Around Corner

Let $\theta$ be the (acute) angle the pipe makes with the horizontal. Dividing the pipe into two sections at the point of contact with the corner of the hallway, we see that the length $L$ of the pipe is $L = 12.8 \sec \theta + 5.4 \csc \theta$ . Differentiating with respect to $\theta$ gives us that $dL/d\theta = 12.8 \sec \theta \tan \theta - 5.4 \csc \theta \cot \theta$ , which equals $0$ when

$\dfrac{12.8 \sin \theta}{\cos^{2} \theta} = \dfrac{5.4 \cos \theta}{\sin^{2} \theta} \Longrightarrow \tan^{3} \theta = \dfrac{5.4}{12.8} = \dfrac{27}{64} \Longrightarrow \tan \theta = \dfrac{3}{4}$ .

As $L$ can be made as long as we'd like this extrema must represent a minimum. So as $\tan \theta = \dfrac{3}{4} \Longrightarrow \sec \theta = \dfrac{5}{4}, \csc \theta = \dfrac{5}{3}$ , the desired length of the longest pipe that can be carried around the corner is

$L = 12.8 \times \dfrac{5}{4} + 5.4 \times \dfrac{5}{3} = 16 + 9 = \boxed{25}$ .