Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Find the number of units that produces a maximum revenue. The revenue \(R\) is measured in dollars and \(x\) is the number of units produced. $$R=1000 x-0.02 x^{2}$$

Step-by-Step Solution

Answer

The number of units that yields the maximum revenue is 25000 units.

Step 1: Differentiate the Revenue function

Revenue function is given as \(R = 1000x - 0.02x^2\). We need to differentiate \(R\) with respect to \(x\). Using the power rule for differentiation, we have \[R'(x) = 1000 - 0.04x\] where \( R'(x)\) represents the derivative of \(R\) with respect to \(x\). The derivative gives us the slopes of the tangent lines to the points on the graph of \(R\).

Step 2: Set the Derivative Equals to Zero

To find the maximum revenue, we must find the point where the derivative equals zero, as this is where the slope of the tangent line to the graph of \(R\) is horizontal. Solving \[R'(x) = 1000 - 0.04x = 0\] gives \(x = 25000\).

Step 3: Confirm Maximum Point

To confirm that this is the maximum point and not minimum, we need to compute the second derivative of \(R\) and evaluate at \(x = 25000 \). The second derivative \(R''(x)\) is a constant \(-0.04\). Because \(R''(25000) < 0 \), according to the second derivative test, there is a maximum at \(x = 25000\).

Key Concepts

Revenue Function

A revenue function in calculus helps us understand how changes in production affect the overall earnings of a business. The revenue function for this exercise is given by \[R = 1000x - 0.02x^2\], where \(R\) represents the revenue and \(x\) is the number of units produced.

• The term \(1000x\) represents the income directly proportional to each unit produced.
• The term \(-0.02x^2\) accounts for reductions in revenue at high production levels, likely due to inefficiencies or increased costs.

This quadratic function models optimal production strategies by balancing income against diminishing returns. Knowing how to analyze such functions helps us identify production levels that maximize revenue.

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. It provides information about the rate of change of a quantity. Here, differentiating the revenue function helps us find where revenue is at its highest. For the revenue function \[ R = 1000x - 0.02x^2, \]we find the derivative using the power rule: \[ R'(x) = 1000 - 0.04x. \]

• The derivative \( R'(x) \) gives us the slope of the tangent line at any point \( x \) on the original function.
• By setting \( R'(x) \) to zero, we identify critical points where the slope is zero, indicating potential maximum or minimum values.

Differentiation facilitates understanding of the function's behavior and finding the optimal points.

Second Derivative Test

The second derivative test further refines our understanding of whether a critical point is a maximum or minimum. After finding a point where the first derivative is zero, the second derivative test confirms its nature.Here's how it works:

• Compute the second derivative of the revenue function: \[ R''(x) = -0.04. \]
• The second derivative \( R''(x) \) reveals if the graph of \( R(x) \) is concave up or down at critical points.
• If \( R''(x) < 0 \), the graph curves downwards, indicating a maximum. If \( R''(x) > 0 \), the graph is concave upwards, indicating a minimum.

In this case, since \( R''(25000) = -0.04 < 0 \), the revenue function has a local maximum at \( x = 25000 \). This method confirms that producing 25000 units maximizes revenue.