Problem 63 - College Algebra with Modeling and Visualization

Question

Maximizing Profit\(\quad\) A business manufactures two parts, \(\mathbf{X}\) and \(\mathbf{Y}\). Machines \(\mathbf{A}\) and \(\mathbf{B}\) are needed to make each part. To make part \(\mathbf{X},\) machine \(\mathbf{A}\) is needed for 4 hours and machine \(B\) is needed for 2 hours. To make part Y, machine A is needed for 1 hour and machine B is needed for 3 hours. Machine \(A\) is available for 40 hours each week and machine \(\mathbf{B}\) is available for 30 hours. The profit from part \(\mathbf{X}\) is \(\$ 500\) and the profit from part \(\mathbf{Y}\) is S600. How many parts of each type should be made to maximize weekly profit?

Step-by-Step Solution

Answer

Produce 7 parts of X and 6 parts of Y for maximum profit.

Step 1: Define Variables

Let's define the variables: Let \( x \) be the number of part X made per week, and \( y \) be the number of part Y made per week.

Step 2: Write the Objective Function

The profit from part \( X \) is \(500, and from part \( Y \) is \)600. Thus, the objective function to maximize is the total profit: \( P = 500x + 600y \).

Step 3: Write the Constraints

We have constraints based on the availability of machines. \( \mathbf{A} \) is available for 40 hours: \( 4x + y \leq 40 \). \( \mathbf{B} \) is available for 30 hours: \( 2x + 3y \leq 30 \). Additionally, we must have \( x \geq 0 \) and \( y \geq 0 \) as non-negativity constraints.

Step 4: Graph the Constraints

Plot the inequalities on a graph. The feasible region is determined by the intersection of the constraint areas: \( 4x + y \leq 40 \), \( 2x + 3y \leq 30 \). Also, account for the non-negativity constraints \( x \geq 0 \), \( y \geq 0 \).

Step 5: Identify the Vertices of the Feasible Region

Find the points of intersection of the boundary lines of the constraints to determine the vertices of the feasible region. The vertices are the potential solutions to maximize profit.

Step 6: Evaluate the Objective Function at Each Vertex

Calculate the profit \( P = 500x + 600y \) at each vertex to determine which one gives the maximum profit. The points of intersection to consider are \((0,10)\), \((5,0)\), \((7,6)\), and \((9.5, 0)\). Evaluate \( P \) at each to find the maximum.

Step 7: Determine the Optimal Solution

After evaluating, the vertex \((7,6)\) gives the highest profit of $6900. Hence, the solution is to produce 7 parts of X and 6 parts of Y.

Key Concepts

Objective Function

In linear programming, the objective function is essentially the heart of the optimization problem. It represents the equation you want to maximize or minimize. In this exercise, we aim to maximize the weekly profit by determining the right mix of two parts: X and Y.

The objective function is crafted from the profits of each part. For each unit of part X, the profit is \(500, and for each unit of part Y, it is \)600. This leads us to the function:

• Maximize: \( P = 500x + 600y \)

Here, \( P \) is the total profit, and \( x \) and \( y \) represent the number of parts X and Y produced, respectively.

The goal is straightforward: find the combination of \( x \) and \( y \) that causes \( P \) to hit its highest possible value. Keeping the objective function clear helps streamline the entire linear programming process.

Constraint Equations

Constraint equations are the rules that restrict the possible solutions for an optimization problem. They essentially define the limits within which the solution must lie. In this business example, constraints come from the limited availability of machines A and B.

Each part, X and Y, requires a certain number of machine hours:

• Machine A: 4 hours for X and 1 hour for Y. Total available: 40 hours.
• Machine B: 2 hours for X and 3 hours for Y. Total available: 30 hours.

From this, we can write the constraint equations as:

• \( 4x + y \leq 40 \) for machine A
• \( 2x + 3y \leq 30 \) for machine B
• \( x \geq 0 \) and \( y \geq 0 \) because you can't produce a negative number of parts

These constraints must be satisfied in any solution. They ensure that the usage of the machines does not exceed their availability.

Feasible Region

The feasible region is the set of all possible points that satisfy the constraint equations. In graphical terms, it represents the area on a graph where all the constraint inequalities overlap. It provides the boundary within which the optimal solution for the linear programming problem exists.

To identify the feasible region, you would typically graph the constraint inequalities:

• \( 4x + y \leq 40 \)
• \( 2x + 3y \leq 30 \)
• Non-negativity constraints keep \( x \) and \( y \) within the first quadrant.

The feasible region will appear as a polygon, bounded where these inequalities intersect.

Finding the feasible region helps in pinpointing the vertices—which are potential candidates for maximizing our objective function. In this exercise, the vertices were identified as \((0,10)\), \((5,0)\), \((7,6)\), and \((9.5, 0)\).

Optimization Problem

An optimization problem is at the core of linear programming. It involves finding the best solution from a set of feasible solutions. In our scenario, the goal is optimizing the production to get the maximum weekly profit.

Once the feasible region is determined, the next step is to evaluate the objective function (profit function, in this case) at each vertex of the feasible region:

• Vertex \((0,10)\): Profit \(= 500(0) + 600(10) = 6000\)
• Vertex \((5,0)\): Profit \(= 500(5) + 600(0) = 2500\)
• Vertex \((7,6)\): Profit \(= 500(7) + 600(6) = 6900\)
• Vertex \((9.5, 0)\): Profit \(= 500(9.5) + 600(0) = 4750\)

The vertex \((7,6)\) yields the maximum profit, confirming the optimal production mix is 7 units of X and 6 units of Y. Such structured approaches transform complex decisions into clear, calculable steps.