A steel pipe is being carried down a hallway 9 ft wide.

$L^{2/3} = x^{2/3} + y^{2/3}$
 

 
$L^{2/3} = 9^{2/3} + 6^{2/3}$

$L = 21.07 ~ \text{ft}$

$a = 9\sec \theta$

$b = 6\csc \theta$
 

 

$L = a + b$

$L = 9\sec \theta + 6\csc \theta$
 

Differentiate and equate to zero for longest $L$.
$\dfrac{dL}{d\theta} = 9\sec \theta \tan \theta - 6\csc \theta \cot \theta = 0$

$9\sec \theta \tan \theta = 6\csc \theta \cot \theta$

$9 \cdot \dfrac{1}{\cos \theta} \cdot \dfrac{\sin \theta}{\cos \theta} = 6 \cdot \dfrac{1}{\sin \theta} \cdot \dfrac{\cos \theta}{\sin \theta}$

$\dfrac{\sin^3 \theta}{\cos^3 \theta} = \dfrac{6}{9}$

$\tan^3 \theta = \dfrac{2}{3}$

$\tan \theta = \dfrac{\sqrt[3]{2}}{\sqrt[3]{3}}$
 

 

Hence,
$\sec \theta = \dfrac{\sqrt{3^{2/3} + 2^{2/3}}}{\sqrt[3]{3}}$

$\csc \theta = \dfrac{\sqrt{3^{2/3} + 2^{2/3}}}{\sqrt[3]{2}}$

The longest $L$
$L = 9\left( \dfrac{\sqrt{3^{2/3} + 2^{2/3}}}{\sqrt[3]{3}} \right) + 6\left( \dfrac{\sqrt{3^{2/3} + 2^{2/3}}}{\sqrt[3]{2}} \right)$

$L = 21.07 ~ \text{ft}$