Respuesta :
Generally a 45 degree angle is the best angle. this yields the smallest triangle with a 6' fence vertical side. As the ladder reaches the wall, two similar triangles are resolved. the 45 degree angle that the ladder makes with the ground yields a 6x6 legged right triangle. Adding 5' to the ground makes the larger 45 degree triangle with a base of 5+6=11' side. Using the formula for Special Triangles, 45-45-90 gives the ladder a 11 times the Squareroot of 2 for its length.
Looks too ugly to me, but this is what I get:
First, draw the triangle of the ladder, with the fence and the building.
There are two similar right triangles: the one from the ground to the fence, and the one from the ground to the building. Let's call L the length of the ladder and l, the length of the ladder from the ground to the fence, then:
L/(x+d) = l/x, where x is the horizontal distance from the point where the ladder touches the ground to the base of the fence and 'd' is the distance between the fence and the building (d=5 ft).
L = (x+d)/x*l, now l = sqrt( x^2+h^2), where h is the height of the fence (h=6 ft).
Then, one usually squares it, because the point where the maximum or minimum is attained is the same when optimizing a distance.
So that, one could look for the minimum of L^2 = (x+d)^2/x^2*(x^2+h^2), where d = 5 ft, and h = 6ft.
What I don;t like, is that L' = dL/dx is a bit lengthy to calculate and then you have to set it = 0. I think all this is right, but I expected a somewhat simpler expression, unless you're doing AP or similar.
First, draw the triangle of the ladder, with the fence and the building.
There are two similar right triangles: the one from the ground to the fence, and the one from the ground to the building. Let's call L the length of the ladder and l, the length of the ladder from the ground to the fence, then:
L/(x+d) = l/x, where x is the horizontal distance from the point where the ladder touches the ground to the base of the fence and 'd' is the distance between the fence and the building (d=5 ft).
L = (x+d)/x*l, now l = sqrt( x^2+h^2), where h is the height of the fence (h=6 ft).
Then, one usually squares it, because the point where the maximum or minimum is attained is the same when optimizing a distance.
So that, one could look for the minimum of L^2 = (x+d)^2/x^2*(x^2+h^2), where d = 5 ft, and h = 6ft.
What I don;t like, is that L' = dL/dx is a bit lengthy to calculate and then you have to set it = 0. I think all this is right, but I expected a somewhat simpler expression, unless you're doing AP or similar.
