I'm self-studying Brownian motion and encountered the following problem.
I understand the author's solution, and it is clear why maximizing the right-hand side of the inequality provides such $t$ that maximizes $A(t) < -1$.
My question is, if we want to maximize $$N\left(\frac{-1 - 0.4t}{\sqrt{0.36}t}\right),$$ could we not set
$$N'\left(\frac{-1 - 0.4t}{\sqrt{0.36}t}\right) = 0,$$
giving us $$\frac{e^{-x^2/2}}{\sqrt{2\pi}} = 0,$$
where $x = \frac{-1 - 0.4t}{\sqrt{0.36}t}$ ?
However, trying to solve for $t$ would appear to be taking the natural log of 0, which is impossible.
What am I overlooking here?