Name: pipe length around a corner
Uploaded: 2026-01-12T12:24:22.000Z
Duration: 1 min 26 s
Channel: JoVE

Un ejemplo práctico de optimización consiste en determinar la longitud máxima de una varilla que puede llevarse alrededor de una esquina en ángulo recto formada por un pasillo de 3 metros de ancho y otro de 2 metros de ancho, sin inclinarla verticalmente.Para resolver esto, imagina un segmento de línea que pasa por la esquina interior y toca las paredes exteriores. Este segmento representa el espacio libre disponible en un ángulo específico.Esta longitud L se divide en dos componentes, L1 y L2, que pueden escribirse en términos de los anchos del pasillo y el seno y coseno del ángulo.Aunque el objetivo es encontrar la longitud máxima, esta longitud está limitada por la parte más cerrada de la curva.Así que se deriva la función de longitud para encontrar dónde la pendiente es cero, identificando el espacio libre mínimo que actúa como cuello de botella para la barra.La ecuación resultante puede resolverse reescribiendo los términos secante y cosecante como senos y cosenos. A continuación, reorganizando los términos a lados opuestos de la ecuación para agrupar los senos y cosenos da una expresión simplificada que involucra el cubo tangente.Sustituir este ángulo de nuevo en la ecuación original de longitud proporciona la longitud máxima de la varilla que puede superar la esquina con seguridad.

problemas de optimización

Optimization Problems

Los problemas de optimizaci&#xF3;n consisten en determinar valores m&#xE1;ximos o m&#xED;nimos bajo restricciones espec&#xED;ficas. Un ejemplo conocido es determinar la tuber&#xED;a horizontal m&#xE1;s larga que puede moverse alrededor de una esquina en &#xE1;ngulo recto, donde un pasillo de 3 metros de ancho se encuentra con otro de 2 metros. Este escenario, com&#xFA;n en el dise&#xF1;o arquitect&#xF3;nico y el transporte industrial, puede comprenderse conceptualmente mediante razonamiento geom&#xE9;trico y trigonom&#xE9;trico.
Para visualizar el problema, considere la tuber&#xED;a como una l&#xED;nea recta que toca la esquina interior del &#xE1;ngulo y se extiende hacia afuera hasta tocar las paredes opuestas de cada pasillo. La longitud total de la tuber&#xED;a depende de su orientaci&#xF3;n, definida por el &#xE1;ngulo que forma con las paredes. Para cualquier &#xE1;ngulo dado, la tuber&#xED;a debe atravesar simult&#xE1;neamente ambos pasillos, y su longitud queda limitada por la secci&#xF3;n m&#xE1;s estrecha de la esquina que atraviesa.
En lugar de intentar determinar directamente la longitud m&#xE1;xima posible, el problema se reformula considerando el espacio libre m&#xE1;s corto que puede tomar la tuber&#xED;a. Este espacio libre m&#xED;nimo corresponde a la posici&#xF3;n m&#xE1;s restrictiva en la que la tuber&#xED;a a&#xFA;n puede atravesar la esquina. Se aplica el c&#xE1;lculo para identificar este punto cr&#xED;tico analizando c&#xF3;mo cambia la longitud total de la trayectoria con el &#xE1;ngulo. Si bien los pasos detallados implican diferenciaci&#xF3;n e identidades trigonom&#xE9;tricas, la idea central es localizar el &#xE1;ngulo que produce la menor holgura, lo que a su vez determina la longitud m&#xE1;xima permitida de la tuber&#xED;a. Para encontrar la longitud de tuber&#xED;a que es v&#xE1;lida para todos los &#xE1;ngulos, minimizamos L(&#x3B8;). Esto garantiza la identificaci&#xF3;n del m&#xED;nimo de las longitudes m&#xE1;s largas posibles, es decir, la longitud m&#xE1;xima de tuber&#xED;a que se ajusta independientemente del &#xE1;ngulo de aproximaci&#xF3;n.
Este enfoque ilustra c&#xF3;mo minimizar una funci&#xF3;n, en lugar de maximizar directamente la magnitud de inter&#xE9;s, puede proporcionar una soluci&#xF3;n en entornos de optimizaci&#xF3;n con restricciones. El resultado final proporciona un valor preciso para la tuber&#xED;a m&#xE1;s larga que puede sortear la esquina correctamente sin inclinarla en vertical.

Problemas de optimización

Los problemas de optimizaci&#xF3;n consisten en determinar valores m&#xE1;ximos o m&#xED;nimos bajo restricciones espec&#xED;ficas. Un ejemplo conocido es ...

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Calculus

Cálculo

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Unidad

Applications of Differentiation

Aplicaciones de la derivación

KEY TERMS AND CONCEPTS

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pipe length around a corner

Pipe length around a corner is a classic constrained optimization problem. It shows how calculus can find a maximum or minimum value when space is limited by fixed walls and angles.

The example used here asks for the longest horizontal pipe that can move around a right-angled turn. One hallway is 3 meters wide and the other is 2 meters wide. The pipe is shown as a straight line that touches the inner corner and reaches the opposite walls of both hallways.

The key idea is that the pipe’s length changes with its angle. For each angle, the pipe must still clear both hallways. That means the tightest part of the corner limits how long the pipe can be.

Instead of trying to maximize the pipe length directly, the problem is rewritten as a search for the shortest clearance path. The minimum clearance gives the most restrictive position where the pipe can still pass. Calculus then helps find this critical angle by looking at how the length function L(θ) changes with θ.

The solution uses differentiation and trigonometric identities, but the main idea is simple. Find the angle that gives the least clearance, and that angle sets the maximum pipe length. This approach shows how minimizing a function can solve a constrained optimization problem.

Pipe Length Around a Corner

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optimization problems

Pipe length around a corner is a classic constrained optimization problem. It shows how calculus can find a maximum or minimum value when space is limited by fixed walls and angles. The example used here asks for the longest horizontal pipe that can move around a right-angled turn. One hallway is 3 meters wide and the other is 2 meters wide. The pipe is shown as a straight line that touches the inner corner and reaches the opposite walls of both hallways. The key idea is that the pipe’s...

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key terms and concepts

Learn how calculus and trigonometry solve a pipe-length optimization problem by finding the longest horizontal pipe that can turn a 3 m by 2 m hallway corner.

Pipe Length Around a Corner | JoVE K12

The JoVE Core Calculus collection provides a comprehensive visual guide to the entire calculus sequence, seamlessly connecting mathematical theory to real-world applications in physics and economics. Through dynamic animated lessons, the course transitions smoothly from foundational single-variable concepts to advanced multivariable topics. Students will explore everything from standard differentiation and integration to spatial vectors, multi-dimensional derivatives, and complex vector fields. By breaking down abstract coordinate systems, infinite series, and spatial motion, this structured approach builds the robust analytical skills and advanced mastery essential for STEM success.

calculus

Pipe Length Around a Corner

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