Two hallways, one 8 feet wide and the other 6 feet wide, meet at right angles. Determine the length of the longest ladder that can be carried horizontally from one hallway into the other.

Answer 1

Make a sketch showing the ladder going around the corner and touching the side walls.
let the part of the ladder in the wider hallway be L1, and the other part of the ladder L2
You should see two similar right-angled triangles.
Let the angle be Ø so that in the larger triangle
sinØ = 8/L1 , and in the smaller
cosØ = 6/L2

L1 = 8/sinØ = 8(sinØ)^-1
L2 = 6/cosØ = 6(cosØ)^-1

L = L2 + L1
= 6(cosØ)^-1 + 8(sinØ)^-1
dL/dØ = -6(cosØ)^-2(-sinØ) - 8sinØ)^-2 (cosØ)
= 6sinØ/cos^2 Ø - 8cosØ/sin^2 Ø
= 0 for a max of L

6sinØ/cos^2 Ø = 8cosØ/sin^2 Ø
sin^3 Ø/cos^3 Ø = 8/6
tan^3 Ø ‚ = 4/3

tanØ = (4/3)^(1/3) = 1.10064...
Ø = 47.7429...°

L = 8/sinØ + 6/cosØ
= appr 19.73 ft