The given question asks us to find expressions related to bicycle sales and revenue.

Step 1: Given Information

• The price function per bicycle $p(x)$ is given by: $p(x) = 150 - \frac{x}{4}$ where $x$ is the number of bicycles sold, and $p(x)$ is the price per bicycle.

Part a: Expression for Total Revenue

Total revenue $R(x)$ is found by multiplying the price per bicycle by the number of bicycles sold, i.e., $R(x) = p(x) \cdot x$ Substitute the expression for $p(x)$ into the equation: $R(x) = \left( 150 - \frac{x}{4} \right) x$ Simplify this: $R(x) = 150x - \frac{x^2}{4}$ Thus, the total revenue function is: $R(x) = 150x - \frac{x^2}{4}$

Part b: Maximizing Revenue

To find the number of bicycle sales that leads to maximum revenue, we need to maximize the revenue function $R(x)$. This involves finding the critical points by taking the derivative of $R(x)$ and setting it equal to zero.

First, take the derivative of $R(x)$: $R'(x) = 150 - \frac{2x}{4}$ Simplify: $R'(x) = 150 - \frac{x}{2}$ Now, set $R'(x) = 0$ to find the critical points: $150 - \frac{x}{2} = 0$ $\frac{x}{2} = 150$ $x = 300$

Conclusion

• The total revenue function is $R(x) = 150x - \frac{x^2}{4}$.
• The number of bicycle sales that leads to maximum revenue is $x = 300$.

Would you like more details or have any other questions?

Follow-up Questions:

1. How would the revenue function change if the price formula had a different constant?
2. What is the maximum possible revenue with $x = 300$?
3. How does the revenue change as the number of bicycles sold increases beyond 300?
4. Can you verify the concavity of the function at $x = 300$ to confirm it's a maximum?
5. What happens to the price per bicycle when 300 units are sold?

Tip:

When finding maximum or minimum points for revenue or profit functions, always check the second derivative to confirm if it’s a maximum or minimum!