Beautiful Question!

Hinzugefügt: 2026-05-05

[00:00:00]This is a nice problem in calculus related to optimization. We are given a situation where a long pipe has to be carried through a hallway that takes a right angle turn.

[00:00:11]The first part of the hallway is 18 ft wide, and after the turn, the hallway becomes narrower with width 12 ft. The pipe must always remain horizontal, meaning it cannot be bent, and our job is to find the maximum possible length of the pipe that can pass through this turn.

[00:00:29]To solve this, we need to think about the most critical position of the pipe.

[00:00:34]That is the position where it is just able to pass through the turn without getting stuck.

[00:00:39]That critical position happens when the pipe is simultaneously touching both walls of the hallway while turning, because any longer pipe would hit the walls and not pass through.

[00:00:51]So, to find the maximum length, imagine the pipe is touching both walls while turning.

[00:00:57]At that moment, the pipe makes an angle theta with the hallway like this. We break the pipe into two parts, one part in the first hallway, and the other in the second hallway.

[00:01:08]Let the total length be L, and we write it as L = L1 + L2, where L1 is the length of this part, and L2 is the length of this part.

[00:01:22]Now, we focus on each part separately.

[00:01:25]In the narrower hallway of width 12 ft, the pipe makes angle theta. So, using basic trigonometry, L1 becomes 12 / cosine theta, or 12 * secant theta.

[00:01:39]In the wider hallway of width 18 ft, the pipe makes the same angle. So, L2 becomes 18 / sine theta, or 18 * cosecant theta.

[00:01:51]So, now total length becomes L = 12 secant theta + 18 cosecant theta.

[00:01:59]So, we now have a function L in terms of theta.

[00:02:03]To find the maximum value of L, we simply take the derivative with respect to theta, and then set it equal to zero.

[00:02:11]The derivative of secant theta is secant * tangent, and the derivative of cosecant theta is - cosecant * cotangent. This is the derivative of L with respect to theta.

[00:02:25]Now, we set this derivative equal to zero to find the critical point. So, we get this equals this.

[00:02:32]Now, this cotangent is simply 1 over tangent. Then, secant is 1 over cosine, and cosecant is 1 over sine theta.

[00:02:42]Take all these to the left-hand side, and this 12 to the right-hand side. Now, sine over cosine is also equal to tangent theta. So, we now have tangent cubed = 18 over 12, or 3 over 2. Take cube root on both sides to get this. So, theta = tangent inverse of the cube root of 3 over 2, which is approximately 0.8528 radians.

[00:03:11]Now, let us check the behavior of L based on theta.

[00:03:16]As theta approaches zero, [snorts] the length becomes very large as the pipe becomes vertical. And as theta approaches 90°, again, the length becomes very large as the pipe becomes horizontal, which cannot pass through the hallway. So, this middle value of theta gives the maximum length of pipe that can easily pass through this hallway. So, we substitute this value of theta back into our length expression. This gives us L approximately 42.14 ft. So, the maximum length of the pipe that can pass through the corner is approximately 42.14 ft. Like, share, and subscribe. So good.