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This appears to be a problem related to revenue and cost functions. The given equations are:

- Revenue function: \( R = 48Q - 2Q^2 \)
- Total cost function: \( TC = 300 - 18Q + Q^2 \)

The problem seems to ask for the value of \( Q \) that maximizes the revenue and also to determine the profit.

Let me solve this step by step:

### a) **Maximizing Revenue**  
Revenue is maximized when the derivative of the revenue function with respect to \( Q \) is zero.

The first derivative of the revenue function \( R = 48Q - 2Q^2 \) with respect to \( Q \) is:

\[
\frac{dR}{dQ} = 48 - 4Q
\]

Now, set the derivative equal to zero to find the critical point:

\[
48 - 4Q = 0
\]

Solving for \( Q \):

\[
4Q = 48 \quad \Rightarrow \quad Q = 12
\]

So, \( Q = 12 \) maximizes the revenue.

Would you like me to proceed with the other parts of the question, or would you like more details on this part?

The given revenue function R = 48Q - 2Q^2 and total cost function TC = 300 - 18Q + Q^2. Find the value of Q that maximizes the revenue and the profit.

This problem involves maximizing revenue and calculating profit using the given revenue and cost functions. Learn how to use derivatives to find the optimal quantity (Q) and analyze the corresponding profit. Suitable for students in Grades 11-12 studying economics or calculus.

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