Problem 66 - Precalculus

Question

In a school district, the board of education has decided on two menus to serve in the school cafeterias. The annual budget for the meal plan is \(\$ 1.2\) million and one of the menus is \(5 \%\) more expensive than the other. What is the annual cost of each menu? Round your answer to the nearest integer.

Step-by-Step Solution

Answer

The annual cost of the cheaper menu is approximately $585,366 and the more expensive menu is $614,634.

Step 1: Define Variables

Let's define two variables. Let \( x \) represent the cost of the cheaper menu, and \( y \) represent the cost of the more expensive menu.

Step 2: Express Relationship Between the Menus

Since the more expensive menu (\( y \)) is 5% more expensive than the cheaper menu (\( x \)), we can express this relationship mathematically as: \[ y = x + 0.05x \] which simplifies to: \[ y = 1.05x \]

Step 3: Total Cost Equation

The total budget for the meal plans is given as \( \$ 1,200,000 \). Therefore, the equation representing the total cost is: \[ x + y = 1,200,000 \]

Step 4: Substitute Relationship into Total Cost Equation

We can substitute the expression for \( y \) from Step 2 into the total cost equation from Step 3: \[ x + 1.05x = 1,200,000 \]

Step 5: Simplify and Solve for x

Combine like terms in the equation: \[ 2.05x = 1,200,000 \] Next, divide both sides by 2.05 to solve for \( x \): \[ x = \frac{1,200,000}{2.05} \] Calculating this gives: \( x \approx 585,366 \) (rounded to the nearest integer).

Step 6: Solve for y

Substitute the value of \( x \) back into the equation \( y = 1.05x \) to find \( y \): \[ y = 1.05 \times 585,366 \approx 614,634 \] (rounded to the nearest integer).

Key Concepts

Linear Relationships

Understanding linear relationships is key to solving problems involving systems of linear equations, much like in our school menu exercise. In this context, a linear relationship is a direct connection between two variables, characterized by a constant rate of change. Here, we see that menu "y" is more expensive than menu "x" by a constant 5% increase. Mathematically, it's expressed as:

• Variable "y" is stated as: \[ y = x + 0.05x \]
• Which simplifies to: \[ y = 1.05x \]

This equation shows a proportional increase, indicating that for any given cost of "x," the cost of "y" will always be 5% more. This forms a simple linear equation where the variables can be clearly determined. Linear relationships like these allow us to determine one variable by knowing the other, simplifying complex real-world problems with consistent patterns.

Percentage Increase

Grasping the idea of percentage increase helps us understand how costs change in proportional terms, crucial for problems involving budget planning. A percentage increase means that a base value has risen by a specific percentage. In the scenario with the school menus, the more expensive menu costs 5% more than the cheaper one. This means:

• Calculate additional cost: \[ 0.05x \]
• Add this to the original cost: \[ y = x + 0.05x = 1.05x \]

Here, "x" represents the initial cost, and "0.05x" represents the 5% increase. Understanding this helps in interpreting how small percentage changes can affect budgets. It becomes particularly important when dealing with large sums, as a small percentage can equate to substantial amounts. Accurate calculation is essential to ensure budget viability.

Budget Constraints

Managing budget constraints is often the crux of financial decision-making, as seen in this exercise. Budget constraints define the limitations within which one must operate when allocating financial resources. The total budget allocated for the school meal plans is \$1,200,000, requiring efficient use to fit within this bound.
The problem here involves:

• Finding cost values for each menu that do not exceed this total: \[ x + y = 1,200,000 \]
• Implementing given relationships and constraints to meet budgetary needs

By substituting the equation for "y": \[ x + 1.05x = 1,200,000 \],we find each menu's cost while adhering to the budget. Computation rules allow us to solve:

• Simplify the expression: \[ 2.05x = 1,200,000 \]
• Divide to find "x": \[ x \approx 585,366 \]
• Calculate "y": \[ y \approx 614,634 \]

Keeping within budget constraints ensures that financial planning is realistic and effective.