challenger calculas question?

Question

lets say there was a corrider stuck to wall of a building. shaped like a rectanglular cube 2 meters wide and 3 meters high
what would be the shortest ladder posibble that if it leaned on that corrider would reach the wall

hint it doesnt have to be 3d when your imagining it just imagine a rectangul 2m wide and 3 meters high in a ritgh angel triangle what could be the shortest hypotenous that could maintian that triangle with that rectangul but still keep it ritgh angled?

Answer 1

just using graph paper...I have my ladder outside the corridor forming two triangles (it's easier INSIDE to show the longest and shortest ladders possible...but that's not my interpretation of what you're asking)

45 degree angle gives you pretty close to the shortest distance...but it's not the case if you change the dimensions of the corridor...so you should end up with pretty close to

(floor to corner of corridor) = sqrt ( 3 squared + 3 squared)
(corner to wall) = sqrt (2 squared = 2 squared)
you want the sum of the two hypotenuses to be as small as possible, and the angles have to match on the two triangles ( m = tan angle) might help...

I think 7.07 is pretty close..but it's not calculus...I just found it interesting to play with awhile.

you can set up third triangle...and use the same variables

hypotenuse of big triangle = sqrt ( (x+2)squared + (y+3)squared = hypotenuse of little triangle (sqrt (x squared + 3 squared))+ hypotenuse of second little triangle (sqrt (y squared + 2 squared)

Answer 2

Well, intuitively, the longest ladder would stretch from corner to corner. Pythagorean tells us that this would be sqrt(2² + 3²) = 3.6 m.

The shortest ladder would lay flat on the ground, wall to wall. This would have the same length as the width, or 2 m.


You want me to show you that using calculus? Ok.

Let's let the ladder have length L. If we set up a coordinate system with the origin at the lower left corner of the room, and lay the ladder at the corner above the origin, we get the following points...

The ladder starts at (0, 3)
The ladder ends at (2, y)

So the length of the ladder is

L² = 2² + (y - 3)²

Now we need to take the derivative and set to zero. This can be done implicitly quite easily.

2L dL = 2(y - 3) dy
dL / dy = (y - 3) / L

Setting dL / dy = 0...

(y - 3) / L = 0 if and only if y = 3.

L² = 2² + (3 - 3)²
L² = 2²
L = 2 m


That's actually the shortest ladder than can be fit in.



I'm not sure if this answers your question. You honestly didn't need calculus for it.