Ok I just need to do this and end this time consuming back and forth. I'm sorry I haven't been that helpful.

. I just noticed a problem. In your original post you wrote $\displaystyle 4\pi ^2 x^2$ but later on you said it was $\displaystyle 4\pi ^2 ax^2$. This will definitely make a difference.

Assuming you mean the second one, this is how to solve for x.

$\displaystyle

f'(x)=(2\pi+8a{\pi}^2)x-(28a\pi)=0

$

$\displaystyle (2\pi+8a{\pi}^2)x=(28a\pi)$

$\displaystyle x=\frac{28a\pi}{2\pi+8a{\pi}^2}$

Clear up this last bit of info and I'll finish the problem!