To solve this, let's analyze the problem step by step.

Given:

The demand function is: $D(p) = \sqrt{100 - 3p}$

We need to:

1. Find the elasticity of demand at $p = 15$.
2. Classify the demand as inelastic, elastic, or unitary.
3. Decide whether to raise or lower the price to increase revenue based on the elasticity.

1. Formula for Elasticity of Demand:

The elasticity of demand, $E(p)$, is given by: $E(p) = \left| \frac{dD(p)}{dp} \cdot \frac{p}{D(p)} \right|$

Step 1: Differentiate $D(p)$ with respect to $p$:

$D(p) = \sqrt{100 - 3p}$ Using the chain rule: $\frac{dD(p)}{dp} = \frac{1}{2\sqrt{100 - 3p}} \cdot (-3) = \frac{-3}{2\sqrt{100 - 3p}}$

Step 2: Substitute $p = 15$ into $D(p)$:

$D(15) = \sqrt{100 - 3(15)} = \sqrt{100 - 45} = \sqrt{55}$

Step 3: Substitute $p = 15$ into $\frac{dD(p)}{dp}$:

$\frac{dD}{dp} = \frac{-3}{2\sqrt{100 - 3(15)}} = \frac{-3}{2\sqrt{55}}$

Step 4: Compute $E(p)$:

$E(15) = \left| \frac{\frac{-3}{2\sqrt{55}} \cdot 15}{\sqrt{55}} \right|$ Simplify: $E(15) = \left| \frac{-3 \cdot 15}{2\sqrt{55} \cdot \sqrt{55}} \right| = \left| \frac{-45}{2 \cdot 55} \right| = \frac{45}{110}$ $E(15) = 0.409$

2. Classification of Demand:

• If $E(p) < 1$, demand is inelastic.
• If $E(p) = 1$, demand is unitary.
• If $E(p) > 1$, demand is elastic.

Since $E(15) = 0.409 < 1$, the demand is inelastic.

3. Recommendation for Revenue:

When demand is inelastic, increasing the price will increase total revenue. Therefore, the correct choice is Raise Prices.

Let me know if you'd like more details on this!

Here are some follow-up questions to expand on this:

1. What is the economic intuition behind price elasticity of demand?
2. How would $E(p)$ change if the price were $p = 20$?
3. Can we determine the revenue-maximizing price from this information?
4. How does elasticity affect government taxation policies?
5. What happens if the demand function changes to a different form, such as $D(p) = 100 - 5p$?

Tip: When calculating elasticity, always double-check that units and scale align correctly to avoid errors.