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Largest right circular cone that can be inscribed within a sphere

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  • Last Updated : 17 Mar, 2021

Given sphere of radius r  . The task is to find the radius of base and height of the largest right circular cone that can be inscribed within it.
Examples
 

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

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

 

 

Approach:
Let the radius of the cone = r 
height of the cone = h
From the diagram it is clear that: 
x = √(R^2 – r^2) and h=x+R 
Now using these values we get, 
 

To maximize the volume of the cone(V): 
V = (πr2h)/3
From the diagram, 
V = (πr2R)/3 + πr2√(R2 – r2)/3
Taking first derivative of V with respect to r we get, 
dV/dr = 2\pi r(\sqrt(R^2 - r^2) + R)/3 - \frac{\pi r^3}{\sqrt(R^2 - r^2)}/3
Now, setting dV/dr = 0 we get, 
\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
Squaring both sides and solving we get, 
9r^4=8r^2 R^2
r=\frac{2\sqrt{2}R}{3}
since, h = R + √(R2 – r2)
Now calculating the second derivative we get 

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

Thus r=(2R√2)/3 is point of maxima 
So, h = 4R/3 
 

Below is the implementation of the above approach: 
 

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>

Output: 

r = 9.42809, h = 13.3333

 


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