The problem of the ladder!

**uruleo** · April 28th 2011, 10:48 AM

I hope you can help me on to the next problem!

I need to enter a ladder to a tower (not considered wide). What is the minimum height of the door (h) that makes this possible?

How I can solve this problem?

Thanks!!

**earboth** · April 28th 2011, 11:32 PM

Originally Posted by uruleo

I hope you can help me on to the next problem!

I need to enter a ladder to a tower (not considered wide). What is the minimum height of the door (h) that makes this possible?

How I can solve this problem?

Thanks!!

1. I've drawn a sketch of the worst case. (see attachment)

2. You get 2 similar triangles. Use proportions:

$\dfrac hy = \dfrac x{27}~\implies~y = \dfrac{27 h}{x}$

3. Use the Pythagorean theorem:

$(h+x)^2+(27+y)^2=125^2$

4. Replace y by the term of #2 and solve for h:

$h(x)=\dfrac{x(125-\sqrt{x^2+729})}{\sqrt{x^2+729}}$

5. Differentiate h wrt x. Solve $\frac{dh}{dx} = 0$ for x. You should come out with x = 36. Which means that the minimum value of h = 64.

Spoiler:

**uruleo** · April 29th 2011, 06:29 AM

Thanks!!

But i did not understand the step 4..

**earboth** · April 29th 2011, 10:45 AM

Originally Posted by uruleo

Thanks!!

But i did not understand the step 4..

I'm not sure what you mean ...(?)

1. You have $y = \dfrac{27h}x$ and $(h+x)^2+(27+y)^2=125^2$ which becomes:

$(h+x)^2+\left(27 + \frac{27h}x\right)^2=125^2$

$(h+x)^2+\left(\frac{27}{x}\right)^2 (x+h)^2=125^2$

$(h+x)^2=\dfrac{125^2x^2}{x^2+27^2}$

2. Can you take it from here?

**uruleo** · May 3rd 2011, 06:51 PM

I think I got it, but I would appreciate if you specify each of the steps.

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