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When a customer buys multiple items, a discount function often comes into play. This function defines how the discount percentage is calculated based on the quantity purchased. In our example, the auto dealer offers a discount equal to the number of cars bought. This means if you buy 5 cars, you get a 5% discount per car.

The discount function can be expressed mathematically. Here, the discount is calculated as \(D(n) = n\% = 0.01n\), where \(n\) is the number of cars. The idea behind the discount function is to encourage bulk buying by reducing costs based on quantity. It becomes a linear model, where the discount increases directly with the number of items purchased.

This type of discount function is an effective marketing strategy not only to entice fleet buyers but also to boost overall sales.

The revenue function represents the total income generated from sales. For the auto dealer's problem, it's necessary to express revenue in terms of the number of cars sold. To do this, use the price per car after applying the discount against the number of cars purchased.

In this scenario, the preincentive price of a car is \$14,400. After adding the discount, the price per car becomes: \(P(n) = 14,400(0.99 - 0.01n)\). The revenue from selling \(n\) cars is then determined by multiplying the after-discount price by the number of cars:\[R(n) = n \times 14,400(0.99 - 0.01n)\]

Expanded, this formula becomes:\[R(n) = 14,256n - 144n^2\] This quadratic equation shows how revenue changes as the number of cars sold changes.

For businesses, maximizing revenue is a critical goal. When modeled mathematically, revenue functions often take the form of quadratic equations. The challenge becomes finding the peak of this curve, as it represents the maximum revenue.

Our revenue function \(R(n) = 14,256n - 144n^2\) is a downward-opening parabola due to the negative coefficient in front of \(n^2\). The maximum point, or vertex, of the parabola gives us the optimal number of cars to sell for maximizing revenue.

The formula for the vertex \(-\frac{b}{2a}\) is used where \(a = -144\) and \(b = 14,256\). Solving this gives \(n = 49.5\). Since you can't sell half a car, calculate revenues for \(n = 49\) and \(n = 50\) to determine the highest possible revenue. Testing these values, selling 50 cars yields the maximum revenue of \$352,800.

Mathematical modeling involves formulating real-world problems into mathematical expressions to analyze and find solutions. In our auto dealer example, mathematical models translate the relationship between discounts and sales revenue into comprehensible functions.

We begin by defining the discount function as \(D(n) = 0.01n\), modeling how discounts change with the quantity purchased. Next, the after-incentive price is calculated with \(P(n) = 14,400(0.99 - 0.01n)\), formulating how price decreases with discounts.

The revenue function \(R(n) = 14,256n - 144n^2\) shows the resultant total earnings, guiding strategic decision-making. Mathematical modeling breaks down complex processes like revenue optimization into simple, manageable steps, allowing businesses to leverage data-driven decisions efficiently.