MHB What is the Longest Steel Pipe That Can Be Carried Horizontally Around a Corner?

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The discussion focuses on determining the longest steel pipe that can be carried horizontally around a corner formed by a 9 ft wide hallway leading into a 6 ft wide hallway. The formula provided, L(θ) = 9/sin(θ) + 6/cos(θ), calculates the length of the pipe but does not account for the corner's constraints. A critical point in the analysis involves finding the derivative of L(θ) and setting it to zero, leading to a solution of approximately -61.979. Substituting this value back into the length formula yields a maximum pipe length of about 21.0704 ft. The problem is noted as a university "Problem of the Week," with additional solutions available online.
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A steel pipe is being carried down a hallway $9\text{ ft}$ wide.
At the end of the hall is a right angled turn into a narrower hallway $6\text { ft}$ wide.
What is the length of the longest pipe that can be carried horizontally around the cornerView attachment 1673
assume that $\displaystyle L(\theta)=\frac{9}{\sin{\theta}} + \frac{6}{\cos{\theta}}$

will give length of pipe but this doesn't take in account the constraints of the hall corner?
 
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You may be interested to know that this problem was given as a university POTW, and a couple of solutions can be found here:

http://mathhelpboards.com/potw-university-students-34/problem-week-59-may-13th-2013-a-4764.html
 
$d/dx L(\theta)=0$ one ans is $-61.979$

then $L(-61.979)=21.0704$

this seems to be the same ans to all given $d/dx L(\theta)=0$ if it is abs

I did look at the solutions to POTW
 
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