Let's solve this step-by-step.

Given Demand Function

The demand function is: $q = D(p) = \sqrt{384 - p}$

(a) The Elasticity of Demand

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

First, let's find $\frac{dq}{dp}$ by differentiating $q = \sqrt{384 - p}$ with respect to $p$.

1. Differentiate $q$ with respect to $p$: $q = (384 - p)^{\frac{1}{2}}$ Using the chain rule: $\frac{dq}{dp} = \frac{1}{2}(384 - p)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{2\sqrt{384 - p}}$

2. Substitute into the elasticity formula: Now, we substitute $q = \sqrt{384 - p}$ and $\frac{dq}{dp} = -\frac{1}{2\sqrt{384 - p}}$: $E(p) = \frac{p}{\sqrt{384 - p}} \cdot \left(-\frac{1}{2\sqrt{384 - p}}\right)$ Simplify: $E(p) = -\frac{p}{2(384 - p)}$

So, the elasticity of demand is: $E(p) = -\frac{p}{2(384 - p)}$

(b) Elasticity at $p = 103$

Now, substitute $p = 103$ into $E(p)$: $E(103) = -\frac{103}{2(384 - 103)} = -\frac{103}{2 \times 281} = -\frac{103}{562} \approx -0.1833$

Since $|E(103)| < 1$, the demand is inelastic at $p = 103$.

(c) The Value(s) of $p$ for Maximum Total Revenue

Total revenue $R(p)$ is given by: $R(p) = p \cdot q = p \cdot \sqrt{384 - p}$

To maximize revenue, we need to find the critical points of $R(p)$ by differentiating it with respect to $p$ and setting the derivative to zero.

1. Differentiate $R(p)$: $R(p) = p \cdot \sqrt{384 - p} = p(384 - p)^{\frac{1}{2}}$ Using the product rule and chain rule: $\frac{dR}{dp} = \sqrt{384 - p} + p \cdot \left(-\frac{1}{2\sqrt{384 - p}}\right)$ Simplifying: $\frac{dR}{dp} = \frac{384 - p - \frac{p}{2}}{\sqrt{384 - p}}$ Setting