Problem 34 - Finite Mathematics and Applied Calculus

Question

Podunk Institute of Technology's Math Department offers two courses: Finite Math and Applied Calculus. Each section of Finite Math has 60 students, and each section of Applied Calculus has \(50 .\) The department is allowed to offer a total of up to 110 sections. Furthermore, no more than 6,000 students want to take a math course. (No student will take more than one math course.) Draw the feasible region that shows the number of sections of each class that can be offered. Find the corner points of the region. HINT [See Example 4.]

Step-by-Step Solution

Answer

The feasible region is defined by the constraints x + y ≤ 110 and 60x + 50y ≤ 6000, where x represents the number of sections of Finite Math and y represents the number of sections of Applied Calculus. By graphing these inequalities and finding the intersections, we identify the corner points of the feasible region as (40, 70), (110, 0), (0, 120), and (0, 0).

Step 1: Define the Variables and Translate the Information into Inequalities

Let x represent the number of sections of Finite Math and y represent the number of sections of Applied Calculus. Given constraints: 1. Total sections allowed: x + y ≤ 110 2. Total students allowed: 60x + 50y ≤ 6000

Step 2: Graph the Constraints

To graph the inequalities, first plot the corresponding equations and then determine the regions satisfying the inequalities. 1. Plot the equation x + y = 110 (Edge of total sections allowed) 2. Plot the equation 60x + 50y = 6000 (Edge of total students allowed) Now, determine the feasible region by shading the area satisfying both inequalities: - The feasible region must satisfy x + y ≤ 110, which lies below the line x + y = 110. - Also, the feasible region must satisfy 60x + 50y ≤ 6000, which lies below the line 60x + 50y = 6000.

Step 3: Identify the Feasible Region and Corner Points

The feasible region is the area lying below both lines x + y = 110 and 60x + 50y = 6000. To find the corner points of the feasible region, we need to find the intersections of these lines: 1. Intersection point of x + y = 110 and 60x + 50y = 6000 (top-right corner) 2. Intersection point of x + y = 110 and the x-axis (y = 0) (top-left corner) 3. Intersection point of 60x + 50y = 6000 and the y-axis (x = 0) (bottom-right corner) 4. Origin (0,0) (bottom-left corner) By solving the system of equations, we can find the coordinates of these corner points: 1. Intersection point of x + y = 110 and 60x + 50y = 6000: (40, 70) 2. Intersection point of x + y = 110 and the x-axis (y = 0): (110, 0) 3. Intersection point of 60x + 50y = 6000 and the y-axis (x = 0): (0, 120) 4. Origin (0,0) The corner points of the feasible region are (40, 70), (110, 0), (0, 120), and (0, 0).

Key Concepts

Linear Inequalities

Linear inequalities are mathematical expressions that involve linear equations where the variable terms are not exactly equal but instead have inequality signs like \( \leq, \geq, <, \) or \( > \). These inequalities help us define constraints in many real-world problems, such as determining limits or restrictions for given resources.

In this exercise, the constraints are represented by two inequalities:

• \( x + y \leq 110 \) - This ensures the total number of math sections is at most 110.
• \( 60x + 50y \leq 6000 \) - This ensures that the total number of students enrolled does not exceed 6,000.

Both equations determine the boundaries of our feasible region by providing the limits within which the values of \( x \) and \( y \), representing the number of sections, must lie.

Understanding these inequalities is crucial in setting up problems involving constraints, especially when planning for resources like classes or materials.

Corner Points

Corner points are the vertices of the feasible region, where the lines representing the linear inequalities intersect. These points are significant because, in the context of optimization problems, the solution often occurs at one of these points.

In this exercise, once the feasible region is plotted on a graph, identifying the corner points becomes an essential task. The corner points can be found where:

• The line \( x + y = 110 \) intersects with the line \( 60x + 50y = 6000 \).
• The line \( x + y = 110 \) intersects the x-axis with \( y = 0 \).
• The line \( 60x + 50y = 6000 \) intersects the y-axis with \( x = 0 \).
• The origin (0, 0), where both \( x \) and \( y \) are zero.

Here, these intersection points provide us with the edges of the allowable section combinations for classes. The corner points for this situation are (40, 70), (110, 0), (0, 120), and (0, 0). These coordinates tell us potential combinations of how many sections of each course can be offered without exceeding the constraints.

Graphical Method

The graphical method is a visual approach to solving linear programming problems, particularly helpful for problems with two variables. By graphing the linear inequalities on a coordinate plane, you can visually identify the feasible region and its boundaries.

In the given exercise, the graphical method involves:

• Plotting each constraint as a linear equation on an x-y axis.
• Identifying the region on the graph that satisfies all inequalities (where the shaded areas overlap, called the feasible region).
• Highlighting or pinpointing the corner points of this feasible region.

The graphical method provides a clear visual representation of the solution space and allows you to see the possible combinations of variables that meet all conditions. Importantly, in optimization scenarios, this approach also outlines where to locate the best possible solution by examining the values at each corner point. The feasible region in this exercise shows all the viable ways in which the math department can offer their courses within the given constraints.