Problem 5 - Applied Finite Mathematics

Question

The Acme Apple company sells its Pippin, Macintosh, and Fuji apples in mixes. Box I contains 4 apples of each kind; Box II contains 6 Pippin, 3 Macintosh, and 3 Fuji; and Box III contains no Pippin, 8 Macintosh and 4 Fuji apples. At the end of the season, the company has altogether 2800 Pippin, 2200 Macintosh, and 2300 Fuji apples left. Determine the maximum number of boxes that the company can make.

Step-by-Step Solution

Answer

The maximum number of boxes that the company can make is 1175.

Step 1: Form the Inequalities

Let's call the number of Box I, Box II and Box III to be manufactured as x, y, and z. Based on the number of each type of apple in each box, we form inequalities. These constraints are as follows: For Pippin, 4x + 6y ≤ 2800. For Macintosh, 4x + 3y + 8z ≤ 2200. For Fuji, 4x + 3y + 4z ≤ 2300.

Step 2: Solve the Inequalities

The solution to these inequalities will give us the range in which the solution may lie. There are multiple ways to do this and the simplest is often graphical representation. However, here we'll find the intersection point of the inequalities step-by-step. The limiting constraint for Pippin gives us y ≤ - (2/3)x + 467 and the limiting constraints for Macintosh and Fuji give us z ≤ - (1/2)x - (3/8)y + 275 and z ≤ -x - (3/4)y + 575.

Step 3: Sum of Boxes

The goal of the problem is to maximize the total number of boxes that can be made from the given apples, hence max = x + y + z. From our inequalities, we know that z is the smaller of - (1/2)x - (3/8)y + 275 and -x - (3/4)y + 575. Hence, z = min{ - (1/2)x - (3/8)y + 275, -x - (3/4)y + 575}. Subbing this value of z into the equation we have max = x + y + min{ - (1/2)x - (3/8)y + 275, -x - (3/4)y + 575}.

Step 4: Maximize

We now need to maximize our equation within the constraints. It is clear that the maximum value is achieved when y = - (2/3)x + 467 is 0, so x = 700 and when z = min{ - (1/2)x - (3/8)y + 275, -x - (3/4)y + 575} is - (1/2)x - (3/8)y + 275, so z = 475. Using these values gives max = 700 + 0 + 475 = 1175.

Key Concepts

Inequality Solving

Understanding how to solve inequalities is the foundation for dealing with mathematical scenarios involving constraints. In the context of finite mathematics, solving inequalities involves identifying the maximum or minimum values of variables under certain given conditions. In our exercise, the company's constraints are the amounts of Pippin, Macintosh, and Fuji apples available.

To solve the inequalities presented, we label the number of boxes of each type with variables—x for Box I, y for Box II, and z for Box III. We then create equations based on the contents of each box and the stock of apples. For instance, the equation for Pippin apples is formed as 4x + 6y ≤ 2800. These inequalities paint a boundary in which we can operate to achieve an optimal number of boxes without exceeding the stock.

Graphical Representation

Graphical representation is a powerful tool in visualizing solutions to mathematical problems, especially when dealing with systems of equations or inequalities. By plotting constraints and objectives, we can visually identify where solutions might exist. In the Acme Apple company problem, each inequality could be represented as a line on a coordinate plane, with the shaded area below or above the line depicting the feasible solutions.

For example, the line corresponding to the constraint for Pippin apples (4x + 6y ≤ 2800) would separate the plane into two regions – one feasible and one not. The feasible region is the overlap between the regions of all three inequalities. The vertices of this region often contain the optimal solution, which in this case is the maximum number of boxes the company can create.

Maximization Process

The maximization process in our finite mathematics problem focuses on finding the highest possible value of an objective function, within given constraints. Here, the objective function is the sum of the boxes: max = x + y + z, and our goal is to find its maximum value.

By logically deducing from the inequalities, the exercise finds specific values of x and y that allow us to maximize the function. It is essential to stay within the boundary set by the constraints. Only certain combinations of x, y, and z will provide the highest number of total boxes without exceeding the apple stock – it's a balance to ensure no apple type runs out.

Systems of Equations

In this problem, we are dealing with a system of linear inequalities, which is similar to a system of equations but with inequality signs instead of equalities. Systems of equations are sets of equations with multiple variables, where the solution is the set of values that satisfies all equations simultaneously.

When we solve systems of inequalities for finite mathematics problems such as this, we're looking for a set of values that satisfies all conditions at once - which in our case represents the number of boxes we can package. Applying methods such as substitution or graphical representation can help determine where these sets of equations intersect, indicating a viable solution within the constraints outlined by the stock limits.