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) = 46(4)^-x/ 5, find the revenue if 15 units are sold. Round to two decimal places.

Given the revenue function \( R(x) = x \cdot p(x) \), where \( p(x) = 46 \cdot 4^{-x/5} \), we need to find the revenue when 15 units are sold, i.e., when \( x = 15 \).

Substituting \( x = 15 \) into the revenue function:

1. First, calculate the unit price when \( x = 15 \):

\[
p(15) = 46 \cdot 4^{-15/5} = 46 \cdot 4^{-3}
\]

Simplify the exponent:

\[
p(15) = 46 \cdot \frac{1}{4^3} = 46 \cdot \frac{1}{64}
\]

\[
p(15) = \frac{46}{64} = 0.71875
\]

2. Now, calculate the revenue:

\[
R(15) = 15 \cdot p(15) = 15 \cdot 0.71875
\]

\[
R(15) = 10.78125
\]

Rounding to two decimal places, the revenue is:

\[
R(15) = 10.78 \text{ dollars}
\]

Would you like more details, or do you have any questions?

Here are some related questions to explore further:

1. How would the revenue change if the price function was different, say \( p(x) = 50 \cdot 3^{-x/4} \)?
2. What is the break-even point for a company if their cost function is also provided?
3. How does the revenue function behave as the number of units sold increases?
4. What is the elasticity of the revenue function with respect to the price?
5. How can the revenue function be optimized for maximum profit?

**Tip:** When working with exponents, make sure to simplify step-by-step to avoid errors, especially when dealing with fractions and negative exponents.

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) = 46(4)^-x/5, find the revenue if 15 units are sold. Round to two decimal places.

Learn how to calculate the revenue for 15 units sold using an exponential price function. The revenue function R(x) = x • p(x) is solved step-by-step, where p(x) = 46 • 4^(-x/5). Suitable for students in Grades 9-12, this example covers key concepts in exponential functions and revenue calculation.

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