# 26-27 Horizontal rod entering into a room from a perpendicular corridor

**Problem 26**

A corridor 4 ft wide opens into a room 100 ft long and 32 ft wide, at the middle of one side. Find the length of the longest thin rod that can be carried horizontally into the room.

**Solution 26**

$a = 4 \sec \theta$

$b = 32 \csc \theta$

Total length of rod:

$L = a + b$

$L = 4 \sec \theta + 32 \csc \theta$

$\dfrac{dL}{d\theta} = 4 \sec \theta \tan \theta - 32 \csc \theta \cot \theta = 0$

$\sec \theta \tan \theta - 8 \csc \theta \cot \theta = 0$

$\sec \theta \tan \theta = 8 \csc \theta \cot \theta$

$\dfrac{1}{\cos \theta} \left( \dfrac{\sin \theta}{\cos \theta} \right) = 8 \left( \dfrac{1}{\sin \theta} \right) \left( \dfrac{\cos \theta}{\sin \theta} \right)$

$\dfrac{\sin^3 \theta}{\cos^3 \theta} = 8$

$\tan^3 \theta = 8$

$\tan \theta = 2$

$L = 4 \sec \theta + 32 \csc \theta$

$L = 4 \left( \dfrac{\sqrt{5}}{1} \right) + 32 \left( \dfrac{\sqrt{5}}{2} \right)$

$L = 20\sqrt{5} = 44.72 \, \text{ ft}$ *answer*

**Problem 27**

Solve Problem 26 if the room is 56 feet long.

**Solution 27**

From the figure:

$L = \sqrt{30^2 + 32^2}$

$L = 43.86 \,\, \text{ ft }$ *answer*