For the demand function $q=D(p)=\sqrt{471-p}$, find the following
a) The elasticity
b) The elasticity at $p=125$, 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)
a) Find the equation for elasticity.
\[
E(p)=
\]
b) Find the elasticity at the given price, stating whether the demand is elastic, inelastic or has unit elasticity. $E(125)=\square$ (Simplify your answer. Type an integer or a fraction.) Is the demand elastic, inelastic, or does it have unit elasticity?
elastic
unit elasticity
inelastic
c) Find the value(s) of $p$ for which total revenue is a maximum (assume that $p$ is in dollars).
$\$ \square$
(Round to the nearest cent. Use a comma to separate answers as needed.)

Step 1 ：First, we need to find the revenue function. The revenue is given by the price times the quantity sold, or \(pD(p)=p\sqrt{471-p}\).

Step 2 ：We want to maximize this expression. To do this, we can take the derivative of the revenue function with respect to \(p\) and set it equal to zero.

Step 3 ：The derivative of \(p\sqrt{471-p}\) is \(\sqrt{471-p}-\frac{p}{2\sqrt{471-p}}\).

Step 4 ：Setting this equal to zero, we get \(\sqrt{471-p}=\frac{p}{2\sqrt{471-p}}\).

Step 5 ：Squaring both sides to eliminate the square root, we get \(471-p=\frac{p^2}{4(471-p)}\).

Step 6 ：Multiplying both sides by \(4(471-p)\) to clear the fraction, we get \(4(471-p)^2=p^2\).

Step 7 ：Solving this quadratic equation, we get \(p=2\sqrt{471-p}\).

Step 8 ：Substituting \(p=2\sqrt{471-p}\) back into the equation, we get \(2\sqrt{471-2\sqrt{471-p}}=p\).

Step 9 ：Solving this equation, we get \(p=\sqrt{471}\).

Step 10 ：However, since the price cannot be greater than 471, the maximum revenue is achieved when \(p=\boxed{\sqrt{471}}\) dollars.