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Chapter 9: Problem 3

Find the number of units \(x\) that produces a maximum revenue \(R\). $$ R=400 x-x^{2} $$

Short Answer

Expert verified
The maximum revenue is achieved when 200 units are produced.

Step by step solution

01

Differentiate the Revenue Function

Find the first derivative of the revenue function to determine the rate of change of the revenue with respect to the number of units. The derivative of \(R = 400x - x^{2}\) with respect to \(x\) is \[R' = 400 - 2x\].
02

Find the Critical Point

Set the derivative equal to zero and solve for \(x\). This yields \[400 - 2x = 0 \Rightarrow x = 200\]. The critical point is \(x = 200\).
03

Determine Whether it is a Maximum

Since the coefficient of the \(x^2\) term in the original equation is negative, the parabola opens downwards, implying that the graph has a maximum point. Thus, \(x = 200\) produces the maximum revenue.

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus that helps us understand how a function changes at any given point. When working with a revenue function, like in the example, the derivative allows us to find how the revenue changes as the number of units, \(x\), changes. By differentiating the function \(R = 400x - x^2\), we get \(R' = 400 - 2x\). This derivative tells us the rate of change of the revenue with respect to \(x\).

Finding derivatives is crucial because:
  • They help identify important points where the function changes behavior.
  • They provide insight into increasing or decreasing trends of the function.
By examining the derivative, we can further identify whether these changes lead to a maximum or minimum value of the function, which is intrinsic to determining maximum revenue in this context.
Critical Point
Critical points are specific values of \(x\) in a function where the first derivative is zero or undefined. These points are essential because they can indicate where maximum or minimum values may occur. For the given revenue function, the critical point is found by setting the derivative to zero: \(400 - 2x = 0\).

Solving this gives us \(x = 200\). This is a critical point because:
  • The derivative at \(x = 200\) is zero, implying no change in revenue at that unit count.
  • Critical points often highlight where functions can stop increasing or decreasing, marking the potential for peaks or valleys.
In our scenario, the critical point \(x = 200\) is vital as it indicates where the maximum revenue may occur, given the function鈥檚 behavior around those points.
Parabola Opens Downwards
When the graph of a quadratic function opens downwards, it means that the function鈥檚 leading term鈥攖he \(x^2\) term鈥攈as a negative coefficient. In our example, the revenue function \(R = 400x - x^2\) has a \(x^2\) term with a negative coefficient, \(-1\). This negative sign indicates that the parabola opens downward.

Understanding this concept highlights:
  • The shape of the graph and the general trend of the revenue function.
  • That the critical point \(x = 200\) corresponds to a maximum, not a minimum, due to the downward opening.
This insight is crucial on a visual and mathematical level as it confirms that the function reaches a high point at the critical point, reinforcing the knowledge that \(x = 200\) produces the maximum revenue.

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Most popular questions from this chapter

Marginal Analysis, use differentials to approximate the change in cost, revenue, or profit corresponding to an increase in sales of one unit. For instance, in Exercise 29, approximate the change in cost as \(x\) increases from 12 units to 13 units. Then use a graphing utility to graph the function, and use the trace feature to verify your result. \(P=-x^{2}+60 x-100 \quad x=25\)

The side of a square is measured to be 12 inches, with a possible error of \(\frac{1}{64}\) inch. Use differentials to approximate the possible error and the relative error in computing the area of the square.

Marginal Analysis, use differentials to approximate the change in cost, revenue, or profit corresponding to an increase in sales of one unit. For instance, in Exercise 29, approximate the change in cost as \(x\) increases from 12 units to 13 units. Then use a graphing utility to graph the function, and use the trace feature to verify your result. \(R=50 x-1.5 x^{2} \quad x=15\)

Sketch a graph of a function \(f\) having the given characteristics. (There are many correct answers.) $$ \begin{aligned} &f(-2)=f(0)=0 \\ &f^{\prime}(x)>0 \text { if } x<-1 \\ &f^{\prime}(x)<0 \text { if }-10 \text { if } x>0 \\ &f^{\prime}(-1)=f^{\prime}(0)=0 \end{aligned} $$

Use a graphing utility to graph the function. Choose a window that allows all relative extrema and points of inflection to be identified on the graph. \(y=\frac{x^{2}}{x^{2}+3}\)
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