1. ## Decide shortest length of ladder

Hi

I have attached a picture which help describe the problem.

The problem is: A fence is located 3m from a wall, and the fence is 2m high.
Decide the shortest length of a ladder (in green in the picture) placed on the ground and over the fence, touching the wall.

There are a lot of congruent right triangles, will this be used in the solution?
If I name the distance from the fence to the position where the ladder touches the ground a, and the height where the ladder touches the wall for $(h+2)$ . (The $+ 2$ comes from the fence height).

Won´t the length of the ladder be given by the pythagoran theorem?

Like

$\sqrt{(a+3)^{2}+(h+2)^{2}}$

Thx

2. Absolutely! And so use your congruent triangles idea to get either one of a or h as a function of the other. Then you can see how the ladder length varies as a function of a or h...

Ouch! (below...)
Quite right

3. Not "congruent" triangles- similar triangles. In particular, $\frac{a}{2}= \frac{h+3}{a+3}$

4. Hi

Yes, similar , thats what I meant =) Just didn´t come up with the english word

I think I got it now, but the answer becomes really messy..
I get a minima for $a=\sqrt[3]{12}$

Now I have $h = 2\cdot \frac{(a+3)}{a}-2$

So I plug these into $\sqrt{(a+3)^{2}+(h+2)^{2}}$ and ugh, I don´t care to write it out, it´s not important. If the solution is correct then I am satisfied.

Thx

5. Surely,

$\frac{h + 2}{a + 3} = \frac{2}{a}$

giving

$h = \frac{6}{a}$

?

6. Hi Tom

Yes, that is also correct.

Differentiating with $h=\frac{6}{a}$ gives $a=\sqrt[3]{12}$

And here it is obv easier to obatin a value for $h$

7. Well done.