Problem 61 - Algebra and Trigonometry

Question

Revenue A manufacturer finds that the revenue generated by selling \(x\) units of a certain commodity is given by the function \(R(x)=80 x-0.4 x^{2},\) where the revenue \(R(x)\) is measured in dollars. What is the maximum revenue, and how many units should be manufactured to obtain this maximum?

Step-by-Step Solution

Answer

The maximum revenue is $4000 at 100 units manufactured.

Step 1: Identify the Revenue Function and Reformat

The given revenue function is \(R(x) = 80x - 0.4x^2\). Notice this is a quadratic function of the form \(ax^2 + bx + c\), where \(a = -0.4\), \(b = 80\), and \(c = 0\).

Step 2: Recognize the Parabola Shape

Since the coefficient of \(x^2\) is \(-0.4\) (negative), the parabola opens downwards. This indicates that the vertex of the parabola represents the maximum point.

Step 3: Calculate the Vertex x-value for Maximum Revenue

The x-coordinate of the vertex for a quadratic function \(ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). Here, \(a = -0.4\) and \(b = 80\), so \(x = -\frac{80}{2(-0.4)}\).

Step 4: Solve for x

Calculate \(x = -\frac{80}{-0.8} = 100\). Thus, 100 units should be manufactured to obtain maximum revenue.

Step 5: Calculate Maximum Revenue

Substitute \(x = 100\) into the revenue function to find the maximum revenue. \(R(100) = 80(100) - 0.4(100)^2\).

Step 6: Compute Revenue Calculation

Calculate \(R(100) = 8000 - 0.4(10000) = 8000 - 4000 = 4000\). Hence, the maximum revenue is $4000.

Key Concepts

Quadratic function

A quadratic function is a mathematical expression which can be represented in the form \( ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants. In our revenue optimization problem, the quadratic function given is \( R(x) = 80x - 0.4x^2 \). This showcases the classic \( ax^2 + bx + c \) format, identifying it as a downward-opening parabola due to the negative coefficient of \( x^2 \).

Quadratic functions are highly valuable in optimizing scenarios like this one, where we aim to find the point of maximum revenue. The role of each coefficient is key:

• \( a \): Determines the direction and curve of the parabola. Negative \( a \) values indicate the parabola opens downward, signifying a maximum value at the vertex.
• \( b \): Influences the placement of the parabola on the x-axis, affecting where the vertex and intercepts occur.
• \( c \): Represents the y-intercept, showing where the graph intersects the y-axis when \( x = 0 \).

Understanding quadratic functions can greatly aid in predicting and understanding patterns within data, especially in contexts such as revenue or cost analysis. They allow us to model realistic business scenarios and make informed decisions based on mathematical predictions.

Vertex of a parabola

The vertex of a parabola is a crucial point that represents either a maximum or minimum value of the function. Since our parabola is downward-opening, the vertex is the location of the maximum revenue.

To find the vertex, we use the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \), applicable to any standard quadratic function \( ax^2 + bx + c \). In the revenue equation \( R(x) = 80x - 0.4x^2 \):

• \( a = -0.4 \)
• \( b = 80 \)

The calculation yields \( x = -\frac{80}{2(-0.4)} = 100 \).

This means producing 100 units will maximize the revenue. Plugging \( x = 100 \) back into the function helps determine the maximum revenue point. Recognizing this vertex and understanding its implications empowers businesses to make decisions with greater financial acumen, ensuring optimal product output strategies.

Maximum revenue point

The maximum revenue point is the pinnacle where the revenue reaches its highest possible value for the business. This occurs at the vertex of a downward-opening parabola formed by the quadratic revenue function. Once the x-coordinate of the vertex is determined (in our case, 100 units), we can maximize profit by focusing production efforts precisely at this point.

Substituting \( x = 100 \) back into the revenue function \( R(x) = 80x - 0.4x^2 \) results in the computation:

• \( R(100) = 80(100) - 0.4(100)^2 \)
• \( R(100) = 8000 - 4000 = 4000 \)

So, the maximum revenue achievable is $4000.

This number isn't just a theoretical exercise. It’s actionable information that can steer business strategies toward efficiency and maximum profitability. By manufacturing precisely 100 units, the business can ensure it captures the optimal revenue without overproducing, saving on costs and maximizing returns.