Thank you for the clarification! The corrected revenue function is:

$R(q) = -0.3q^2 + 30q + 12$

Let's calculate the maximum revenue using this updated function.

We'll follow the same steps as before:

1. Find the $q$-coordinate of the vertex (i.e., the value of $q$ that maximizes the revenue).
2. Substitute this $q$-value back into the function to find the maximum revenue $k$.With the updated revenue function $R(q) = -0.3q^2 + 30q + 12$, the $q$-coordinate of the vertex (which represents the number of units sold to achieve maximum revenue) is $q = 50$. Substituting $q = 50$ back into the revenue function, the maximum expected revenue $k$ is:

$k = 762$

The correct answer is:

(B) 752

Would you like more details or have any further questions?

Here are five related questions to expand this topic:

1. What would happen to the maximum revenue if the quadratic term coefficient changes?
2. How does the initial constant $12$ in the revenue function affect the maximum revenue?
3. How can you use derivatives to confirm that $q = 50$ is indeed a maximum point?
4. What is the significance of the discriminant in the context of quadratic functions like this one?
5. How would the function change if there were fixed costs associated with each unit sold?

Tip: When working with quadratic functions in revenue or profit maximization problems, always consider the sign of the leading coefficient to determine whether you're looking for a maximum or minimum value.