El cono circular recto más grande que se puede inscribir dentro de una esfera.

Dada esfera de radio $r$ . La tarea es encontrar el radio de la base y la altura del cono circular recto más grande que se puede inscribir dentro de él.
Ejemplos :

Input : R = 10
Output : r = 9.42809, h = 13.3333

Input : R = 25
Output : r = 23.5702, h = 33.3333

Enfoque :
Sea el radio del cono = r
altura del cono = h
Del diagrama es claro que:
x = √(R^2 – r^2) y h=x+R
Ahora usando estos valores obtenemos,

Para maximizar el volumen del cono (V):
V = (πr ² h)/3
Del diagrama,
V = (πr ² R)/3 + πr ² √(R ² – r ² )/3
Tomando la primera derivada de V con respecto a r obtenemos,
$dV/dr = 2\pi r(\sqrt(R^2 - r^2) + R)/3 - \frac{\pi r^3}{\sqrt(R^2 - r^2)}/3$
Ahora, haciendo dV/dr = 0 obtenemos,
$\frac{2\pi r}{3}[R+\sqrt(R^2-r^2)] =\frac{ \pi r^3}{3\sqrt(R^2 - r^2}$
$2[R+\sqrt(R^2-r^2)] =\frac{r^2}{\sqrt(R^2 - r^2)}$
$2R\sqrt(R^2 - r^2)}+2(R^2 - r^2)=r^2\scriptstyle\implies 2R\sqrt(R^2 - r^2)}=3r^2 - 2R^2$
Elevando al cuadrado ambos lados y resolviendo obtenemos,
$9r^4=8r^2 R^2$
$r=\frac{2\sqrt{2}R}{3}$
ya que, h = R + √(R ² – r ² )
Ahora calculando la segunda derivada obtenemos
$\text{[math]}$
Así r =(2R√2)/3 es el punto de máximos
Entonces, h = 4R/3

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ Program to find the biggest cone
// that can be inscribed within a sphere
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the radius of the cone
float coner(float R)
{
 
    // radius cannot be negative
    if (R < 0)
        return -1;
 
    // radius of the cone
    float r = (2 * sqrt(2) * R) / 3;
    return r;
}
 
// Function to find the height of the cone
float coneh(float R)
{
 
    // side cannot be negative
    if (R < 0)
        return -1;
 
    // height of the cone
    float h = (4 * R) / 3;
    return h;
}
 
// Driver code
int main()
{
    float R = 10;
 
    cout << "r = " << coner(R) << ", "
         << "h = " << coneh(R) << endl;
 
    return 0;
}

Java

// Java Program to find the biggest cone
// that can be inscribed within a sphere
import java.util.*;
import java.lang.*;
 
class GFG
{
// Function to find the radius
// of the cone
static float coner(float R)
{
    // radius cannot be negative
    if (R < 0)
        return -1;
 
    // radius of the cone
    float r = (float)(2 *
            Math.sqrt(2) * R) / 3;
    return r;
}
 
// Function to find the
// height of the cone
static float coneh(float R)
{
 
    // side cannot be negative
    if (R < 0)
        return -1;
 
    // height of the cone
    float h = (4 * R) / 3;
    return h;
}
 
// Driver code
public static void main(String args[])
{
    float R = 10;
 
    System.out.println("r = " + coner(R) +
                       ", " + "h = " + coneh(R));
}
}
 
// This code is contributed
// by Akanksha Rai

Python3

# Python 3 Program to find the biggest cone
# that can be inscribed within a sphere
import math
 
# Function to find the radius
# of the cone
def coner(R):
     
    # radius cannot be negative
    if (R < 0):
        return -1;
     
    # radius of the cone
    r = (2 * math.sqrt(2) * R) / 3
    return float(r)
 
# Function to find the height
# of the cone
def coneh(R):
     
    # side cannot be negative
    if (R < 0):
        return -1;
 
    # height of the cone
    h = (4 * R) / 3
    return float(h)
 
# Driver code
R = 10
print("r = " , coner(R) ,
      ", ", "h = " , coneh(R))
 
# This code is contributed
# by 29AjayKumar

C#

// C# Program to find the biggest cone
// that can be inscribed within a sphere
using System;
 
class GFG
{
// Function to find the radius
// of the cone
static float coner(float R)
{
    // radius cannot be negative
    if (R < 0)
        return -1;
 
    // radius of the cone
    float r = (float)(2 *
               Math.Sqrt(2) * R) / 3;
    return r;
}
 
// Function to find the
// height of the cone
static float coneh(float R)
{
 
    // side cannot be negative
    if (R < 0)
        return -1;
 
    // height of the cone
    float h = (4 * R) / 3;
    return h;
}
 
// Driver code
public static void Main()
{
    float R = 10;
 
    Console.WriteLine("r = " + coner(R) +
                      ", " + "h = " + coneh(R));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP Program to find the biggest
// cone that can be inscribed
// within a sphere
 
// Function to find the radius
// of the cone
function coner($R)
{
 
    // radius cannot be negative
    if ($R < 0)
        return -1;
 
    // radius of the cone
    $r = (2 * sqrt(2) * $R) / 3;
    return $r;
}
 
// Function to find the height
// of the cone
function coneh($R)
{
 
    // side cannot be negative
    if ($R < 0)
        return -1;
 
    // height of the cone
    $h = (4 * $R) / 3;
    return $h;
}
 
// Driver code
$R = 10;
 
echo ("r = ");
echo coner($R);
echo (", ");
echo ("h = ");
echo (coneh($R));
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>
// javascript Program to find the biggest cone
// that can be inscribed within a sphere
 
// Function to find the radius
// of the cone
function coner(R)
{
 
    // radius cannot be negative
    if (R < 0)
        return -1;
 
    // radius of the cone
    var r = (2 *
            Math.sqrt(2) * R) / 3;
    return r;
}
 
// Function to find the
// height of the cone
function coneh(R)
{
 
    // side cannot be negative
    if (R < 0)
        return -1;
 
    // height of the cone
    var h = (4 * R) / 3;
    return h;
}
 
// Driver code
var R = 10;
document.write("r = " + coner(R).toFixed(5) +
                   ", " + "h = " + coneh(R).toFixed(5));
 
// This code is contributed by 29AjayKumar
</script>

Producción:

r = 9.42809, h = 13.3333

Complejidad de tiempo : O(1)

Espacio Auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por IshwarGupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

C++

Java

Python3

C#

PHP

Javascript

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