Hi,

Just wanting to check that I have taken the correct approach with this question. Its a two part question, the first part the standard longest rod/beam that can be carried horizontally through a hallway of $\displaystyle a$ meters by $\displaystyle b$ meters which is equal to:

$\displaystyle L_{1} = l_{1} + l_{2}$

$\displaystyle L = \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\fra c{3}{2}} $

Or $\displaystyle L = a\csc(\theta) + b\sec(\theta)$

Where $\displaystyle \theta = \tan^{-1}\sqrt[3]{a/b}$

The second part of the question which is the problem is...'If the height of both hallways in part (a) is c metres and if the beam need not be

carried horizontally, what is the length of longest beam that can be carried around the corner? [Hint: This is easy if you use the result of part (a).]

My apprach is, to say you lay the rod/beam on the ground at point between the hallways that allows for the longest rod. If you then tilt the rod upward from one corner toward the ceiling you can 'draw' the rod (ie. the longest rod will go from the floor to the cieling at the point where the intial beam was the longest ie. it forms a right angled triangle), so to calculate this length is pythagoras theorem

$\displaystyle L_{2} = \sqrt{L_{1}^2+c^2}$