*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.