Random Calculus Question

**xwrathbringerx** · April 24th 2009, 11:27 PM

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

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

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

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

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

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 $b^4 - 25b^2 + 37.20 = 0$ and then equating it to 0 like similar questions but it doesn't seem to work...

**Twig** · April 24th 2009, 11:33 PM

hi

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

**xwrathbringerx** · April 24th 2009, 11:46 PM

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).

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