For the following demand function, find $\mathbf{a}$. $\mathrm{E}$, and $\mathbf{b}$. the values of $\mathrm{q}$ (if any) at which total revenue is maximized.
\[
p=500 e^{-0.5 q}
\]
a. Find the elasticity of demand $(E)$ in terms of $q$.
\[
E(q)=\square
\]
b. Find the values of q (if any) at which total revenue is maximized. Select the correct choice below, and if necessary, fill in the answer box within your choice.
A. Revenue is maximized at $q=\square$.
(Type an integer or a simplified fraction.)
B. No values of q maximize revenue.

Step 1 ：The elasticity of demand (E) is defined as the percentage change in quantity demanded divided by the percentage change in price. We can find it by taking the derivative of the natural logarithm of the price function with respect to q, and then multiplying by q/p.

Step 2 ：First, let's find the derivative of \(\ln(p)\) with respect to q: \(\ln(p) = \ln(500) - 0.5q\), so \(\frac{d[\ln(p)]}{dq} = -0.5\).

Step 3 ：Then, we multiply by q/p: \(E(q) = q * (-0.5) / (500e^{-0.5q}) = -0.5q / (500e^{-0.5q}) = -0.5q * e^{0.5q} / 500 = -qe^{0.5q} / 1000\).

Step 4 ：\(\boxed{E(q) = -qe^{0.5q} / 1000}\) is the elasticity of demand.

Step 5 ：Total revenue (R) is defined as the price (p) times the quantity (q). To find the values of q at which total revenue is maximized, we need to find the derivative of the revenue function with respect to q, set it equal to zero, and solve for q.

Step 6 ：R = p*q = 500qe^{-0.5q}. Taking the derivative of R with respect to q gives: \(\frac{dR}{dq} = 500e^{-0.5q} - 250qe^{-0.5q}\).

Step 7 ：Setting this equal to zero and solving for q gives: \(500e^{-0.5q} - 250qe^{-0.5q} = 0\), which simplifies to \(e^{-0.5q} (2 - q) = 0\).

Step 8 ：This equation is satisfied when q = 2. So, revenue is maximized at q = 2.

Step 9 ：\(\boxed{q = 2}\) is the quantity at which total revenue is maximized.