Given here is a cube of side length a, which inscribes a cone which in turn inscribes a right circular cylinder. The task is to find the largest possible volume of this cylinder.
Examples:
Input: a = 5 Output: 232.593 Input: a = 8 Output: 952.699
Approach:
From the figure, it is very clear, height of cone, H = a and radius of the cone, R = a?2, please refer Largest cone that can be inscribed within a cube.
and, radius of the cylinder, r = 2R/3 and height of the cylinder, h = 2H/3, please refer Largest right circular cylinder that can be inscribed within a cone.
So, radius of cylinder with respect to cube, r = 2a?2/3 and height of cylinder with respect to cube, h = 2a/3.
So, volume of the cylinder, V = 16?a^3/27.
Below is the implementation of the above approach:
C++
// C++ Program to find the biggest right circular // cylinder that can be inscribed within a right // circular cone which in turn is inscribed // within a cube#include <bits/stdc++.h>using namespace std;// Function to find the biggest// right circular cylinderfloat cyl(float a){ // side cannot be negative if (a < 0) return -1; // radius of right circular cylinder float r = (2 * a * sqrt(2)) / 3; // height of right circular cylinder float h = (2 * a) / 3; // volume of right circular cylinder float V = 3.14 * pow(r, 2) * h; return V;}// Driver codeint main(){ float a = 5; cout << cyl(a) << endl; return 0;} |
Java
// Java Program to find the biggest right circular // cylinder that can be inscribed within a right // circular cone which in turn is inscribed // within a cubeimport java.lang.Math;class cfg {// Function to find the biggest// right circular cylinderstatic float cyl(float a){ // side cannot be negative if (a < 0) return -1; // radius of right circular cylinder float r = (2 * a *(float)(Math.sqrt (2)) / 3); // height of right circular cylinder float h = (2 * a) / 3; // volume of right circular cylinder float V =(3.14f *(float)(Math.pow(r, 2) * h)); return V;}// Driver codepublic static void main(String[] args){ float a = 5; System.out.println(cyl(a));}}// This code is contributed by Mukul Singh. |
Python3
# Python3 Program to find the biggest # right circular cylinder that can be # inscribed within a right circular # cone which in turn is inscribed # within a cubeimport math as mt# Function to find the biggest# right circular cylinderdef cyl(a): # side cannot be negative if (a < 0): return -1 # radius of right circular cylinder r = (2 * a * mt.sqrt(2)) / 3 # height of right circular cylinder h = (2 * a) / 3 # volume of right circular cylinder V = 3.14 * pow(r, 2) * h return V# Driver codea = 5print(cyl(a))# This code is contributed by# Mohit kumar 29 |
C#
// C# Program to find the biggest // right circular cylinder that can// be inscribed within a right circular// cone which in turn is inscribed // within a cube using System;class GFG { // Function to find the biggest // right circular cylinder static float cyl(float a) { // side cannot be negative if (a < 0) return -1; // radius of right circular cylinder float r = (2 * a * (float)(Math.Sqrt (2)) / 3); // height of right circular cylinder float h = (2 * a) / 3; // volume of right circular cylinder float V =(3.14f * (float)(Math.Pow(r, 2) * h)); return V; } // Driver code public static void Main() { float a = 5; Console.Write(cyl(a)); } } // This code is contributed by Rajput-Ji |
PHP
<?php// PHP Program to find the biggest right // circular cylinder that can be inscribed // within a right circular cone which in // turn is inscribed within a cube// Function to find the biggest// right circular cylinderfunction cyl( $a ){ // side cannot be negative if ($a < 0) return -1; // radius of right circular cylinder $r = (2 * $a * sqrt(2)) / 3; // height of right circular cylinder $h = (2 * $a) / 3; // volume of right circular cylinder $V = 3.14 * pow($r, 2) * $h; return $V;}// Driver code$a = 5;echo cyl($a);// This code is contributed by Mahadev99?> |
Javascript
<script>// javascript Program to find the biggest right circular // cylinder that can be inscribed within a right // circular cone which in turn is inscribed // within a cube// Function to find the biggest// right circular cylinderfunction cyl(a){ // side cannot be negative if (a < 0) return -1; // radius of right circular cylinder var r = (2 * a *(Math.sqrt (2)) / 3); // height of right circular cylinder var h = (2 * a) / 3; // volume of right circular cylinder var V =(3.14 *(Math.pow(r, 2) * h)); return V;}// Driver codevar a = 5;document.write(cyl(a).toFixed(5));// This code contributed by Princi Singh </script> |
Output:
232.593
Time Complexity: O(1)
Auxiliary Space: O(1)
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