To find the quantity when total revenue is maximized, we first need to derive the total revenue function and then determine the quantity at which this revenue is maximized.

Step 1: Express the Total Revenue Function

The demand function is given by:

$P = \frac{1154}{2} - \frac{7}{2}Q = 577 - 3.5Q$

Total Revenue (TR) is given by:

$TR = P \times Q$

Substituting the demand function into the total revenue equation:

$TR = (577 - 3.5Q) \times Q$

This simplifies to:

$TR = 577Q - 3.5Q^2$

Step 2: Differentiate the Total Revenue Function

To find the quantity that maximizes total revenue, we need to take the derivative of the total revenue function with respect to $Q$ and set it equal to zero:

$\frac{d(TR)}{dQ} = 577 - 7Q$

Step 3: Set the Derivative Equal to Zero and Solve for $Q$

$577 - 7Q = 0$

$7Q = 577$

$Q = \frac{577}{7} \approx 82.429$

So, the quantity that maximizes total revenue is 82.429 units.

Would you like further details or have any other questions?

Here are 5 relative questions to expand on the topic:

1. How do fixed and variable costs affect the profit function?
2. What is the relationship between marginal revenue and marginal cost at the profit-maximizing quantity?
3. How would a change in the unit cost affect the profit-maximizing quantity?
4. How do you calculate the break-even quantity for a firm?
5. What is the significance of elasticity of demand in revenue maximization?

Tip: The quantity that maximizes total revenue occurs where marginal revenue equals zero.