
Random Calculus Question
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...

hi
$\displaystyle \frac{d}{db} (b^{4}  25b^{2} +37.20) = 4b^{3}  50b = 4b(b^{2}  \frac{50}{4}) = 0$

That's what I did but the answer I got wasn't the same as the one in the solutions (which I've checked to be right). (Headbang)