Answer
$x=\frac{a}{2}$, $v(\frac{a}{2})=\frac{1}{4}ka^2$
Work Step by Step
Step 1. Given the rate equation $v=kax-kx^2$, we need to find the $x$ value to give a maximum of $v$.
Step 2. Take the derivative to get $v'=ka-2kx$ and let $v'=0$ to get $x=\frac{a}{2}$ which gives an extreme value of $v(\frac{a}{2})=ka(\frac{a}{2})-k(\frac{a}{2})^2=\frac{1}{2}ka^2-\frac{1}{4}ka^2=\frac{1}{4}ka^2$
Step 3. Check $v''=-2k\lt0$ (because $k\gt0$), indicating a concave down region with a local maximum.
