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?

Answer 1

Maximum girder length is #35.1# ft

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 #x#, #y# and #l# being strictly positive; #0 lt alpha lt pi/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 #(dL)/dx=0 => (15+x)(2 -3000/x^3 ) = 0 #

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 #alpha#, which invokes less algebra

From the large #Delta# we get;

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

From the small #Delta# we get;

# 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 #l# wrt #alpha# we get:

# (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

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Answer 2

The longest girder that can be turned in the corner is the diagonal of the rectangle formed by the two corridors. Using the Pythagorean theorem, the length of the diagonal can be calculated as √(15^2 + 9^2) = √(225 + 81) = √306 ≈ 17.49 feet. Therefore, the longest girder that can be turned in the corner is approximately 17.49 feet long.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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