I am sorry mikau, you were right. I found my earlier mistake that made me put my foot in my mouth.

Here is exactly how I did the problem:

A = πr √[r² + (300/πr²)²]

A = πr √[r² + (90000/π²r^4)] ; squared the inside term as indicated

A = πr √[(π²r^6 + 90000) / (π²r^4)] ; made the denominators the same

A = (1/r) √(π²r^6 + 90000) ; moved the denominator outside the radical sign

We know that the derivative of this function is; f(x)f'(g) + f'(x)f(g)

f(x) = 1/r; f(g) = √(π²r^6 + 90000);

(1/r) (1/[2√(π²r^6 + 90000)]) (6π²r^5) = f(x)f'(g)

f(x)f'(g) = 3π²r^4/(√(π²r^6 + 90000)

f'(x)f(g) = (-1/r²)(√(π²r^6 + 90000)) = -√(π²r^6 + 90000)/r²

So the derivative is;

3π²r^4 _ √(π²r^6 + 90000)

√(π²r^6 + 90000) r²

Making the denominators the same and simplifying gives;

2π²r^6 - 90000

r²√(π²r^6 + 90000)

So for this to equal zero r^6 = 90000/ (2π²)

You are correct, my mistake. (they are becoming more frequent with age.)

For some reason or other I had a positive sign before the constant when I integrated before. Nice move, your calculations seemed much neater.