Ladder

A 20 foot ladder is standing on the ground, leaning against a wall, and touching a 6 foot high, 6 foot wide shed built against the the wall's base. (Forming a right angled triangle, 20' diagonal, enclosing a 6' square)
What is the maximum height of the top of the ladder?

Originally posted by CauselessOne
A 20 foot ladder is standing on the ground, leaning against a wall, and touching a 6 foot high, 6 foot wide shed built against the the wall's base. (Forming a right angled triangle, 20' diagonal, enclosing a 6' square)
What is the maximum height of the top of the ladder?

20 feet.

you must have very flexible feet

Originally posted by CauselessOne
you must have very flexible feet

I see no restriction on placing the ladder vertically against the wall.

But perhaps that's me.

Originally posted by TheMaster37
I see no restriction on placing the ladder vertically against the wall.

But perhaps that's me.

Except maybe the shed in the way. 😛

14

Er - - - - no.

"Maximum"? You already told us exactly how the ladder is positioned. Or did you mean we were allowed to move it?

20/sqrt2

Originally posted by AThousandYoung
"Maximum"? You already told us exactly how the ladder is positioned. Or did you mean we were allowed to move it?

20/sqrt2

I'd rather it stayed put, thanks. You can help me move it later, if you like (It will be quite heavy).
'Maximum' selects the higher of the two possible answers.

Originally posted by CauselessOne
A 20 foot ladder is standing on the ground, leaning against a wall, and touching a 6 foot high, 6 foot wide shed built against the the wall's base. (Forming a right angled triangle, 20' diagonal, enclosing a 6' square)
What is the maximum height of the top of the ladder?

Here's my solution:

We want a line passing through the point (6,6). This is the same as moving a line passing through the origin up 6 units and right 6 units, so we write the equation of the line as:

(y-6) = m(x-6)

Rearranging, we get:

y = mx + 6(1-m)

Solving for the x- and y-intercepts (the foot and top of the ladder, respectively) we get:

y=0, x = -6(1-m)/m; and x=0, y = 6(1-m)

The distance between these points will be 20, so we write the distance equation accordingly:

(6(1-m))^2 + (-6(1-m)/m)^2 = 20^2

Rearranging, we get:

m^4 - 2m^3 + (2-(20/6)^2)m^2 - 2m + 1 = 0

With a little help from Excel, I got three solutions: m1=-1.9734, m2=-0.5068, and m3=0.2356. m1 is the correct slope. m2 gives the minimum height, and m3 gives a slope going through the shed.

Plugging m1 back into the equation for the y-intercept, we get:

y = 6(1-m) = 6(1-(-1.9734)) = 17.840

This is the maximum height of the ladder against the wall.

Originally posted by PBE6
Here's my solution:

We want a line passing through the point (6,6). This is the same as moving a line passing through the origin up 6 units and right 6 units, so we write the equation of the line as:

(y-6) = m(x-6)

Rearranging, we get:

y = mx + 6(1-m)

Solving for the x- and y-intercepts (the foot and top of the ladder, respectively) we get:

y ...[text shortened]... 6(1-m) = 6(1-(-1.9734)) = 17.840

This is the maximum height of the ladder against the wall.

..... is the correct answer! As far as I know, there is no neat mathematical solution to this one - some form of linear regression or 'goal seek' method is the only way. If anyone can prove me wrong I would be interested.

Originally posted by CauselessOne
..... is the correct answer! As far as I know, there is no neat mathematical solution to this one - some form of linear regression or 'goal seek' method is the only way. If anyone can prove me wrong I would be interested.

I believe it's possible to analytically solve polynomials up to degree 5 (and only some of degree 5), but the solutions look worse than the original problems!! Case in point:

http://mathworld.wolfram.com/QuinticEquation.html

Originally posted by CauselessOne
..... is the correct answer! As far as I know, there is no neat mathematical solution to this one - some form of linear regression or 'goal seek' method is the only way. If anyone can prove me wrong I would be interested.

Oops. I assumed the ladder was at a 45 degree angle for some reason.

I still don't understand the problem 🙁