Problem 71 - College Algebra

Question

Suppose that the price per unit in dollars of a cell phone production is modeled by p = $45 ? 0.0125x, where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ? p. Find the production level that will maximize revenue.

Step-by-Step Solution

Answer

The production level to maximize revenue is 1,800,000 phones.

Step 1: Set up the Revenue Function

The revenue function is given by \( R = x \cdot p \). Substitute the expression for \( p \) into this equation, \( p = 45 - 0.0125x \). Therefore, \( R = x \cdot (45 - 0.0125x) \). Simplify it to get \( R = 45x - 0.0125x^2 \).

Step 2: Rewrite the Revenue Function

The revenue function in terms of \( x \) is now \( R(x) = -0.0125x^2 + 45x \). This is a quadratic equation of the form \( ax^2 + bx + c \), where \( a = -0.0125 \) and \( b = 45 \).

Step 3: Find the Vertex of the Parabola

The revenue function is a downward-opening parabola because the coefficient of \( x^2 \) is negative. To find the maximum point, use the vertex formula \( x = -\frac{b}{2a} \), where \( a = -0.0125 \) and \( b = 45 \).

Step 4: Calculate Maximum Production Level

Using the vertex formula, substitute the values: \( x = -\frac{45}{2(-0.0125)} = \frac{45}{0.025} = 1800 \). Therefore, the production level that maximizes revenue is 1800 thousand phones.

Step 5: Confirm the Context and Units

Recall that \( x \) is in thousands of units. Thus, 1800 represents 1,800,000 phones. This is the production quantity that will maximize revenue.

Key Concepts

Revenue Maximization

Revenue maximization is crucial when determining the optimal level of production. The goal is to produce a quantity of goods that will generate the most revenue possible. This involves understanding how changes in production levels impact the revenue function, which combines both the price per unit and the number of units sold. Here, revenue is modeled with a quadratic function derived from the given price equation.

• Start by identifying the revenue function: Revenue, \( R \), is the product of price per unit and the number of units.
• Seek to find the production level that maximizes \( R \): Often, this involves completing the square or finding the vertex of a parabola, representing the peak of revenue.

By calculating exactly where this peak occurs, businesses can produce the optimal number of units to achieve maximum revenue.

Vertex of a Parabola

The vertex of a parabola plays a significant role when dealing with quadratic functions, especially those modeling revenue. You can think of the vertex as the highest or lowest point on a parabola.

• For revenue functions, parameters often lead to a downward-opening parabola. This is because an increase in quantity grows revenue up to a point, after which it decreases due to diminishing price per unit or other factors.
• To locate the vertex: Use the formula \( x = -\frac{b}{2a} \) where \( a \) and \( b \) are coefficients from the quadratic function.

This vertex calculation tells us the optimal production level, marking the maximum revenue point before revenues begin to decrease.

Quadratic Functions

Quadratic functions often emerge in revenue analyses due to their squared term, \( ax^2 \). This kind of function forms a parabola when graphed, offering a clear visual representation of how changes in one variable - usually the production level - affect another variable, like revenue.

• Standard form is \( ax^2 + bx + c \), with specific coefficients altering the graph's orientation and steepness.
• The sign of \( a \) tells direction: negative for a downward-opening parabola or positive for upward.
• Understanding properties of quadratic functions helps predict and explain changes in revenue with varying production levels, highlighting points of interest like maximums or minimums.

In revenue maximization, these functions provide valuable insights into efficient production and pricing strategies.

Revenue Function

A revenue function begins with defining the price per unit and multiplying by the number of units sold. In this case, the price equation \( p = 45 - 0.0125x \) was provided, where \( x \) represents thousands of units.

• Integrate price into the revenue formula: Substitute \( p \) into \( R = x \cdot p \). With \( p = 45 - 0.0125x \), we obtain \( R(x) = 45x - 0.0125x^2 \).
• This expression is a quadratic function, capturing how revenue changes with production quantity.
• Calculate various points of interest: Use algebraic methods like finding the vertex to understand production levels that bring maximum revenue.

Through such mathematical formulations, revenue functions ensure a comprehensive understanding of how pricing and production intersect to impact overall revenue.