Solving a Canal Turn Problem: Finding the Longest Barge

[allergic]

alright, so i have this problem, and i must have gone wrong somewhere, because i am now lost.
the problem is this:
you have a canal that makes a right angle turn. the width of the canal for the incoming leg is a, and the width for the outgoing is b. what is the longest narrow barge that can be moved around the turn?

we were given this hint:
let x be the angle formed by a line segment touching the inside corner and terminating at the outside walls of the canal; here the angle is formed by the line segment and the outside wall of the leg of width b. the length of the line segment is
L(x)=a sec x + b csc x
as x approaches zero or pi/2, the length of this line segment approaches positive infinity. the maximum length of a barge that can make the turn is the minimum value of L(x).

so this is what i did:
i said L'(x) = a sec x tan x - b csc x cot x and set that = 0
I ended up getting b/a = tan ^3 (x) and then, x = tan ^-1 [(b/a)^(1/3)]
that could be where i screwed up, but who knows. i then wanted to plug that all back into the L(x) equation. but there i am stuck, because if i make (b/a)^(1/3) = p, i end up with L = a sec (cot p) + b csc (cot p) and i have NO idea where to go from there.
anyone got any ideas?

 

[Tide]

Don't forget that

$$a^2 + b^2 = L^2$$

 

[HallsofIvy]

Don't forget that

$$a^2 + b^2 = L^2$$

Why would that be true? a and b are not legs of a right triangle in this problem.

allergic (to what? trig?), your result that tan³x= b/a looks good to me. That is, of course, the same as saying tan x= b^1/3/a^1/3.

Now, imagine a right triangle set up with legs of length b^1/3 (opposite angle x) and a^1/3 (next to angle x). The hypotenuse is given by $\sqrt{b^{2/3}+a^{2/3}}$. You can calculate sec x and csc x from that.

 

[allergic]

ah-ha! that would be it. i always forget that kinda stuff.
thanks

 

[Tide]

Halls,

Sorry about that - I was looking at the condition for the ship to just fit the corner then misread my own sketch!