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ahhhh.. hard one.. .mainly because my differentiation and trig. skills haven't been used in YEARS!!!
lets see....(highlight to read.... )
Spoiler:
define x to be the angle between the ladder and the vertical wall. (that is, the wall that belongs to corridor B.
Let L be the maximum length of ladder than can be placed in the corner such that it has angle x with the B wall.
doing some simple trig,
L = A /cos(x) + B/sin(x) equation(1)
Now we find the angle x, for which L is minimized. (this restricts the maximum length of ladder....
dL/dx = A secx tanx -B cosecx tan x = 0
solving,
tan^3(x) = B/A
solve for x, plug this into the equation(1) to get the maximum length L.
Now, the above solution assumes the ladder is parallel to the floor. we can increase the length further by tilting the ladder so it touches the ceiling and floor.
looking from the side, it looks like a right triangle, with L and K forming the right angle.
By Pythagora's Theorem,
L" = sqrt(L^2 + K^2)
Last edited by dimbulb; 12-03-2003 at 04:28 PM..
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