Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

A small firm manufactures bookshelves and desks for microcomputers. For each product it is necessary to use a table saw and a power router. To manufacture each bookshelf, the saw must be used for \(\frac{1}{2}\) hour and the router for 1 hour. A desk requires the use of each machine for 2 hours. The profits are $$\$ 20$$ per bookshelf and $$\$ 50$$ per desk. If the saw can be used for 8 hours per day and the router for 12 hours per day, how many bookshelves and desks should be manufactured each day to maximize the profit?

Step-by-Step Solution

Answer

Produce 4 bookshelves and 4 desks per day for maximum profit.

Step 1: Understand the Objective

Our task is to maximize the profit from manufacturing bookshelves and desks, given the constraints on machine usage time and profit per item.

Step 2: Define Variables

Let \( x \) represent the number of bookshelves manufactured per day, and \( y \) represent the number of desks manufactured per day.

Step 3: Formulate the Objective Function

The profit function we want to maximize is given by: \( P = 20x + 50y \), where \( P \) represents the total profit.

Step 4: Establish the Constraints

Based on machine usage, the constraints are: \( \frac{1}{2}x + 2y \leq 8 \) for the table saw, and \( x + 2y \leq 12 \) for the router. Additionally, \( x \geq 0 \) and \( y \geq 0 \), since negative production isn't possible.

Step 5: Graph the Constraints

Plot the constraints on a graph. The inequality \( \frac{1}{2}x + 2y \leq 8 \) becomes \( x + 4y \leq 16 \), and \( x + 2y \leq 12 \) lines are plotted. The feasible region is where these constraints intersect and include the axes (because \( x \) and \( y \) must be non-negative).

Step 6: Identify the Vertices of the Feasible Region

Determine where the constraint lines intersect each other and the axes to find the vertices of the feasible region. These are typically where the potential maximum values of \( x \) and \( y \) occur, which need evaluation.

Step 7: Evaluate the Objective Function at Each Vertex

Calculate the total profit at each vertex by substituting these points into \( P = 20x + 50y \). Compare these profits to identify which gives the highest value.

Step 8: Optimal Solution

Determine which combination of \( x \) and \( y \) at these vertices maximizes the profit while satisfying all constraints. This will be our solution.

Key Concepts

Objective Function

In linear programming, the objective function is what you're trying to optimize, which is usually either maximizing or minimizing something. In this exercise, the target is to maximize profit from manufacturing two products: bookshelves and desks. The objective function can be expressed as a mathematical equation, where you assign variables to quantities you want to find.

Here, let’s define:

• Variable \( x \): Number of bookshelves produced per day.
• Variable \( y \): Number of desks produced per day.

Our objective function becomes \( P = 20x + 50y \),
where \( P \) is the total profit. The numbers 20 and 50 represent the profit per bookshelf and desk, respectively.

By maximizing this function under given constraints, you find the most profitable production plan.

Constraints

Constraints in linear programming are limitations or requirements that the solution must satisfy. They often represent resources, time, or other boundaries that restrict possibilities.

For this problem, we have two main constraints based on the machine hours available:

• The time constraint for the table saw: \( \frac{1}{2}x + 2y \leq 8 \). This represents the available time for the table saw, in hours.
• The time constraint for the power router: \( x + 2y \leq 12 \). It limits the available hours for the power router.

Additionally, the constraints include:

• Non-negativity constraints: \( x \geq 0 \) and \( y \geq 0 \), meaning you can't produce a negative number of products.

Constraints are translated into inequalities in linear programming, guiding the feasible region within which the best possible solution exists.

Feasible Region

The feasible region is the area in the graph where all constraints overlap. It represents all possible solutions that satisfy all constraints at once. This region is typically bounded by the lines you've plotted from the constraints, forming a polygonal shape on a graph.

In this exercise, after plotting the equations \( x + 4y \leq 16 \) and \( x + 2y \leq 12 \), along with adding the non-negativity constraints, the feasible region emerges as the portion where these lines intersect and the solutions remain positive for both variables.

The vertices of this region are critical because the optimal solution often exists at one of these corners. To find these vertices, you solve the equations where the lines intersect each other and the axes.

Graphical Method

The graphical method is a way to solve linear programming problems visually by plotting constraints on a graph and identifying the feasible region. This approach is suitable for problems with two variables, like the one in this exercise.

To use the graphical method effectively:

• First, rewrite the constraints in terms of equalities to draw straight lines on a graph.
• Plot each line on the coordinate plane. The area where all shaded regions from inequalities overlap is the feasible region.
• Next, identify the vertices of the feasible region. These points are potential candidates for the optimal solution.
• Finally, evaluate the objective function at each vertex to find which point gives the maximum profit.

This method provides a visual understanding of how constraints influence the solution, making it easier to grasp the concept of optimization in linear programming.