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Chapter 8: Problem 94

Maximizing Revenue. When priced at \(\$ 30\) each, a toy has annual sales of \(4,000\) units. The manufacturer estimates that each S1 increase in price will decrease sales by 100 units. Find the unit price that will maximize total revenue. (Hint: Total revenue \(=\) price \(\cdot\) the number of units sold.)

Short Answer

Expert verified
The unit price that maximizes total revenue is $35.

Step by step solution

01

Define Variables

Let the unit price be \( p \) dollars and the number of units sold be \( n \). Initially, \( p = 30 \) and \( n = 4000 \).
02

Establish Relationships

We know that for each 1 dollar increase in price, sales decrease by 100 units. Therefore, if the price increases by \( x \) dollars, the number of units sold changes to \( 4000 - 100x \). Similarly, the new price will be \( 30 + x \).
03

Express Revenue as a Function

Total revenue \( R \) is given by the product of price and the number of units sold. Therefore, \[ R = (30 + x)(4000 - 100x) \].
04

Expand the Revenue Function

Expand the revenue function: \[ R = 30(4000 - 100x) + x(4000 - 100x) \]. This simplifies to \[ R = 120000 - 3000x + 4000x - 100x^2 \].
05

Simplify the Revenue Function

Combine like terms to simplify \[ R = 120000 + 1000x - 100x^2 \].
06

Find the Maximum Revenue

The revenue function is a downward opening parabola (since the \( x^2 \) term is negative). The vertex of this parabola gives the maximum revenue. Use the vertex formula for a parabola \( x = -\frac{b}{2a} \). Here, \( a = -100 \), \( b = 1000 \), so \( x = -\frac{1000}{2(-100)} = 5 \).
07

Find the Optimal Unit Price

Substitute \( x = 5 \) into the expression for price: \( p = 30 + x = 30 + 5 = 35 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Functions
Quadratic functions are fascinating mathematical expressions that describe special types of relationships. They often appear in the form of \[ ax^2 + bx + c \] where \( a \), \( b \), and \( c \) are constants. These functions are crucial as they capture the essence of many real-world situations, such as projectile paths, areas, and financial calculations.
In this exercise, we use a quadratic function to express how the revenue, which depends on the price and sales volume, behaves. This quadratic form allows us to explore how changes in the price affect the total revenue, making it a powerful tool for analysis and decision-making.
Parabolas
A parabola is a unique geometric curve that is the visual representation of a quadratic function. You might recognize its "U" shape, which can open upwards or downwards. In the context of this problem, the parabola opens downwards because the leading coefficient (the \( a \) value) of our quadratic function is negative.
This behavior indicates that, over the range of possible prices, there is a single maximum point for the revenue. The top of the parabola, or its highest point, represents the maximum revenue we can achieve. This is essential for determining the optimal pricing strategy that would yield the best financial results.
Vertex Formula
The vertex formula is a powerful tool used to find the top point of a parabola defined by a quadratic equation. When the equation is given in the standard form \( ax^2 + bx + c \), the formula to find the \( x \)-coordinate of the vertex is \[ x = -\frac{b}{2a} \].
Why is this important? The vertex coordinates help us find the maximum (or minimum) point of the parabola. In price optimization, this is crucial because it tells us the price that will maximize revenue. In our exercise, we used this formula to find that the price adjustment \( x \) should be 5, meaning the optimal price is originally \(30 + 5 = \)35.
Price Optimization
Price optimization involves finding the right balance between price and sales volume to maximize revenue. In business, it鈥檚 important to set a price that maximizes revenue while remaining competitive.
In our scenario, we initially have a price of \(30 and are told that increasing the price decreases sales. By setting up a quadratic function to represent revenue as \[ R = (30 + x)(4000 - 100x) \], we were able to simplify and analyze it to find the optimal price. The vertex form of our parabola revealed that adjusting the price to \)35 gave the highest possible revenue. This process showcases how mathematics aids in making smart financial decisions.

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Most popular questions from this chapter

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