Problem 91 - A Graphical Approach to College Algebra

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 gallon, 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 for maximum revenue.

Step 1: Define Variables

Let \( x \) represent the number of gallons of gasoline produced per day and \( y \) represent the number of gallons of fuel oil produced per day. We need to determine the values of \( x \) and \( y \) that maximize revenue.

Step 2: Write the Objective Function

The revenue function \( R \) can be expressed in terms of \( x \) and \( y \) as follows: \[ R = 1.90x + 1.50y \] This is because each gallon of gasoline brings in \( \\(1.90 \), and each gallon of fuel oil brings in \( \\)1.50 \).

Step 3: Set Up the Constraints

The problem provides several constraints: 1. At least 2 gallons of gasoline for each gallon of fuel oil: \( x \geq 2y \)2. At least 3 million gallons of fuel oil must be produced: \( y \geq 3 \text{ million} \)3. Gasoline demand is no more than 6.4 million gallons: \( x \leq 6.4 \text{ million} \) 4. Non-negativity: \( x \geq 0 \), \( y \geq 0 \)

Step 4: Identify the Feasibility Region

Graph the constraints on a coordinate plane where \( x \) represents gasoline (in millions of gallons) and \( y \) represents fuel oil (in millions of gallons). The solution lies within the intersection of these inequalities.

Step 5: Determine Vertices of Feasibility Region

Solve the constraints intersection equations to find the vertices: - Intersection of \( x = 2y \) and \( y = 3 \): \( x = 6 \), \( y = 3 \)- Intersection of \( x = 2y \) and \( x = 6.4 \): \( x = 6.4 \), \( y = 3.2 \) Check that these satisfy all constraints.

Step 6: Evaluate Revenue at Vertices

Substitute the vertex points into the revenue function to determine which one provides the maximum revenue:- For (6, 3): \( R = 1.90(6) + 1.50(3) = 15.9 \text{ million dollars} \)- For (6.4, 3.2): \( R = 1.90(6.4) + 1.50(3.2) = 16.48 \text{ million dollars} \)

Step 7: Conclusion

The maximum revenue is achieved at \((6.4, 3.2)\) meaning 6.4 million gallons of gasoline and 3.2 million gallons of fuel oil should be produced daily.

Key Concepts

Constraints

In linear programming, constraints are the conditions or limitations that define the feasible solutions for a problem. These constraints are expressed as linear inequalities that restrict the range of possible values for the decision variables involved, which in this case are the amounts of gasoline and fuel oil to be produced.
In the exercise, we encounter several constraints:

• First, the production of gasoline must be at least twice the production of fuel oil, formulated as the inequality: \( x \geq 2y \).
• Secondly, the daily production of fuel oil should not be less than 3 million gallons, represented by: \( y \geq 3 \text{ million} \).
• Additionally, the demand for gasoline dictates a maximum production of 6.4 million gallons daily, captured as: \( x \leq 6.4 \text{ million} \).
• Lastly, non-negativity constraints are applied to ensure that negative values for production are not considered, thus: \( x \geq 0 \), \( y \geq 0 \).

Each constraint narrows down the feasible region, which will be explored further in the sections below.

Objective Function

The objective function in linear programming represents the formula that needs to be maximized or minimized. It connects the decision variables to an outcome that we want to optimize, like cost or, as in our problem, revenue.
In this exercise, the objective function is centered around maximizing the revenue from selling gasoline and fuel oil:

• The revenue from gasoline is described by \( 1.90x \), where \( x \) is the quantity of gasoline produced.
• For fuel oil, the revenue is \( 1.50y \), where \( y \) is the quantity of fuel oil produced.
• The combined revenue equation becomes: \( R = 1.90x + 1.50y \).

The aim is to find the values of \( x \) and \( y \) within the defined constraints that will maximize this revenue equation.

Feasibility Region

The feasibility region, also known as the feasible set, is the graphical representation of all the possible combinations of the decision variables that satisfy all constraints. In our linear programming problem, it is crucial to identify this region as it contains all potential solutions for maximizing revenue.
By plotting each constraint line onto a coordinate graph, areas that meet all conditions outlined by the constraints emerge:

• The line \( x \geq 2y \) implies that points on or above this line are viable given the gasoline to fuel oil production rule.
• Similarly, \( y \geq 3 \text{ million} \) indicates that any acceptable solution must lie above or on the line at \( y = 3 \).
• The constraint \( x \leq 6.4 \text{ million} \) establishes that the region of interest is to the left of the line at \( x = 6.4 \).
• Intersecting all these inequalities reveals the feasible region. Only within this polygon do the potential solutions that meet every constraint exist.

This region is essential for determining where vertices lie, which are key in identifying the best solution for our objective function.

Graphical Method

The graphical method in linear programming is an approach used to solve optimization problems by visually identifying the feasible region and evaluating the objective function at its vertices. This technique gives a clear visual representation of how constraints shape the solution space and where optimality is achieved.
In applying the graphical method:

• First, we plot each constraint on a coordinate plane, thus outlining the region where all conditions are satisfied, known as the feasible region.
• Next, we identify the vertices or corner points of this region, since a linear objective function will achieve its maximum or minimum value at one of these points.
• As explained in the solution, the vertices are located by calculating the intersection of constraint lines, such as \( x = 2y \) and \( y = 3 \), yielding the point (6, 3), and \( x = 2y \) with \( x = 6.4 \), resulting in the point (6.4, 3.2).
• Finally, by evaluating the revenue function \( R = 1.90x + 1.50y \) at these vertices, we determine the maximum possible revenue, which provides the optimal production solution.

This visual approach not only determines the ideal production levels but also enhances understanding by showing how each constraint limits or guides the outcome. It is especially helpful for problems involving two decision variables.
Through the graphical method, students learn not just how to find solutions, but why the solutions work as they do within the defined constraints.