05 Jul 06
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
Originally posted by CauselessOne20 feet.
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
Originally posted by AThousandYoungI'd rather it stayed put, thanks. You can help me move it later, if you like (It will be quite heavy).
"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
Originally posted by CauselessOneHere's my solution:
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
Originally posted by PBE6..... 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.
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
Originally posted by CauselessOneI 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:
..... 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
Originally posted by CauselessOneOops. I assumed the ladder was at a 45 degree angle for some reason.
..... 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.