corresponds to a line on a graph. The area where all these lines overlap and are bounded is the feasible region.
• Non-negativity: Since non-negative quantities are required, the feasible region is within the first quadrant of the graph, which means both coordinates are positive.
• Intersection points: The corners of this region are points where two or more constraints meet. These are termed corner points and hold potential solutions.

By defining a feasible region, one can find which combinations of variables—in this instance, the number of basic and deluxe units—are possible for consideration in optimizing revenue.

Revenue Function

A revenue function in linear programming is an equation that represents the total income earned from selling or renting products, given certain variables. In this contractor's problem, the revenue function calculates the total monthly income from constructing and renting out storage units.

Essential elements of the revenue function include:

• Basic units contribution: Each basic unit rented brings in \(75 per month, represented as a term in the function: \(75x\).
• Deluxe units contribution: Each deluxe unit rented contributes \)120 monthly, forming another term in the equation: \(120y\).
• Total revenue calculation: The function tops up to the complete equation \( R = 75x + 120y \), which shows how different numbers of each unit type impact total revenue.

Understanding the revenue function is crucial for calculating potential monthly earnings, allowing us to predict which unit combinations will be most profitable.

Optimization

Optimization in linear programming is about finding the best possible solution, within given constraints, that maximizes or minimizes a particular objective. In this exercise, the contractor's goal is to optimize monthly revenue by deciding on the best quantity of basic and deluxe storage units to construct.

Here's how optimization works with the given problem:

• Maximize revenue: The purpose is to find the values of \(x\) and \(y\) that yield the greatest potential revenue, as defined by the revenue function \(R = 75x + 120y\).
• Evaluating corner points: Since the solution lies at the corners of the feasible region, calculate the revenue function at these points to determine the maximum possible revenue.
• Comparing values: By plugging these corner values into the revenue equation, compare results to see which yields the highest revenue.
• Optimal solution selection: Choose the point (number of basic and deluxe units) that gives the maximum revenue, ensuring it respects all constraints.

Optimization ensures the most effective and efficient use of resources, providing the contractor with the best strategy to maximize profitability.