Given a right circular cylinder which is inscribed in a cone of height h and base radius r. The task is to find the largest possible volume of the cylinder.
Examples:
Input: r = 4, h = 8
Output: 119.087
Input: r = 5, h = 9
Output: 209.333
Approach: The volume of a cylinder is V = ?r^2h
In this problem, first derive an equation for volume using similar triangles in terms of the height and radius of the cone. Once we have the modified the volume equation, we’ll take the derivative of the volume and solve for the largest value.
Let x be the radius of the cylinder and y be the distance from the top of the cone to the top of the inscribed cylinder. Therefore, the height of the cylinder is h – y
The volume of the inscribed cylinder is V = ?x^2(h-y).
We use the method of similar ratios to find a relationship between the height and radius, h-y and x.
y/x = h/r
y = hx/r
Substitute the equation for y into the equation for volume, V.
V = ?x^2(h-y)
V = ?x^2(h-hx/r)
V = ?x^2h – ?x^3h/r
now, dV/dx = d(?x^2h – ?x^3h/r)/dx
and setting dV/dx = 0
we get, x = 0, 2r/3
So, x = 2r/3
and, y = 2h/3
So, V = ?8r^2h/27
Below is the implementation of the above approach:
// C++ Program to find the biggest // right circular cylinder that can // be fit within a right circular cone #include <bits/stdc++.h> using namespace std;
// Function to find the biggest right circular cylinder float cyl( float r, float h)
{ // radius and height cannot be negative
if (r < 0 && h < 0)
return -1;
// radius of right circular cylinder
float R = (2 * r) / 3;
// height of right circular cylinder
float H = (2 * h) / 3;
// volume of right circular cylinder
float V = 3.14 * pow (R, 2) * H;
return V;
} // Driver code int main()
{ float r = 4, h = 8;
cout << cyl(r, h) << endl;
return 0;
} |
// Java Program to find the biggest // right circular cylinder that can // be fit within a right circular cone import java.io.*;
class GFG {
// Function to find the biggest right circular cylinder static double cyl( double r, double h)
{ // radius and height cannot be negative
if (r < 0 && h < 0 )
return - 1 ;
// radius of right circular cylinder
double R = ( 2 * r) / 3 ;
// height of right circular cylinder
double H = ( 2 * h) / 3 ;
// volume of right circular cylinder
double V = 3.14 * Math.pow(R, 2 ) * H;
return V;
} // Driver code public static void main (String[] args) {
double r = 4 , h = 8 ;
System.out.println (cyl(r, h));
}
//This code is contributed by ajit } |
# Python 3 Program to find the biggest # right circular cylinder that can # be fit within a right circular cone import math
# Function to find the biggest # right circular cylinder def cyl(r, h):
# radius and height cannot
# be negative
if (r < 0 and h < 0 ):
return - 1
# radius of right circular cylinder
R = ( 2 * r) / 3
# height of right circular cylinder
H = ( 2 * h) / 3
# volume of right circular cylinder
V = 3.14 * math. pow (R, 2 ) * H
return V
# Driver code r = 4 ; h = 8 ;
print (cyl(r, h), "\n" )
# This code is contributed # by Akanksha Rai |
// C# Program to find the biggest // right circular cylinder that // can be fit within a right circular cone using System;
class GFG
{ // Function to find the biggest // right circular cylinder static double cyl( double r, double h)
{ // radius and height cannot
// be negative
if (r < 0 && h < 0)
return -1;
// radius of right circular cylinder
double R = (2 * r) / 3;
// height of right circular cylinder
double H = (2 * h) / 3;
// volume of right circular cylinder
double V = 3.14 * Math.Pow(R, 2) * H;
return V;
} // Driver code static public void Main ()
{ double r = 4, h = 8;
Console.WriteLine(cyl(r, h));
} } // This code is contributed by jit_t |
<script> // javascript Program to find the biggest // right circular cylinder that can // be fit within a right circular cone // Function to find the biggest right circular cylinder function cyl(r , h)
{ // radius and height cannot be negative
if (r < 0 && h < 0)
return -1;
// radius of right circular cylinder
var R = (2 * r) / 3;
// height of right circular cylinder
var H = (2 * h) / 3;
// volume of right circular cylinder
var V = 3.14 * Math.pow(R, 2) * H;
return V;
} // Driver code var r = 4, h = 8;
document.write(cyl(r, h).toFixed(5)); // This code is contributed by shikhasingrajput </script> |
<?php // PHP Program to find the biggest // right circular cylinder that can // be fit within a right circular cone // Function to find the biggest // right circular cylinder function cyl( $r , $h )
{ // radius and height cannot
// be negative
if ( $r < 0 && $h < 0)
return -1;
// radius of right circular cylinder
$R = (int)(2 * $r ) / 3;
// height of right circular cylinder
$H = (int)(2 * $h ) / 3;
// volume of right circular cylinder
$V = 3.14 * pow( $R , 2) * $H ;
return $V ;
} // Driver code $r = 4;
$h = 8;
echo cyl( $r , $h );
// This code is contributed by ajit ?> |
119.087
Time Complexity: O(1)
Auxiliary Space: O(1)