For the following demand function, find (a) $\mathrm{E}$, and (b) values of $q$ (if any) at which total revenue is maximized.
\[
q=50-\frac{p}{5}
\]
(a) $E=$
(b) Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. No values of q maximize the total revenue.
B. The total revenue is maximized at $q=$

Step 1 ：First, we need to understand the problem. The demand function is given by \(q=50-\frac{p}{5}\), where \(q\) is the quantity demanded and \(p\) is the price. We are asked to find the price elasticity of demand \(E\) and the quantity \(q\) that maximizes total revenue.

Step 2 ：Total revenue \(R\) is given by the product of the price \(p\) and the quantity demanded \(q\), or \(R=pq\). Substituting the demand function into this equation, we get \(R=p(50-\frac{p}{5})=50p-\frac{p^2}{5}\).

Step 3 ：We want to maximize this expression by completing the square. We can factor out a \(-\frac{1}{5}\) to get \(-\frac{1}{5}(p^2-250p)\).

Step 4 ：To complete the square, we add \((250/2)^2=62500\) inside the parenthesis and subtract \(-\frac{1}{5}\cdot62500=-12500\) outside. We are left with the expression \(-\frac{1}{5}(p^2-250p+62500)+12500=-\frac{1}{5}(p-125)^2+12500\).

Step 5 ：Note that the \(-\frac{1}{5}(p-125)^2\) term will always be nonpositive since the perfect square is always nonnegative. Thus, the revenue is maximized when \(-\frac{1}{5}(p-125)^2\) equals 0, which is when \(p=125\).

Step 6 ：Substitute \(p=125\) into the demand function, we get \(q=50-\frac{125}{5}=25\).

Step 7 ：So, the total revenue is maximized at \(q=\boxed{25}\).

Step 8 ：For the price elasticity of demand \(E\), it is defined as the percentage change in quantity demanded divided by the percentage change in price. In this case, \(E=\frac{dq/dp}{q/p}=-\frac{1/5}{50-\frac{p}{5}}\cdot p=-\frac{p}{250-p}\).

Step 9 ：Substitute \(p=125\) into the elasticity function, we get \(E=-\frac{125}{250-125}=-1\).

Step 10 ：So, the price elasticity of demand \(E=\boxed{-1}\).