To find the quantity that maximizes total revenue, we can follow these steps:

1. Express the Total Revenue Function:

The total revenue (TR) is given by the product of the price (P) and quantity (Q):

$TR = P \times Q$

Given the demand function:

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

So, the total revenue function becomes:

$TR = Q \times (577 - 3.5Q) = 577Q - 3.5Q^2$

2. Find the Quantity that Maximizes Total Revenue:

To maximize the total revenue, take the derivative of the $TR$ function with respect to $Q$ and set it equal to zero:

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

Set the derivative equal to zero to find the critical point:

$577 - 7Q = 0$

$7Q = 577$

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

So, the quantity $Q$ that maximizes total revenue is 82.429 (rounded to three decimal places).

Would you like further details or explanations?

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Here are some related questions:

1. What is the maximum total revenue?
2. How do we calculate the profit at the revenue-maximizing quantity?
3. What is the price at the revenue-maximizing quantity?
4. How would fixed costs impact profit at this quantity?
5. What is the elasticity of demand at the revenue-maximizing quantity?

Tip: The revenue-maximizing quantity occurs where the marginal revenue equals zero, not necessarily where profit is maximized.