To solve the problem, let's first outline the solution:

The demand function is given as:

$p = 75e^{-0.000025x}$

Where $p$ is the price per unit, and $x$ is the number of units.

Part (a): Maximizing Revenue

Revenue ($R$) is calculated as:

$R = p \cdot x = 75xe^{-0.000025x}$

To maximize revenue, we take the derivative of $R$ with respect to $x$, set it equal to zero, and solve for $x$.

Part (b): Maximum Revenue

Once $x$ is found in part (a), substitute it back into the revenue function $R$ to compute the maximum revenue.

Let me calculate the solution. sympy as sp

Define variables

x = sp.Symbol('x', positive=True, real=True)

Define price function and revenue function

p = 75 * sp.exp(-0.000025 * x) R = x * p

Derivative of revenue with respect to x

R_derivative = sp.diff(R, x)

Solve R' = 0 to find critical points

critical_points = sp.solve(R_derivative, x)

Calculate the maximum revenue

x_max = max(critical_points) # Assuming positive x is the solution max_revenue = R.subs(x, x_max)

Calculate the price at maximum revenue

price_at_max = p.subs(x, x_max)

float(x_max), float(max_revenue), float(price_at_max)### Results:

1. Price that yields maximum revenue: $27.59 (rounded to the nearest cent).
2. Maximum revenue: $1,103,638 (rounded to the nearest dollar).

If you need detailed steps or have any further questions, let me know!

Follow-up Questions:

1. How is the derivative used to determine maximum or minimum values?
2. Can you explain why setting the derivative to zero finds critical points?
3. What happens to the revenue if the price is set too high or too low?
4. Can this method work for any type of demand function?
5. How can constraints, such as a limited number of units, affect the solution?

Tip: Always double-check calculations when solving for critical points to ensure accuracy.