Problem 24 - Finite Mathematics and Applied Calculus

Question

Enormous State University's Business School is buying computers. The school has two models from which to choose, the Pomegranate and the iZac. Each Pomegranate comes with \(400 \mathrm{MB}\) of memory and \(80 \mathrm{~GB}\) of disk space; each iZac has \(300 \mathrm{MB}\) of memory and \(100 \mathrm{~GB}\) of disk space. For reasons related to its accreditation, the school would like to be able to say that it has a total of at least \(48,000 \mathrm{MB}\) of memory and at least \(12,800 \mathrm{~GB}\) of disk space. If the Pomegranate and the iZac cost \(\$ 2,000\) each, how many of each should the school buy to keep the cost as low as possible? HINT [See Example 4.]

Step-by-Step Solution

Answer

The feasible region is given by the following inequalities: \(x \geq 0\) \(y \geq 0\) \(y \geq \frac{48,000 - 400x}{300}\) \(y \geq \frac{12,800 - 80x}{100}\) By analyzing the graph, we find the point (x, y) = (160, 0) minimizes the cost function. Therefore, the school should buy 160 Pomegranate computers and no iZac computers. The minimum cost is \(C = 2000(160) + 2000(0) = \$320,000\).

Step 1: Identify the variables

Let x be the number of Pomegranate computers and y be the number of iZac computers.

Step 2: Set up the equations

According to the problem, the school needs at least \(48,000 \mathrm{MB}\) of memory and \(12,800 \mathrm{~GB}\) of disk space. We can set up two inequalities to represent these requirements: Memory requirement: \(400x + 300y \geq 48,000\) Disk space requirement: \(80x + 100y \geq 12,800\) The goal is to minimize the cost, where each computer costs \(\$ 2,000\). So, the cost function to minimize is: Cost: \(C = 2000x + 2000y\)

Step 3: Solve the inequalities for y

To find the region in the x-y plane that satisfies both requirements, first, we'll solve the inequalities for y: Memory requirement: \(y \geq \frac{48,000 - 400x}{300}\) Disk space requirement: \(y \geq \frac{12,800 - 80x}{100}\)

Step 4: Find the feasible region

The feasible region must satisfy both inequalities. Also, as we are dealing with computers which have to be positive integers, we can say x and y are both non-negative integers: \(x \geq 0\) \(y \geq 0\) From the inequalities above, we can graph the feasible region in the x-y plane.

Step 5: Find the minimum cost

Since the cost function is given by \(C = 2000x + 2000y\), we're looking for the smallest integer values of x and y within the feasible region. By analyzing the graph, we find the point (x, y) where the cost is minimized, and then we can calculate the minimum cost at that point.

Step 6: Calculate the minimum number of computers for each model and the cost

Assume we find the point (x, y) in the feasible region that minimizes the cost function. Then the minimum number of Pomegranate and iZac computers the school should buy is x and y, respectively. Multiply these numbers by the price, and you have the minimum cost.

Key Concepts

Inequalities

In linear programming, inequalities are essential for defining a set of conditions or constraints that a solution must satisfy. In the context of the exercise about choosing computers, inequalities help us establish the boundaries within which a valid solution lies. Specifically, we use inequalities to ensure the computer resources meet specific criteria like memory and disk space.

We have two main inequalities derived from the problem requirements:

• Memory: The combined memory of Pomegranates and iZacs must be at least 48,000 MB, hence the inequality: \(400x + 300y \geq 48,000\).
• Disk Space: Similarly, the total disk space must meet or exceed 12,800 GB, leading to \(80x + 100y \geq 12,800\).

These inequalities help us determine viable combinations of computers that fulfill the basic needs of the university while staying within budgetary constraints.

Cost Minimization

The goal of cost minimization is to achieve the lowest possible expenditure while meeting all requirements set by the constraints. This problem requires minimizing the cost of purchasing computers, with each computer costing \( \$2000\).

In mathematical terms, we express the cost function as \(C = 2000x + 2000y\), where:

• \(x\) is the number of Pomegranates.
• \(y\) is the number of iZacs.

To minimize the cost, we look for the smallest integer values of \(x\) and \(y\) that satisfy the constraints outlined by our inequalities. This involves finding the point in the feasible region where the total cost \(C\) is the lowest. It often helps to use optimization techniques like graphing or using linear programming software to identify the optimal point.

Feasible Region

The feasible region represents all possible combinations of Pomegranate and iZac computers that meet the problem's constraints. It is crucial in linear programming as it visually maps out which solutions are viable given the constraints derived from our inequalities.

To define this area, consider:

• Both inequalities must be satisfied simultaneously.
• Values of \(x\) and \(y\) must be non-negative as you cannot purchase a negative amount of computers.

Graphically, the feasible region is located at the intersection of all constraints, displaying a polygonal area on the x-y plane. This region indicates the range of choices available for purchasing computers, constrained by memory and disk space requirements. From here, we select the point that offers the least cost while fulfilling all criteria.

Integer Solutions

In linear programming, integer solutions are solutions where the variables take whole number values. This is important when dealing with quantities that cannot be fractioned, like purchasing individual computers.

For the given exercise, finding integer solutions is essential because:

• The university cannot buy a fraction of a computer; hence, \(x\) and \(y\) must be integers.
• Even when the feasible region contains viable solutions, only those with integer values for \(x\) and \(y\) are practical.

Achieving integer solutions may require additional optimization steps, such as rounding or utilizing integer programming techniques. This ensures the final solution is not only cost-effective but also implementable in real-world terms.