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**7.** A business forms a model of its widget sales via a pricing function $ p(x) = 400 - \frac{1}{70}x $.
Here, $ x $ is the number of widgets sold, and $ p(x) $ is the sales price (in dollars per widget).

**Questions:**
(a) Find the revenue function $ R(x) $ for this business (revenue is total sales).

(b) Find the number of widgets $ x $ sold that will maximize revenue. (This will be an unrealistic fraction, but do not round.)

(c) What is the maximum revenue?

(d) What is the price $ p(x) $ that yields maximum revenue?

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Let's go through each part systematically.

### Part (a) - Revenue Function
The revenue function $ R(x) $ is given by:
\[
R(x) = x \cdot p(x)
\]
Substituting $ p(x) = 400 - \frac{1}{70}x $, we get:
\[
R(x) = x \left( 400 - \frac{1}{70}x \right)
\]
\[
R(x) = 400x - \frac{1}{70}x^2
\]

### Part (b) - Number of Widgets for Maximum Revenue
To maximize the revenue, we need to take the derivative of $ R(x) $ and set it equal to 0.

\[
R'(x) = 400 - \frac{2}{70}x
\]
Setting $ R'(x) = 0 $:
\[
400 - \frac{2}{70}x = 0
\]
\[
\frac{2}{70}x = 400
\]
\[
x = \frac{400 \times 70}{2} = 14000
\]

So, 14,000 widgets will maximize the revenue.

### Part (c) - Maximum Revenue
Substitute $ x = 14000 $ into the revenue function $ R(x) $:

\[
R(14000) = 400(14000) - \frac{1}{70}(14000)^2
\]
\[
R(14000) = 5600000 - \frac{1}{70} \times 196000000
\]
\[
R(14000) = 5600000 - 2800000 = 2800000
\]

The maximum revenue is \$2,800,000.

### Part (d) - Price that Yields Maximum Revenue
Substitute $ x = 14000 $ into the price function $ p(x) $:

\[
p(14000) = 400 - \frac{1}{70}(14000)
\]
\[
p(14000) = 400 - 200 = 200
\]

The price that yields maximum revenue is \$200 per widget.

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Do you want any further details or have any questions?

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### Relative Questions:
1. How does the shape of the revenue function relate to the quadratic nature of its formula?
2. Why is the derivative of the revenue function important for finding the maximum?
3. Could the price function be used to model other types of business scenarios?
4. What happens to revenue if the number of widgets sold is significantly higher or lower than 14,000?
5. How would this analysis change if the price per widget decreased more rapidly with an increase in the number of widgets?

### Tip:
When finding maximums or minimums in revenue or cost functions, always take the first derivative to identify critical points where slopes change from positive to negative!

A business forms a model of its widget sales via a pricing function p(x) = 400 - (1/70)x. Here, x is the number of widgets sold, and p(x) is the sales price (in dollars per widget).

Learn how to maximize revenue using a pricing function in this business-related problem. This step-by-step guide walks through calculating the revenue function, finding the number of widgets sold for maximum revenue, and the price that yields maximum revenue. Suitable for Grades 11-12.

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