Problem 107 - A Graphical Approach to Precalculus with Limits

Question

A manufacturing process requires that oil refineries manufacture at least 2 gallons of gasoline for each gallon of fuel oil. To meet winter demand for fuel oil, at least 3 million gallons a day must be produced. The demand for gasoline is no more than 6.4 million gallons per day. If the price of gasoline is \(\$ 1.90\) per gallon and the price of fuel oil is \(\$ 1.50\) per gal. Ion, how much of each should be produced to maximize revenue?

Step-by-Step Solution

Answer

Produce 6.4 million gallons of gasoline and 3.2 million gallons of fuel oil.

Step 1: Define Variables

Let \( x \) be the number of gallons of gasoline produced per day. Let \( y \) be the number of gallons of fuel oil produced per day. We aim to maximize revenue by determining the values of \( x \) and \( y \).

Step 2: Establish Constraints

The first constraint comes from the gasoline-to-fuel oil ratio: \( x \geq 2y \). The second constraint is to produce at least 3 million gallons of fuel oil: \( y \geq 3 \). The third constraint limits gasoline production to a maximum: \( x \leq 6.4 \). All values are in millions of gallons.

Step 3: Revenue Equation

The revenue from selling gasoline and fuel oil can be represented by the equation: \( R = 1.90x + 1.50y \), where \( R \) is the total revenue in million dollars.

Step 4: Determine Feasible Region

Plot the constraints on a graph to identify the feasible region. The constraints are: \( x \geq 2y \), \( y \geq 3 \), and \( x \leq 6.4 \). The feasible region will be the area where all conditions overlap.

Step 5: Identify Vertices of the Feasible Region

By finding the intersection points of the boundary lines, we determine the vertices of the feasible region: (6.4, 3.2), (6.4, 3), and (6, 3). These points will be checked to identify which gives the highest revenue.

Step 6: Calculate Revenue at Each Vertex

Compute the revenue for each vertex:- For (6.4, 3.2): \( R = 1.90(6.4) + 1.50(3.2) = 12.16 + 4.8 = 16.96 \, million \, dollars\)- For (6.4, 3): \( R = 1.90(6.4) + 1.50(3) = 12.16 + 4.5 = 16.66 \, million \, dollars\)- For (6, 3): \( R = 1.90(6) + 1.50(3) = 11.4 + 4.5 = 15.9 \, million \, dollars\)

Step 7: Determine the Optimal Production

The highest revenue of 16.96 million dollars occurs at the vertex (6.4, 3.2). Therefore, to maximize revenue, produce 6.4 million gallons of gasoline and 3.2 million gallons of fuel oil per day.

Key Concepts

Constraints

In linear programming, constraints are essentially the rules or limits that define the boundaries of a problem. They are the conditions that any solution must meet.
In our problem, these constraints are:

• There must be at least two gallons of gasoline produced for each gallon of fuel oil. This is expressed as the inequality: \( x \geq 2y \), where \( x \) is the gasoline produced and \( y \) is the fuel oil produced.
• A minimum of 3 million gallons of fuel oil must be produced each day: \( y \geq 3 \).
• A maximum of 6.4 million gallons of gasoline can be produced per day: \( x \leq 6.4 \).

These constraints define the allowable production combinations of gasoline and fuel oil. Understanding these limits is crucial as violations would render the solution infeasible.

Feasible Region

The feasible region is a key concept in solving linear programming problems. It is the graphical area where all the constraints overlap. It represents all the possible solutions that satisfy all the given constraints.
By plotting the constraints on a graph, you can visually identify the feasible region. In the exercise, the feasible region was found by:

• Drawing each constraint as a line on a graph. For instance, \( x = 2y \) or \( y = 3 \).
• Identifying the area where all these lines intersect or overlap. This overlapping area is what we call the feasible region.

Within this region, any point represents a combination of gasoline and fuel oil production that satisfies the constraints. The goal is to explore these points further to find the one that maximizes revenue.

Revenue Maximization

Revenue maximization is the core goal of our linear programming problem. It involves determining the highest possible earnings under given conditions. In this case, it involves maximizing the income from gasoline and fuel oil sales.
The revenue is given by the equation:\[ R = 1.90x + 1.50y \] where \( R \) is the total revenue. The coefficients 1.90 and 1.50 represent the price per gallon for gasoline and fuel oil, respectively.
Maximizing revenue means finding the point within the feasible region that leads to the highest value of \( R \). As seen in the solution, this is done by evaluating the revenue at each vertex of the feasible region, as these represent potential optimal solutions.

Optimization

Optimization is the process of making the best or most effective use of a situation or resource. In linear programming, it refers to finding the best possible outcome within the given constraints.
The optimization goal here is to maximize revenue by choosing the optimal quantities of gasoline and fuel oil to produce. Once the feasible region is determined, the next step is to evaluate it for the highest revenue.
Each vertex of the feasible region is tested against the revenue equation. By calculating the revenue for each vertex, we identify which point achieves the maximal revenue.
Optimization ensures not just any solution, but the best one, which in our example, means the point (6.4, 3.2) yielding the highest revenue of 16.96 million dollars.