Let's work through the problem step-by-step.

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Part 1: Elasticity of Demand (E)

The demand function is given by:
$q = 37,400 - 3p^2$
Elasticity of demand, $E$, is generally defined as: $E = \frac{p}{q} \cdot \frac{dq}{dp}$

1. Find the derivative of $q$ with respect to $p$: $\frac{dq}{dp} = -6p$

2. Substitute the values of $q$ and $\frac{dq}{dp}$ into the formula for $E$: $E = \frac{p}{37,400 - 3p^2} \cdot (-6p) = \frac{-6p^2}{37,400 - 3p^2}$

Thus, the elasticity of demand is: $E = \frac{6p^2}{37,400 - 3p^2}$

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Part 2: Values of $q$ that Maximize Total Revenue

Total revenue is given by: $TR = p \cdot q$ Since $q = 37,400 - 3p^2$, we substitute this into the expression for total revenue: $TR = p \cdot (37,400 - 3p^2) = 37,400p - 3p^3$

Now, to find the value of $p$ that maximizes total revenue, we take the derivative of $TR$ with respect to $p$ and set it equal to 0.

1. Differentiate $TR$ with respect to $p$: $\frac{d(TR)}{dp} = 37,400 - 9p^2$

2. Set the derivative equal to 0: $37,400 - 9p^2 = 0$ $9p^2 = 37,400$ $p^2 = \frac{37,400}{9} = 4,155.56$ $p = \sqrt{4,155.56} \approx 64.5$

3. Find the corresponding value of $q$: Substituting $p = 64.5$ into the demand function: $q = 37,400 - 3(64.5)^2 = 37,400 - 12,495.75 = 24,904.25 \approx 24,904$

So, the value of $q$ that maximizes total revenue is approximately: $q \approx 24,904$

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Final Answers:

Part 1:
$E = \frac{6p^2}{37,400 - 3p^2}$

Part 2:
A. Total revenue is maximized at approximately $q = 24,904$.

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Follow-up Questions:

1. How does elasticity of demand relate to maximizing revenue?
2. What happens to total revenue if $p$ is increased beyond 64.5?
3. Can we use the second derivative test to confirm that this value of $p$ gives a maximum?
4. How would the result change if the demand function had a linear form?
5. How does elasticity change as the price $p$ increases?

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Tip:

The total revenue is maximized when the price elasticity of demand is equal to 1 (unitary elastic).