Problem 61 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Correspond to those in Exercises \(15-27\), Section \(5.1 .\) Use the results of your previous work to help you solve these problems. The management of Hartman Rent-A-Car has allocated $$\$ 1.5$$ million to buy a fleet of new automobiles consisting of compact, intermediate-size, and full- size cars. Compacts cost $$\$ 12,000$$ each, intermediatesize cars cost $$\$ 18,000$$ each, and full-size cars cost $$\$ 24,000$$ each. If Hartman purchases twice as many compacts as intermediate-size cars and the total number of cars to be purchased is 100 , determine how many cars of each type will be purchased. (Assume that the entire budget will be used.)

Step-by-Step Solution

Answer

Hartman Rent-A-Car will purchase 60 compact cars, 30 intermediate-size cars, and 10 full-size cars.

Step 1: Define the variables

Let's denote the number of compact cars as \(x\), the number of intermediate-size cars as \(y\), and the number of full-size cars as \(z\).

Step 2: Write the Equations

We need to write equations based on the given information: 1. The total cost of purchasing the cars is $$1.5 million: \(12,000x + 18,000y + 24,000z = 1,500,000\). 2. The total number of cars to be purchased is 100: \(x + y + z = 100\). 3. Hartman purchases twice as many compacts as intermediate-size cars: \(x = 2y\).

Step 3: Solve the system of equations

We will first rewrite the equation for compacts in terms of intermediate-size cars: \(x = 2y\). Now, let's substitute this expression for x in the other equations. In the total cost equation: \(12,000(2y) + 18,000y + 24,000z = 1,500,000 \Rightarrow 42,000y + 24,000z = 1,500,000\) In the total number of cars equation: \(2y + y + z = 100 \Rightarrow 3y + z = 100\) Let's solve the equation \(3y + z =100\) for y and substitute the expression in the cost equation. So, \(z = 100 - 3y\) Now in the total cost equation: \(42,000y + 24,000(100 - 3y) = 1,500,000 \Rightarrow 42,000y + 2,400,000 - 72,000y = 1,500,000\) \(-30,000y = -900,000\) \(y = 30\) Now substitute the value of y in the total number of cars equation to find z: \(3(30) + z = 100\) \(z = 10\) From the equation for compacts, we find the number of compacts: \(x = 2(30)\) \(x = 60\)

Step 4: Interpret the Solution

The management of Hartman Rent-A-Car will purchase: - 60 compact cars - 30 intermediate-size cars - 10 full-size cars

Key Concepts

Linear Programming

Linear programming is a mathematical method used to find the best possible outcome or solution from a given set of parameters or list of requirements, which are represented as linear relationships. It is widely used in business, economics, and engineering to maximize or minimize a linear function subject to various constraints.

In the context of Hartman Rent-A-Car's problem, linear programming would be the process of determining the optimal number of compact, intermediate-size, and full-size cars to purchase within the constraints of budget and total fleet size. By setting up a linear function that represents the total cost of purchasing different types of cars, and including constraints that reflect the budget limit and required fleet composition, the problem becomes a classic example of linear programming.

Application in the Exercise

Here, we see three variables representing the quantities of each car type to be purchased. The objective function would be to minimize the difference between the amount spent and the budget available, which naturally leads to using the entire budget. The constraints are represented by the budget, the fleet size, and the condition that the number of compacts must be twice the number of intermediate-size cars.

Algebraic Modeling

Algebraic modeling is the process of representing real-world scenarios with mathematical symbols and expressions, typically using algebraic equations. This technique allows for the abstraction and simplification of problems to be solved mathematically.

In our textbook example, algebraic modeling is evident in the step of defining variables for the number of each type of car and translating the given financial and quantitative constraints into algebraic equations. These equations form a system that can subsequently be manipulated and solved.

Equations in the Exercise

Three key equations were formulated in this problem: the budget equation, the total car count equation, and the relationship between the number of compacts and intermediate-size cars. By algebraically modeling the problem this way, it becomes structured for analysis and for the application of solving techniques like substitution or linear combination.

Optimization

Optimization in mathematics is the process of finding the best solution from all feasible solutions. It often involves determining the maximum or minimum value of a particular function under a given set of constraints.

The Hartman Rent-A-Car exercise is aimed at optimizing the allocation of a fixed budget to purchase a fleet of cars that satisfies certain conditions, such as the number of cars of each type. The process of solving the system of equations is essentially an optimization problem: how to maximize the number of cars purchased without exceeding the budget and while adhering to the specific requirement of the number ratio of compacts to intermediate-size cars.

Finding the Optimal Solution

The step-by-step solution provided demonstrates the optimization process, where the variable representing the number of intermediate-size cars is solved first, then used to determine the quantities of the other car types. This process ensures that Hartman Rent-A-Car's purchase strategy uses the full budget in the most efficient way within the given constraints, achieving the optimal purchase plan.