Problem 10 - Algebra and Trigonometry with Analytic Geometry

Question

A manufacturer of cell phones makes a profit of \(\$ 25\) on a deluxe model and \(\$ 30\) on a standard model. The company wishes to produce at least 80 deluxe models and at least 100 standard models per day. To maintain high quality, the daily production should not exceed 200 cell phones. How many of each type should be produced daily in order to maximize the profit?

Step-by-Step Solution

Answer

Produce 80 deluxe and 120 standard models to maximize profit.

Step 1: Identify Variables

Let \( x \) represent the number of deluxe models produced daily, and \( y \) represent the number of standard models produced daily.

Step 2: Define Objective Function

The problem asks for maximizing the profit, which can be represented as \( P = 25x + 30y \).

Step 3: Set Constraints

Based on the problem description, we have three constraints: \( x \geq 80 \), \( y \geq 100 \), and \( x + y \leq 200 \).

Step 4: Graph Constraints

Graph the inequalities on a coordinate plane with \( x \) on the x-axis and \( y \) on the y-axis. The feasible region is determined by the intersection of these inequalities.

Step 5: Identify Feasible Region

Mark the region of the graph where all constraints overlap. This is the feasible set of solutions where the company can operate under these limits.

Step 6: Find Vertices of the Feasible Region

The vertices of the feasible region are determined from the points of intersection of the line equations. These are typically \((80, 100)\), \((80, 120)\), and \((100, 100)\).

Step 7: Calculate Profit for Each Vertex

Evaluate the objective function \( P = 25x + 30y \) at each vertex:- At \((80, 100)\), \( P = 25(80) + 30(100) = 5000 \).- At \((80, 120)\), \( P = 25(80) + 30(120) = 5600 \).- At \((100, 100)\), \( P = 25(100) + 30(100) = 5500 \).

Step 8: Choose the Maximum Profit

Compare profits at each vertex and select the vertex with the highest value. In this case, the maximum profit is 5600 at the point \((80, 120)\).

Key Concepts

Objective Function

The objective function is a mathematical expression that represents the goal of the problem. In the case of linear programming, it typically involves maximizing or minimizing a particular value. For this exercise, the objective function represents the total profit from selling two types of cell phones: deluxe and standard models.
The problem hints at maximizing profit. Therefore, our objective function is defined as \( P = 25x + 30y \), where \( x \) is the number of deluxe models produced daily, and \( y \) is the number of standard models produced daily. Each deluxe phone contributes \(25 to the profit, and each standard phone adds \)30.
In essence, the objective function provides a formula for evaluating different production plans and choosing the one that yields the greatest profit.

Constraints

Constraints are the conditions or limitations imposed on the decision-making process. These constraints ensure that the solution to our problem stays within acceptable and practical limits. In this exercise, the constraints come from production requirements and limits.

• At least 80 deluxe models need to be produced, so \( x \geq 80 \).
• At least 100 standard models are required daily, so \( y \geq 100 \).
• The total number of phones produced should not exceed 200, thus \( x + y \leq 200 \).

After identifying these constraints, they can be graphed as inequalities to visualize the problem's limitations. The feasible solution will lie within the area where all these constraints intersect.

Feasible Region

A feasible region is the area on a graph that represents all possible combinations of decision variables that satisfy the constraints of a linear programming problem. Graphing the constraints allows us to visualize this region.
For the given exercise, the feasible region is determined by plotting the constraints on a coordinate plane with \( x \) on the x-axis and \( y \) on the y-axis. The intersection area where these inequalities overlap defines our feasible region. This area shows all the potential combinations of deluxe and standard models that comply with production constraints.
The feasible points will typically form a polygon, known as the feasible set, and each vertex of this polygon provides a potential solution to evaluate for optimality.

Profit Maximization

Profit maximization is the ultimate goal of this exercise. Once we have identified the feasible region, the next step is to determine which combination of deluxe and standard phones yields the maximum profit. This is achieved by evaluating the objective function at each vertex of the feasible region.
Potential vertices for this exercise include:

• \((80, 100)\)
• \((80, 120)\)
• \((100, 100)\)

By plugging these points into the objective function \( P = 25x + 30y \):

• At \((80, 100)\), the profit \( P = 5000 \).
• At \((80, 120)\), the profit \( P = 5600 \).
• At \((100, 100)\), the profit \( P = 5500 \).

The maximum profit, and thus the optimal solution, is found at the point \((80, 120)\), resulting in a profit of $5600. By reaching this point, the profit is maximized while still adhering to the specified constraints.