To solve this problem step-by-step, we will analyze the demand function $q = D(p) = \sqrt{335 - p}$, and calculate the required components.

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Part 1: Find the equation for elasticity, $E(p)$

The formula for price elasticity of demand is: $E(p) = -\frac{p}{q} \cdot \frac{dq}{dp}$

1. Calculate $\frac{dq}{dp}$: From $q = \sqrt{335 - p}$, differentiate with respect to $p$: $\frac{dq}{dp} = \frac{1}{2} (335 - p)^{-1/2} \cdot (-1) = -\frac{1}{2\sqrt{335 - p}}$

2. Substitute $\frac{dq}{dp}$ and $q$ into $E(p)$: $q = \sqrt{335 - p}, \quad \frac{dq}{dp} = -\frac{1}{2\sqrt{335 - p}}$ Therefore: $E(p) = -\frac{p}{\sqrt{335 - p}} \cdot \left(-\frac{1}{2\sqrt{335 - p}}\right)$ Simplify: $E(p) = \frac{p}{2(335 - p)}$

The equation for elasticity is: $E(p) = \frac{p}{2(335 - p)}$

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Part 2: Find $E(102)$ and determine elasticity type

Substitute $p = 102$ into $E(p) = \frac{p}{2(335 - p)}$: $E(102) = \frac{102}{2(335 - 102)}$ Simplify the denominator: $E(102) = \frac{102}{2 \cdot 233} = \frac{102}{466} = \frac{51}{233}$

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Part 3: Determine elasticity type

Compare $E(102)$ to 1:

• If $E(p) > 1$, the demand is elastic.
• If $E(p) = 1$, the demand has unit elasticity.
• If $E(p) < 1$, the demand is inelastic.

Since $\frac{51}{233} \approx 0.219$, which is less than 1, the demand is inelastic at $p = 102$.

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Part 4: Find the value(s) of $p$ for which total revenue is a maximum

Total revenue, $R(p)$, is given by: $R(p) = p \cdot q = p \cdot \sqrt{335 - p}$

1. Differentiate $R(p)$ with respect to $p$: Using the product rule: $R'(p) = \sqrt{335 - p} + p \cdot \frac{d}{dp}\left(\sqrt{335 - p}\right)$ Substituting $\frac{d}{dp} \sqrt{335 - p} = -\frac{1}{2\sqrt{335 - p}}$: $R'(p) = \sqrt{335 - p} - \frac{p}{2\sqrt{335 - p}}$

2. Set $R'(p) = 0$: $\sqrt{335 - p} = \frac{p}{2\sqrt{335 - p}}$ Multiply through by $2\sqrt{335 - p}$ to eliminate the denominator: $2(335 - p) = p$ Simplify: $670 - 2p = p \quad \implies \quad 670 = 3p \quad \implies \quad p = \frac{670}{3} \approx 223.33$

3. Verify the maximum: The revenue function $R(p)$ increases for $p < 223.33$ and decreases for $p > 223.33$. Thus, $p = 223.33$ maximizes total revenue.

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Final Answers:

(a) Elasticity equation: $E(p) = \frac{p}{2(335 - p)}$

(b) Elasticity at $p = 102$: $E(102) = \frac{51}{233} \quad \text{(approximately 0.219, inelastic)}$

(c) Price for maximum revenue: $p = 223.33 \, \text{(rounded to nearest cent)}$

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Do you want further explanations or detailed derivations?

Here are 5 follow-up questions to expand the topic:

1. How is elasticity related to revenue maximization?
2. Can you explain how elasticity varies for different demand curves?
3. What happens to elasticity as $p \to 0$ or $p \to 335$?
4. Why is the demand inelastic at $p = 102$?
5. How does differentiating total revenue help find its maximum?

Tip: For maximizing or minimizing a function, always confirm your critical points using second derivatives or monotonicity analysis!