A 5 m ladder leans against a wall with a 1 m x 1 m crate between the ladder and the wall. a is the distance from the bottom of the wall to the base of the ladder where it is touching the ground. b is the distance from the bottom of the wall to the point the ladder is touching the wall.

From the previous parts of the question, I've got the following answers:

a + b = ab

$\displaystyle a^2 + b^2 = 25 $ (Pythagoras')

$\displaystyle (ab)^2 - 2ab - 25 = 0$

$\displaystyle a = \frac{\sqrt{26} + 1}{b}$

$\displaystyle b^4 - 25b^2 + 37.20 = 0$

From all the formulas I've found out, how do I find the maximum value of b?

The answer's 4.84.

I've tried using differentiating $\displaystyle b^4 - 25b^2 + 37.20 = 0$ and then equating it to 0 like similar questions but it doesn't seem to work...