Consider the following revenue function, where R is measured in dollars.
R = 52x − 1.5x2
Find the marginal revenue, 
dR
dx
.
dR
dx
 = 
 
Use differentials to approximate the change in revenue corresponding to an increase in sales of one unit when 
x = 15.
 (Round your answer in dollars to the nearest cent.)
$ 
Find the actual change in revenue corresponding to an increase in sales of one unit when 
x = 15.
 (Round your answer in dollars to the nearest cent.)

We are given the revenue function:

\[
R(x) = 52x - 1.5x^2
\]

### 1. Finding the marginal revenue, $ \frac{dR}{dx} $:

To find the marginal revenue, we take the derivative of $ R(x) $ with respect to $ x $:

\[
\frac{dR}{dx} = \frac{d}{dx} \left( 52x - 1.5x^2 \right)
\]

\[
\frac{dR}{dx} = 52 - 3x
\]

Thus, the marginal revenue function is:

\[
\frac{dR}{dx} = 52 - 3x
\]

### 2. Approximate the change in revenue using differentials when $ x = 15 $:

To approximate the change in revenue, we use the marginal revenue at $ x = 15 $. The change in revenue $ dR $ can be approximated by:

\[
dR \approx \frac{dR}{dx} \cdot dx
\]

Here, $ dx = 1 $ (since we are considering an increase of one unit in sales), and we need to find $ \frac{dR}{dx} $ when $ x = 15 $:

\[
\frac{dR}{dx} \Big|_{x = 15} = 52 - 3(15) = 52 - 45 = 7
\]

So, the approximate change in revenue is:

\[
dR \approx 7 \cdot 1 = 7 \text{ dollars}
\]

### 3. Find the actual change in revenue when $ x = 15 $:

The actual change in revenue corresponds to the difference in the revenue function when $ x $ increases from 15 to 16. We calculate $ R(16) $ and $ R(15) $:

\[
R(15) = 52(15) - 1.5(15)^2 = 780 - 337.5 = 442.5
\]

\[
R(16) = 52(16) - 1.5(16)^2 = 832 - 384 = 448
\]

The actual change in revenue is:

\[
\Delta R = R(16) - R(15) = 448 - 442.5 = 5.5 \text{ dollars}
\]

### Summary:
1. **Marginal revenue**: $ \frac{dR}{dx} = 52 - 3x $
2. **Approximate change in revenue** using differentials when $ x = 15 $: $7
3. **Actual change in revenue** when $ x = 15 $: $5.50

Would you like further details or have any questions?

#### Relative questions:
1. How does the marginal revenue change as sales increase?
2. What is the significance of the second derivative of the revenue function?
3. Can marginal revenue be negative? If so, what does it indicate?
4. How does the approximation using differentials compare to the actual change?
5. How does increasing sales beyond a certain point affect total revenue?

#### Tip:
The marginal revenue represents the additional revenue generated by selling one more unit, and it declines if the revenue function is quadratic, indicating diminishing returns.

Consider the following revenue function, where R is measured in dollars. R = 52x − 1.5x^2. Find the marginal revenue, dR/dx. Use differentials to approximate the change in revenue corresponding to an increase in sales of one unit when x = 15. Find the actual change in revenue corresponding to an increase in sales of one unit when x = 15.

Learn how to calculate the marginal revenue for the function R = 52x − 1.5x^2 and use differentials to approximate the change in revenue for an increase in sales by one unit at x = 15. This guide covers differentiation techniques and provides both approximate and actual revenue changes, ideal for students in Grades 11-12 studying calculus.

Math Problem Statement

Consider the following revenue function, where R is measured in dollars. R = 52x − 1.5x2 Find the marginal revenue, dR dx . dR dx

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