Problem 38 - Algebra for College Students

Question

A manufacturer of golf clubs makes a profit of \(\$ 50\) per set on a model A set and \(\$ 45\) per set on a model \(B\) set. Daily production of the model A clubs is between 30 and 50 sets, inclusive, and that of the model B clubs is between 10 and 20 sets, inclusive. The total daily production is not to exceed 50 sets. How many sets of each model should be manufactured per day to maximize the profit?

Step-by-Step Solution

Answer

Produce 40 sets of model A and 10 sets of model B to maximize profit.

Step 1: Define the Variables

Let \( x \) represent the number of model A sets produced, and \( y \) represent the number of model B sets produced. Our objective is to determine the optimal values for \( x \) and \( y \) that maximize profit.

Step 2: Construct the Profit Function

The profit function \( P \) is given by the equation:\[ P = 50x + 45y \] This function needs to be maximized under the given constraints.

Step 3: Establish the Constraints

The problem statement gives us the following constraints:1. Model A production: \( 30 \leq x \leq 50 \)2. Model B production: \( 10 \leq y \leq 20 \)3. Total production: \( x + y \leq 50 \)

Step 4: Graph the Constraints

Draw a coordinate plane with \( x \) on the horizontal axis and \( y \) on the vertical axis. Plot the lines for each constraint to determine the feasible region. The lines will be:- \( x = 30 \), \( x = 50 \)- \( y = 10 \), \( y = 20 \)- \( x + y = 50 \)The feasible region is bounded by the intersection of these lines.

Step 5: Identify Corner Points

Solve for the intersection points of the lines to determine the vertices of the feasible region. Potential vertices can be evaluated:- Intersection of \( x=30 \) and \( y=10 \) => (30, 10)- Intersection of \( x=30 \) and \( y=20 \) => (30, 20)- Intersection of \( x=30 \) and \( x+y=50 \) => (30, 20)- Intersection of \( x=50 \) and \( y=10 \) => (40, 10) - Intersection of \( x+y=50 \) and \( y=10 \) => (40, 10)- Intersection of \( x=50 \) and \( y=20 \) => (30, 20)Checking above, valid vertices given all constraints are (30, 20) and (40, 10).

Step 6: Calculate the Profit at Each Vertex

Substitute each vertex into the profit function to find which configuration yields the highest profit:- For (30, 20): \[ P = 50(30) + 45(20) = 1500 + 900 = 2400 \]- For (40, 10): \[ P = 50(40) + 45(10) = 2000 + 450 = 2450 \]

Step 7: Determine Maximum Profit

The maximum profit is \( 2450 \) and occurs when \( x = 40 \) and \( y = 10 \). Therefore, the manufacturer should produce 40 sets of model A and 10 sets of model B daily to maximize profits.

Key Concepts

Profit Maximization

Profit Maximization is a fundamental goal in business operations, especially in manufacturing settings like the one described. When it comes to linear programming, maximizing profit involves creating equations based on the revenue generated by each product.
In this case, each set of model A golf clubs brings in a profit of \(50, while each set of model B earns \)45. These values help us form the profit function, written as:

• \[ P = 50x + 45y \]

To find profit maximization points, you must calculate profits at various production levels, dictated by constraints. By focusing on the intersection of the allowed production quantities, or vertices of the feasible region, you can evaluate which scenario yields the highest profit. Maximization occurs by finding the vertex within the feasible region that satisfies all conditions while offering the greatest calculated profit.
In this scenario, maximizing profit was achieved with the production of 40 model A sets and 10 model B sets per day, resulting in a $2,450 profit.

Constraint Analysis

An essential aspect of linear programming is analyzing constraints. Constraints refer to the limitations or requirements set on production. In this problem, constraints ensure that production falls within allowable limits. Each constraint forms a linear inequality that shapes the feasible region.
There are three primary constraints:

• Model A production ranges between 30 to 50 sets: \( 30 \leq x \leq 50 \)
• Model B production ranges between 10 to 20 sets: \( 10 \leq y \leq 20 \)
• Total production should not exceed 50 sets: \( x + y \leq 50 \)

Analyzing these constraints helps graphically depict the feasible region on a coordinate plane. Each constraint is a boundary of the feasible region, and by plotting them, you see where the solutions can exist.
Understanding constraints is crucial to ensure production choices are viable and meet all operational requirements, guiding you towards optimal profit-making strategies.

Feasible Region

The feasible region is a critical concept in linear programming as it represents all possible combinations of production that meet the given constraints. Visualizing it involves plotting the established constraints on a graph and identifying where they intersect.
The lines corresponding to each constraint—such as \( x = 30 \), \( y = 10 \), and \( x + y = 50 \)—demarcate the edges of this region. The feasible region is typically a polygon, whose vertices represent potential solutions or production levels.
In this scenario, the valid vertices of the feasible region are (30, 20) and (40, 10). These points are examined to determine which configuration maximizes profit. By assessing each vertex within the feasible region, you can identify the best production plan.
The feasible region is significant because any point within it represents a permissible production level, ensuring that operational constraints are respected while striving for optimal outcomes.