# 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?

##### 1 Answer

Maximum girder length is

#### Explanation:

Let us set up the following variables:

# {(x, "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",) :} #

All of

As the girder is moved through the corner of the corridor, it will get stuck if it is ever longer than the line segment **minimise**

**METHOD 1**

Our aim is to find

By Pythagoras:

# AB^2=OA^2+OB^2 #

# :. l^2=(15+x)^2+y^2 #

By similar triangles:

# OB : OA = 10 : x #

# :. y/(15+x) = 10/x #

# :. y = (10(15+x))/x #

Combining these results we get:

# :. l^2=(15+x)^2+((10(15+x))/x)^2 #

By taking the (positive) square root we have achieved our goal of getting

# L = (15+x)^2 + (100(15+x)^2)/x^2 #

# \ \ = (15+x)^2(1+100/x^2) #

Differentiating wrt

# (dL)/dx = (15+x)^2(-200/x^3)+2(15+x)(1+100/x^2) #

# " " = (15+x){-200/x^3 (15+x) + 2(1+100/x^2) }#

# " " = (15+x)( -3000/x^3 - 200/x^2 + 2+200/x^2 )#

# " " = (15+x)(2 -3000/x^3 )#

At a max/min we have

Either

# 15+x = 0 = > x=-15 # . But#x gt 0#

Or# \ \ \ \ \ \ 2 -3000/x^3 = 0 => 1500/x^3=1 => x=1500^(1/3)#

Thus we have:

# x = 11.44714 ... #

# L = 1233.2326... #

# l \ = 35.11741 ... #

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

From the large

# cos alpha = (15+x)/l #

# :. l = (15+x)/cos alpha #

# :. l = (15+x)sec alpha #

From the small

# tan alpha = 10/x #

# :. x = 10/tan alpha #

# :. x = 10cot alpha #

Combining these results we get:

# l = (15 + 10 cot alpha)sec alpha #

# \ = 15sec alpha + 10 cot alpha sec alpha #

# \ = 15sec alpha + 10 cos alpha /sin alpha 1/cos alpha #

# \ = 15sec alpha + 10 /sin alpha #

# \ = 15sec alpha + 10 csc alpha #

Differentiating

# (dl)/(d alpha) = 15sec alpha tan alpha - 10 csc alpha cot alpha #

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

# :. 15sec alpha tan alpha - 10 csc alpha cot alpha = 0 #

# :. 3/(cos alpha) (sin alpha)/(cos alpha) = 2/(sin alpha) (cos alpha)/(sin alpha) #

# :. 3sin^3alpha=2cos^3alpha #

# :. tan^3alpha=2/3 #

From which we get:

# alpha = 0.71802 ... ^c#

# alpha = 41.1398 ... ^o#

# l \ = 35.1174 ... # , as with Method 1