# A steel girder is taken to a 15ft wide corridor. At the end of the corridor there is a 90° turn, to a 9ft wide corridor. How long is the longest girder than can be turned in this corner?

Mar 22, 2017

Maximum girder length is $35.1$ ft

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{Partial distance along bottom" & x+15=OA \\ y & "total length along side" & y=OB \\ l & "Line Segment AB" & l=AB \\ alpha & angle " between AB and bottom} & \null\end{matrix}\right.$

All of $x$, $y$ and $l$ being strictly positive; $0 < \alpha < \frac{\pi}{2}$

As the girder is moved through the corner of the corridor, it will get stuck if it is ever longer than the line segment $l = A B$. Therefore the longest girder that will go around the corner is length of the shortest line segment $A B$; therefore we should minimise $l$

METHOD 1

Our aim is to find $l \left(x\right)$, (a function of a single variable, $x$) and to minimize $l$ wrt $x$ (equally we could the same with $y$ and we would get the same result). ie we want a critical point of $\frac{\mathrm{dl}}{\mathrm{dx}}$.

By Pythagoras:

$A {B}^{2} = O {A}^{2} + O {B}^{2}$
$\therefore {l}^{2} = {\left(15 + x\right)}^{2} + {y}^{2}$

By similar triangles:

$O B : O A = 10 : x$
$\therefore \frac{y}{15 + x} = \frac{10}{x}$
$\therefore y = \frac{10 \left(15 + x\right)}{x}$

Combining these results we get:

$\therefore {l}^{2} = {\left(15 + x\right)}^{2} + {\left(\frac{10 \left(15 + x\right)}{x}\right)}^{2}$

By taking the (positive) square root we have achieved our goal of getting $l$ a function of $x$ alone. We want to minimize l, but equally we can minimize $L = {l}^{2}$ to avoid needing to work with the square root; So:

$L = {\left(15 + x\right)}^{2} + \frac{100 {\left(15 + x\right)}^{2}}{x} ^ 2$
$\setminus \setminus = {\left(15 + x\right)}^{2} \left(1 + \frac{100}{x} ^ 2\right)$

Differentiating wrt $x$ and plying the product rule we get:

$\frac{\mathrm{dL}}{\mathrm{dx}} = {\left(15 + x\right)}^{2} \left(- \frac{200}{x} ^ 3\right) + 2 \left(15 + x\right) \left(1 + \frac{100}{x} ^ 2\right)$
$\text{ } = \left(15 + x\right) \left\{- \frac{200}{x} ^ 3 \left(15 + x\right) + 2 \left(1 + \frac{100}{x} ^ 2\right)\right\}$
$\text{ } = \left(15 + x\right) \left(- \frac{3000}{x} ^ 3 - \frac{200}{x} ^ 2 + 2 + \frac{200}{x} ^ 2\right)$
$\text{ } = \left(15 + x\right) \left(2 - \frac{3000}{x} ^ 3\right)$

At a max/min we have $\frac{\mathrm{dL}}{\mathrm{dx}} = 0 \implies \left(15 + x\right) \left(2 - \frac{3000}{x} ^ 3\right) = 0$

Either $15 + x = 0 = > x = - 15$. But $x > 0$
Or $\setminus \setminus \setminus \setminus \setminus \setminus 2 - \frac{3000}{x} ^ 3 = 0 \implies \frac{1500}{x} ^ 3 = 1 \implies x = {1500}^{\frac{1}{3}}$

Thus we have:

$x = 11.44714 \ldots$
$L = 1233.2326 \ldots$
$l \setminus = 35.11741 \ldots$

I won't show that this is minimum (as required) but a plot or the second derivative test will show this is the case.

METHOD 2

We can also use a max/min method using trigonometry, considering the angle $\alpha$, which invokes less algebra

From the large $\Delta$ we get;

$\cos \alpha = \frac{15 + x}{l}$
$\therefore l = \frac{15 + x}{\cos} \alpha$
$\therefore l = \left(15 + x\right) \sec \alpha$

From the small $\Delta$ we get;

$\tan \alpha = \frac{10}{x}$
$\therefore x = \frac{10}{\tan} \alpha$
$\therefore x = 10 \cot \alpha$

Combining these results we get:

$l = \left(15 + 10 \cot \alpha\right) \sec \alpha$
$\setminus = 15 \sec \alpha + 10 \cot \alpha \sec \alpha$
$\setminus = 15 \sec \alpha + 10 \cos \frac{\alpha}{\sin} \alpha \frac{1}{\cos} \alpha$
$\setminus = 15 \sec \alpha + \frac{10}{\sin} \alpha$
$\setminus = 15 \sec \alpha + 10 \csc \alpha$

Differentiating $l$ wrt $\alpha$ we get:

$\frac{\mathrm{dl}}{d \alpha} = 15 \sec \alpha \tan \alpha - 10 \csc \alpha \cot \alpha$

At a max/min we have (dl)/(d alpha) #

$\therefore 15 \sec \alpha \tan \alpha - 10 \csc \alpha \cot \alpha = 0$
$\therefore \frac{3}{\cos \alpha} \frac{\sin \alpha}{\cos \alpha} = \frac{2}{\sin \alpha} \frac{\cos \alpha}{\sin \alpha}$
$\therefore 3 {\sin}^{3} \alpha = 2 {\cos}^{3} \alpha$
$\therefore {\tan}^{3} \alpha = \frac{2}{3}$

From which we get:

$\alpha = 0.71802 {\ldots}^{c}$
$\alpha = 41.1398 {\ldots}^{o}$
$l \setminus = 35.1174 \ldots$, as with Method 1