- 05 Jul '06 10:27

20 feet.*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? - 05 Jul '06 20:32

I'd rather it stayed put, thanks. You can help me move it later, if you like (It will be quite heavy).*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

'Maximum' selects the higher of the two possible answers. - 06 Jul '06 18:36

Here's my solution:*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?

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. - 06 Jul '06 19:07

..... 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 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. - 06 Jul '06 19:31

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:*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.**

http://mathworld.wolfram.com/QuinticEquation.html - 06 Jul '06 22:38

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