The revenue function is given by R(x)=x⋅p(x)
 dollars where x is the number of units sold and p(x)
 is the unit price.  If p(x)=35(2)^−x7
, find the revenue if 14
 units are sold.  Round to two decimal places.

The revenue function is given by $ R(x) = x \cdot p(x) $, where $ x $ is the number of units sold and $ p(x) $ is the unit price.

We are also given that the unit price function is $ p(x) = 35 \cdot 2^{-\frac{x}{7}} $.

To find the revenue when 14 units are sold, substitute $ x = 14 $ into both functions.

### Step 1: Compute $ p(14) $
\[
p(14) = 35 \cdot 2^{-\frac{14}{7}} = 35 \cdot 2^{-2} = 35 \cdot \frac{1}{4} = 8.75
\]

### Step 2: Compute the revenue $ R(14) $
\[
R(14) = 14 \cdot p(14) = 14 \cdot 8.75 = 122.50
\]

Thus, the revenue when 14 units are sold is **\$122.50**.

Would you like further details, or do you have any other questions?

Here are 5 related questions to deepen your understanding:
1. What happens to the revenue as the number of units sold increases?
2. How does the exponential decay in the price function affect the revenue?
3. Can you derive a formula for the revenue function $ R(x) $ in terms of $ x $?
4. How would you interpret the significance of the constant 35 in the price function?
5. What would happen to the revenue if the unit price had no exponential term?

**Tip:** When dealing with functions involving exponentials, it's important to carefully follow the rules for exponents, especially when negative or fractional exponents are involved.

The revenue function is given by R(x) = x * p(x) dollars where x is the number of units sold and p(x) is the unit price. If p(x) = 35 * (2)^(-x/7), find the revenue if 14 units are sold. Round to two decimal places.

Find the revenue when 14 units are sold with the unit price given by an exponential function p(x) = 35 * (2)^(-x/7). This problem involves key algebraic concepts like exponential functions and revenue calculation. Suitable for high school students in Grades 10-12.

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