1. ## lenght on a ladder

I have problem with this question :/

a 2 meter high plank stands on the distanse 3 meter from a wall. Decide the shortest lenght a ladder must have to reach from the ground to the wall. (the wall and the plank is vertikal and the ground is horisontel.)

the right answer is (3 + ^3(12^0.5) (1+(4 / 12^2/3))^0.5 but i dont get it :/

i did this way:
i say that the distance from the plank to the ladder is x

L^2 = (3+x)^2 + (6/x + 2)^2
2L= 2(3+X) + 2(6/x + 2)(-6/x^2)
L´= 0 => x4 + 3x^3 - 12x - 36 = 0
i dont know more, help please

2. Originally Posted by paulaa
I have problem with this question :/

a 2 meter high plank stands on the distanse 3 meter from a wall. Decide the shortest lenght a ladder must have to reach from the ground to the wall. (the wall and the plank is vertikal and the ground is horisontel.)

the right answer is (3 + ^3(12^0.5) (1+(4 / 12^2/3))^0.5 but i dont get it :/

i did this way:
i say that the distance from the plank to the ladder is x

L^2 = (3+x)^2 + (6/x + 2)^2
2L= 2(3+X) + 2(6/x + 2)(-6/x^2)
L´= 0 => x4 + 3x^3 - 12x - 36 = 0
i dont know more, help please
This is incomprehensible, is there an associated diagram for this question.

CB

3. no just the information i gave :/

4. Originally Posted by paulaa
no just the information i gave :/
Then the question is either incomplete or the translation is confusing.

CB

5. I think this may be the set up ... goal is to minimize L

6. exacly

7. I interpret this as meaning that the "2 meter high plank" forms a barrier three feet from the wall that the ladder must clear.

Let x be the distance from the plank to the foot of the ladder, and y the height that the ladder reaches on the wall. Assuming the ladder just clears the barrier, we have, by "similar triangles", $\frac{x}{2}= \frac{x+ 3}{h}$.

Essentially, then, you want to minimize $(x+3)^2+ h^2$ the length of the hypotenuse, subject to the condition that $\frac{x}{2}= \frac{x+3}{h}$. I don't see any good way to do that without Calculus.

8. If angle of ladder with ground is 45 degrees L =5rad2=7.071 m.There is a miminum at approx 41 degrees
L=7.023m

bjh

9. A "ye olde" one: