For the demand function q equals Upper D left parenthesis p right parenthesis equals StartRoot 384 minus p EndRootq=D(p)=384− p​, find the following.
​a) The elasticity
​b) The elasticity at pequals=103103​, stating whether the demand is​ elastic, inelastic or has unit elasticity
​c) The​ value(s) of p for which total revenue is a maximum​ (assume that p is in​ dollars)

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

For the demand function q = D(p) = √(384 - p), find the following: a) The elasticity b) The elasticity at p = 103, stating whether the demand is elastic, inelastic or has unit elasticity c) The value(s) of p for which total revenue is a maximum (assume that p is in dollars)

This detailed solution explains how to calculate the elasticity of demand, find its value at p = 103, and determine the price for maximum total revenue for the demand function q = √(384 - p). The problem covers key topics in differentiation, elasticity, and maximization. Suitable for students in Grades 11-12.

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