Problem 73 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. The revenue (in thousands of dollars) from producing \(x\) units of an item is modeled by $$R(x)=10 x-0.002 x^{2}$$. Find the marginal revenue at \(x=1000\).

Step-by-Step Solution

Answer

The marginal revenue at \(x = 1000\) is \(6\) thousand dollars.

Step 1: Understanding Marginal Revenue

Marginal revenue represents the additional revenue that is generated by increasing product sales by one unit. Mathematically, it is the derivative of the revenue function with respect to the number of units, x.

Step 2: Find the Derivative of Revenue Function

The revenue function is given as \(R(x) = 10x - 0.002x^2\). To find the marginal revenue, calculate its derivative with respect to \(x\). The derivative is \(R'(x) = \frac{d}{dx}(10x - 0.002x^2)\).

Step 3: Differentiate the Revenue Function

Apply the power rule of differentiation to each term.\[ \frac{d}{dx}(10x) = 10\] and \[ \frac{d}{dx}(-0.002x^2) = -0.004x\]. Therefore, the derivative \(R'(x) = 10 - 0.004x\).

Step 4: Evaluate the Derivative at x=1000

Once you have the derivative, substitute \(x = 1000\) into \(R'(x)\). This gives us \(R'(1000) = 10 - 0.004(1000)\).

Step 5: Calculate Marginal Revenue

Substitute \(1000\) into the derived function: \(R'(1000) = 10 - 4 = 6\). Thus, the marginal revenue when \(x = 1000\) is \(6\) thousands of dollars.

Key Concepts

Understanding the Revenue Function

The revenue function helps businesses understand the total amount of money generated from selling a certain number of products. In our exercise, the revenue function is given by the formula: \[ R(x) = 10x - 0.002x^2 \]Here, \( R(x) \) represents the revenue earned in thousands of dollars, and \( x \) denotes the number of units sold.
The function combines a linear term \(10x\) and a quadratic term \(-0.002x^2\). This reflects that as the number of units \(x\) sold increases, the revenue increases by $10 for each unit due to the linear term, while the quadratic term indicates a decreasing effect on revenue due to certain factors like increased production costs or market saturation.
Understanding this function is crucial for analyzing how sales volume affects total revenue.

Grasping the Derivative Concept

Understanding derivatives is vital as they measure how a function changes as its input changes. In the context of revenue functions, the derivative, often called the marginal revenue, indicates how the revenue changes with an additional unit sold.Mathematically, if you have a function \(f(x)\), its derivative \(f'(x)\) provides the rate of change of \(f\) concerning \(x\).
The derivative gives an instantaneous rate of change, and in economics, it shows sensitivity in revenue as sales change. It's not just about algebraic manipulation but understanding how small changes in activities translate into changes in outcomes, like revenue.

The Process of Differentiation

Differentiation is the mathematical process of finding a derivative. In this exercise, we are finding the derivative of the revenue function \( R(x) = 10x - 0.002x^2 \).To differentiate this function, you apply certain rules of differentiation. This includes the power rule, which makes it relatively straightforward to compute derivatives of polynomials by analyzing each term separately.
Differentiate each term in the function. Clear understanding of differentiation not only unravels the behavior of the function but also lays down the marginal revenue, helping businesses capitalize on incremental changes in production.

Applying the Power Rule in Differentiation

The power rule is a fundamental tool in differentiation that helps simplify finding derivatives for functions involving powers of \(x\). It states if you have a function \(f(x) = x^n\), its derivative \(f'(x)\) is \(nx^{n-1}\).In our revenue function, applying the power rule involves:

• The derivative of \(10x\) with respect to \(x\) becomes 10. This is because it’s equivalent to applying the power rule on \(x^1\), resulting in \(1 \cdot x^{0} = 1\) and thus simply the constant 10.
• The derivative of \(-0.002x^2\) becomes \(-0.004x\). Following the power rule results in \(-2 \cdot 0.002 \cdot x^{1}\), simplifying to \(-0.004x\).

Grasping how to use the power rule simplifies solving many derivatives and empowers us to handle complex cases where revenue is affected by varying factors. This not only assists in calculations but gives practical insight for business decision-making.