可以外接半径为 R 的球体的最小圆锥体积

给定半径为 R 的球体，任务是找出可以外接的圆锥体的最小体积。

例子：

输入： R = 10输出：圆锥体积 = 8373.33解释：圆锥半径 = 14.14，圆锥高度 = 40，圆锥体积 = $(1/3) * \pi * r^2 * h$所以，体积 = 8373.33输入： R = 4输出：锥体体积 = 535.89

方法：
我们已经给出了一个内接于锥体的半径为 R 的球体。我们需要找出圆锥体的半径和高度来找出圆锥体的体积。

1. 在三角形 AOE 和 ALC 中计算 sin(X) 即对于三角形 AOE $sin(X) = (R/H-R)$对于三角形 ALC $sin(X) = (r/\sqrt{r^2 + H^2})$
2. 现在，通过将两者相等，我们得到$H = ( (-2 (r^2) R) / (R^2 - r^2))$
3. 在 Volume 中插入 H 的值，即$V = (1/3) * (3.14) * (r^2) * (H)$并且音量最小$d(V)/dr = 0$ .
4. 从上面的等式我们得到$r = \sqrt{2} * R$并将这个值放入 H 我们得到$H = 4*R$
5. 因此，应用锥体和推杆的体积公式$r = \sqrt{2} * R$和$H = 4 * R$我们得到了想要的结果。

C++

// C++ program to find the minimum
// volume of the cone that can be 
// circumscribed about a sphere
// of radius R
#include
using namespace std;
 
// Function to find the volume
// of the cone
float Volume_of_cone(float R)
{
     
    // r = radius of cone
    // h = height of cone
    // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h)
    // we get radius of cone from the derivation
    // is root(2) times multiple of R
    // we get height of cone from the derivation
    // is 4 times multiple of R
    float V = (1 / 3.0) * (3.14) * (2 * ( R * R ) ) * (4 * R);
     
    return V;
}
     
// Driver code
int main()
{
    float R = 10.0;
    cout << Volume_of_cone(R);
}
     
// This code is contributed by Samarth

Java

// Java program to find the minimum
// volume of the cone that can be
// circumscribed about a sphere
// of radius R
import java.util.*;
 
class GFG{
 
// Function to find the volume
// of the cone
static double Volume_of_cone(double R)
{
     
    // r = radius of cone
    // h = height of cone
    // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h)
    // we get radius of cone from the derivation
    // is root(2) times multiple of R
    // we get height of cone from the derivation
    // is 4 times multiple of R
    double V = (double)((1 / 3.0) * (3.14) * (2 * (R * R)) *
                                                  (4 * R));
    return V;
}
     
// Driver code
public static void main(String[] args)
{
    double R = 10.0;
    System.out.print(Volume_of_cone(R));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to find the minimum
# Volume of the cone that can be circumscribed
# about a sphere of radius R
 
import math
 
# Function to find the volume
# of the cone
 
def Volume_of_cone(R):
 
    # r = radius of cone
    # h = height of cone
    # Volume of cone = (1 / 3) * (3.14) * (r**2) * (h)
    # we get radius of cone from the derivation
    # is root(2) times multiple of R
    # we get height of cone from the derivation
    # is 4 times multiple of R
     
    V = (1 / 3) * (3.14) * (2 * ( R**2 ) ) * (4 * R)
     
    return V
     
 
# Driver code
if __name__ == "__main__":
     
    R = 10
     
    print(Volume_of_cone(R))

C#

// C# program to find the minimum
// volume of the cone that can be
// circumscribed about a sphere
// of radius R
using System;
class GFG{
 
// Function to find the volume
// of the cone
static double Volume_of_cone(double R)
{
     
    // r = radius of cone
    // h = height of cone
    // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h)
    // we get radius of cone from the derivation
    // is root(2) times multiple of R
    // we get height of cone from the derivation
    // is 4 times multiple of R
    double V = (double)((1 / 3.0) * (3.14) *
                    (2 * (R * R)) * (4 * R));
    return V;
}
     
// Driver code
public static void Main()
{
    double R = 10.0;
    Console.Write(Volume_of_cone(R));
}
}
 
// This code is contributed by Nidhi_biet

Javascript

// Javascript program to find the minimum
// volume of the cone that can be
// circumscribed about a sphere
// of radius R
 
    // Function to find the volume
    // of the cone
    function Volume_of_cone( R)
    {
 
        // r = radius of cone
        // h = height of cone
        // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h)
        // we get radius of cone from the derivation
        // is root(2) times multiple of R
        // we get height of cone from the derivation
        // is 4 times multiple of R
        let V =  ((1 / 3.0) * (3.14) * (2 * (R * R)) * (4 * R));
        return V;
    }
 
    // Driver code
    let R = 10.0;
    document.write(Volume_of_cone(R));
 
// This code is contributed by 29AjayKumar

输出：

8373.333333333332