Question

A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner?

Short answer

The length of the longest pipe that can be carried horizontally around the corner is \(3{\left( {{2^{2/3}} + {3^{2/3}}} \right)^{3/2}}\;ft.\).

Step-by-step solution

Given data

The hallway before the turn is 9 ft. wide and after the turn is 6 ft. wide.

Maximization condition

The conditions for minima of a function is:

\({\frac{{df\left( x \right)}}{{dx}}_{x = {x_{max}}}} = 0\;\;\;\;\;.....\left( 1 \right)\)

Determining the maximum length of the pipe

Label the figure as follows:

The length of the pipe is as follows:

\(\begin{aligned}{c}l &= BC + CE\\ &= \frac{{AC}}{{\sin \theta }} + \frac{{CD}}{{\cos \theta }}\\ &= \frac{9}{{\sin \theta }} + \frac{6}{{\cos \theta }}\end{aligned}\)

From equation (1), the maximum length will be for:

\(\begin{aligned}{c}{\frac{{dl}}{{d\theta }}_{\theta = {\theta _{\max }}}} &= 0\\ - \frac{{9\cos {\theta _{\max }}}}{{{{\sin }^2}{\theta _{\max }}}} + \frac{{6\sin {\theta _{\max }}}}{{{{\cos }^2}{\theta _{\max }}}} &= 0\\{\tan ^3}{\theta _{\max }} = \frac{9}{6}\\\tan {\theta _{\max }} &= \sqrt(3){{\frac{3}{2}}}\end{aligned}\)

\(\sin {\theta _{\max }}\)and \(\cos {\theta _{\max }}\) can be obtained as follows:

\(\begin{aligned}{c}\sin {\theta _{\max }} &= \sqrt {\frac{{{{\tan }^2}{\theta _{\max }}}}{{1 + {{\tan }^2}{\theta _{\max }}}}} \\ &= \sqrt {\frac{{{{\left( {\frac{3}{2}} \right)}^{2/3}}}}{{1 + {{\left( {\frac{3}{2}} \right)}^{2/3}}}}} \\ &= \frac{{{3^{1/3}}}}{{\sqrt {{2^{2/3}} + {3^{2/3}}} }}\end{aligned}\)

\(\begin{aligned}{c}\cos {\theta _{\max }} &= \sqrt {\frac{{{{\cot }^2}{\theta _{\max }}}}{{1 + {{\cot }^2}{\theta _{\max }}}}} \\ &= \sqrt {\frac{{{{\left( {\frac{2}{3}} \right)}^{2/3}}}}{{1 + {{\left( {\frac{2}{3}} \right)}^{2/3}}}}} \\ &= \frac{{{2^{1/3}}}}{{\sqrt {{2^{2/3}} + {3^{2/3}}} }}\end{aligned}\)

Thus, the maximum length is:

\(\begin{aligned}{c}{l_{\max }} &= \frac{9}{{\sin {\theta _{\max }}}} + \frac{6}{{\cos {\theta _{\max }}}}\\ &= \frac{9}{{\frac{{{3^{1/3}}}}{{\sqrt {{2^{2/3}} + {3^{2/3}}} }}}} + \frac{6}{{\frac{{{2^{1/3}}}}{{\sqrt {{2^{2/3}} + {3^{2/3}}} }}}}\\ &= 3\sqrt {{2^{2/3}} + {3^{2/3}}} \left( {\frac{3}{{{3^{1/3}}}} + \frac{2}{{{2^{1/3}}}}} \right)\\ &= 3{\left( {{2^{2/3}} + {3^{2/3}}} \right)^{3/2}}\end{aligned}\)

The maximum length is \(3{\left( {{2^{2/3}} + {3^{2/3}}} \right)^{3/2}}\)ft.