g A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 ft wide into a hallway 4 ft wide. What is the longest length that the pipe can be

The length of the longest pipe that can be carried horizontally is given by the minimum value of the function of the length of the pipe.

The longest length that the pipe can be is approximately 16.648 feet.

Reasons:

The length of the pipe, L

Width of the first hallway = 8 ft.

The width of the second hallway = 4 ft.

We have;

(L - a)·sin(θ) = 8

a·cos(θ) = 4

Therefore;

L - a = 8·csc(θ)

a = 4·sec(θ)

L(θ) = 8·csc(θ) + 4·sec(θ)

At the minimum length, we have;

L'(θ) = 4·sec(θ)·tan(θ) - 8·csc(θ)·cot(θ) = 0

4·sec(θ)·tan(θ) = 8·csc(θ)·cot(θ)

Therefore;

[tex]\displaystyle \frac{sec(\theta) \cdot tan(\theta)}{csc(\theta) \cdot cot(\theta)} = \frac{sin(\theta) \cdot tan(\theta) \cdot tan(\theta) }{cos(\theta) } =tan^3(\theta) = \frac{8}{4}[/tex]

[tex]\displaystyle tan(\theta) =\sqrt[3]{\frac{8}{4} } ={\sqrt[3]{2} }[/tex]

θ = arctan([tex]\sqrt[3]{2}[/tex]) ≈ 51.56°

a = 4×sec(arctan([tex]\sqrt[3]{2}[/tex]))

a ≈ 6.434

L = 8·csc(θ) + a = 8 × csc(arctan([tex]\sqrt[3]{2}[/tex])) + 4×sec(arctan([tex]\sqrt[3]{2}[/tex])) ≈ 16.648

The longest length that the pipe can be, L ≈ 16.648 ft.

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The longest length of the pipe is an illustration of trigonometry functions

The longest length of the pipe is 16.64 ft

The widths of the hallways are given as:

[tex]\mathbf{w_1 = 8}[/tex]

[tex]\mathbf{w_2 = 4}[/tex]

See attachment for the diagram that represents the scenario

From the diagram, we have:

[tex]\mathbf{sin(\theta) = \frac 4x}[/tex]

[tex]\mathbf{cos(\theta) = \frac 8y}[/tex]

Make x and y the subject

[tex]\mathbf{x = \frac 4{sin(\theta)}}[/tex]

[tex]\mathbf{y = \frac 8{cos(\theta)}}[/tex]

Where, the total length (L) of the pipe is:

[tex]\mathbf{L = x + y}[/tex]

So, we have:

[tex]\mathbf{L = \frac 4{sin(\theta)} + \frac 8{cos(\theta)}}[/tex]

Differentiate

[tex]\mathbf{L' = -\frac{4cos(\theta)}{sin^2(\theta)} + \frac{8sin(\theta)}{cos^2(\theta)}}[/tex]

Set to 0

[tex]\mathbf{-\frac{4cos(\theta)}{sin^2(\theta)} + \frac{8sin(\theta)}{cos^2(\theta)} = 0}[/tex]

Rewrite as:

[tex]\mathbf{\frac{4cos(\theta)}{sin^2(\theta)} = \frac{8sin(\theta)}{cos^2(\theta)}}[/tex]

Divide through by 4

[tex]\mathbf{\frac{cos(\theta)}{sin^2(\theta)} = \frac{2sin(\theta)}{cos^2(\theta)}}[/tex]

Cross multiply

[tex]\mathbf{2sin^3(\theta) = cos^3(\theta)}[/tex]

Divide both sides by [tex]\mathbf{2cos^3(\theta)}[/tex]

[tex]\mathbf{tan^3(\theta) = \frac 12}[/tex]

Take cube roots of both sides

[tex]\mathbf{tan(\theta) = \sqrt[3]{\frac 12}}[/tex]

[tex]\mathbf{tan(\theta) = 0.7937}[/tex]

Take arc tan of both sides

[tex]\mathbf{\theta = tan^{-1}(0.7937)}[/tex]

[tex]\mathbf{\theta = 38.4}[/tex]

Substitute [tex]\mathbf{\theta = 38.4}[/tex] in [tex]\mathbf{L = \frac 4{sin(\theta)} + \frac 8{cos(\theta)}}[/tex]

[tex]\mathbf{L = \frac 4{sin(38.4)} + \frac 8{cos(38.4)}}[/tex]

Evaluate

[tex]\mathbf{L = 6.43+ 10.21}}[/tex]

[tex]\mathbf{L = 16.64}}[/tex]

Hence, the longest length of the pipe is 16.64 ft

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