Trig Calculus Corner question

This is my final review question I can't figure out from the trigonometry unit, I would appreciate any and all help. This week should open a whole new can of problems when I start to learn integration :P . I thought I had a solid solution, then I hid my solution and redid the question twice and I came up with three different numbers. This is the one that makes the most sense to me, but I still don't understand one of the steps completely.

"John has to carry a long piece of wood horizontally around the corner from a hallway of width 2 m to a hallway of width 2.5 m. Assuming that the piece of wood has no width, what is the maximum length of this piece of wood (in meters)?"

This is my (possibly incorrect) solution:
http://i19.photobucket.com/albums/b1...Solution-1.jpg

I have a feeling this particular question is going to appear on my final. Are there any different/easier approaches? If so, I'd like to see them.

Hello, turillian!

Your set-up and differentiation is correct.


We have: .L .= .PQ .= .2.5·secθ + 2·cscθ

Then: .L' .= .2.5·secθ·tanθ - 2·cscθ·cotθ .= .0

. . . . . . . . . . . .1 . . sinθ . . . .1 . .cosθ
We have: .2.5·------·------ - 2·-----·------ .= .0
. . . . . . . . . . .cosθ .cosθ . . .sinθ .sinθ


. . Then: .2.5·sin³θ - 2·cos³θ .= .0 . . → . . 2.5·sin³θ .= .2·cos³θ

. - . . . . . .sin³θ. . . . .2
. . . And: .-------- .= .---- . . → . . tan³θ .= .0.8
. . . . . . . .cos³θ . . . 2.5

. . - . . - . . - . . - . . .____
. . Then: .tanθ .= .³√(0.8) .= .0.928317767

. . Hence: .θ .≈ .42.87°


Therefore: .L .= .2.5·sec(42.87°) + 2·csc(42.87°) .≈ .6.35 m