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

Un esempio pratico di ottimizzazione consiste nel determinare la lunghezza massima di una barra che può essere trasportata attorno a un angolo retto formato da un corridoio largo 3 metri e uno largo 2 metri, senza inclinarlo verticalmente.Per risolvere questo problema, immagina un segmento di linea che passa attraverso l'angolo interno e tocca le pareti esterne. Questo segmento rappresenta l'altezza disponibile a un angolo specifico.Questa lunghezza L è divisa in due componenti, L1 e L2, che possono essere scritte in termini di larghezze del corridoio e del seno e coseno dell'angolo.Sebbene l'obiettivo sia trovare la lunghezza massima, questa lunghezza è limitata dalla parte più stretta della curva.Quindi, differenzia la funzione di lunghezza per trovare dove la pendenza è zero, identificando il minimo spazio che funge da collo di bottiglia per l'asta.L'equazione risultante può essere risolta riscrivendo i termini secante e cosecante come seni e cosenosi. Successivamente, riorganizzando i termini ai lati opposti dell'equazione per raggruppare i seni e i cosini si ottiene un'espressione semplificata che coinvolge il cubo tangente.Sostituendo questo angolo nell'equazione originale di lunghezza, si ottiene la lunghezza massima dell'asta che può superare in sicurezza l'angolo.

problemi di ottimizzazione

Optimization Problems

I problemi di ottimizzazione spesso implicano l&#x2019;individuazione di valori massimi o minimi in presenza di vincoli specifici. Un esempio classico consiste nel determinare il tubo orizzontale pi&#xF9; lungo che possa essere fatto passare attorno a un angolo retto, dove un corridoio largo 3 m si congiunge a un corridoio largo 2 m. Questo scenario, frequente nella progettazione architettonica e nel trasporto industriale, pu&#xF2; essere compreso sul piano concettuale attraverso ragionamenti geometrici e trigonometrici.
Per visualizzare il problema, si immagini il tubo come un segmento rettilineo che tocca lo spigolo interno della svolta e si estende verso l&#x2019;esterno fino a toccare le pareti opposte di ciascun corridoio. La lunghezza complessiva del tubo dipende dal suo orientamento, definito dall&#x2019;angolo che esso forma con le pareti. Per un angolo assegnato, il tubo deve riuscire a superare simultaneamente entrambi i corridoi e la sua lunghezza risulta vincolata dal passaggio pi&#xF9; stretto della svolta che deve attraversare.
Anzich&#xE9; cercare direttamente la massima lunghezza ammissibile, il problema viene riformulato considerando il percorso di passaggio con lo spazio libero minimo possibile. Tale spazio libero minimo corrisponde alla configurazione pi&#xF9; restrittiva in cui il tubo riesce comunque a superare l&#x2019;angolo. Si applica quindi il calcolo differenziale per individuare questo punto critico, analizzando come la lunghezza totale del percorso vari al variare dell&#x2019;angolo. Sebbene i passaggi dettagliati richiedano la derivazione e le identit&#xE0; trigonometriche, l&#x2019;idea centrale &#xE8; individuare l&#x2019;angolo che determina lo spazio libero minimo, il quale a sua volta stabilisce la massima lunghezza consentita del tubo. Per trovare la lunghezza del tubo che funzioni per tutti gli angoli, si minimizza L(&#x3B8;). In questo modo si identifica il minimo tra i valori massimi possibili, cio&#xE8; la maggiore lunghezza del tubo che possa passare indipendentemente dall&#x2019;angolo con cui lo si introduce.
Questo metodo mostra come, in contesti di ottimizzazione vincolata, minimizzare una funzione&#x2014;anzich&#xE9; massimizzare direttamente la quantit&#xE0; di interesse&#x2014;possa condurre alla soluzione. Il risultato finale fornisce un valore preciso per il tubo pi&#xF9; lungo che possa superare l&#x2019;angolo senza inclinazione in verticale.

Problemi di ottimizzazione

I problemi di ottimizzazione spesso implicano l&#x2019;individuazione di valori massimi o minimi in presenza di vincoli specifici. Un esempio classico ...

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Calculus

Analisi matematica

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Applications of Differentiation

Applicazioni della differenziazione

KEY TERMS AND CONCEPTS

TERMINI E CONCETTI CHIAVE

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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second derivatives and the shape of a graph

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Using Second Derivatives to Read Curvature

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slant asymptotes

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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...

application of differentiation to business

application-of-differentiation-to-business

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Newton’s Method for Finding Roots

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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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