How to find the maximum of this function on the positive real line?

Question

I need to maximize this function on the positive real line:

$$ \frac{1}{\Gamma(x)^{14}}\cdot\frac{1}{{\frac{323.6}{14x}}^{14x}}\cdot(1.22578*10^{19})^{x-1}e^{-14x} $$ the correct answer should be around $x=514$, but I could not use mathematica to show it. My plot has a peak at $x=680$ instead. I want to ask if anyone has any idea how to get the correct answer.

The code I am using now is:

A = 1/Gamma[x]^{14}*(301.6/(14*x))^{-14 *x}*(1.22578*10^{19})^{x -1} E^{-14 x}

and I plot it by

Plot[A, {x, 513, 700}, PlotPoints -> 1000]

The result is something very strange:

Answer 1

I don't like using approximate numbers when we can use exact ones, therefore with Rationalize and RootApproximant I can rewrite the function this way:

f[x_] := 1/Gamma[x]^14 1/(1618/(70 x))^(14 x) 12257800000000000000^(x - 1) Exp[-14 x]

let's plot this function:

Plot[ f[x], {x, 0, 1500}, PlotStyle -> Thick]

We can use:

NMaximize[{ f[x], 10^5 > x > 0}, x]

{1.82059*10^-9, {x -> 514.198}}

which most likely is a good approximation. However when we use e.g.

NMaximize[{ f[x], 10^6 > x > 0}, x]

{3.92756*10^-30, {x -> 0.327666}}

we get an incorret result. It may depend on the computer system one uses.
I guess this issue is explained with anwswers to this post: Numerical underflow for a scaled error function.

Answer 2

Following Artes' Rationalizeation, why don't you simply find where the first derivative goes to zero?

f[x_] := 1/Gamma[x]^14 1/(1618/(70 x))^(
    14 x) 12257800000000000000^(x - 1) Exp[-14 x]

Plot[f'[x], {x, 100, 1100}]

FindRoot[f'[x] == 0, {x, 500}]

    {x -> 514.198}