可内接圆锥的最大右圆柱
给定一个内接高度为 h 和基础半径为 r 的圆锥体的右圆柱体。任务是找到圆柱体的最大可能体积。 例:
Input: r = 4, h = 8
Output: 119.087
Input: r = 5, h = 9
Output: 209.333
逼近:圆柱体的体积为 V = πr^2h 在这个问题中，首先根据圆锥体的高度和半径，利用相似的三角形推导出体积方程。一旦我们修改了体积方程，我们将得到体积的导数并求解最大值。 设 x 为圆柱体半径， y 为圆锥体顶部到内接圆柱体顶部的距离。因此，圆柱体的高度为h–y 内接圆柱体的体积为 V = πx^2(h-y) 。 我们用相似比的方法找到高度和半径的关系，h-yx。 y/x = h/r y = hx/r 将 y 的方程式代入体积方程式，V.
v = πx^2(h-y) v = πx^2(h-hx/r) v = πx^2h–πx^3h/r 现在，dv/dx = d(πx^2h–πx^3h/r)/dx 和设置 dV/dx = 0 我们得到， x = 0，2r/3 所以， x = 2r/3 和， y
以下是上述方法的实现:
C++
// 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 语言(一种计算机语言，尤用于创建网站)
// 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
# 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
// 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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
?>
java 描述语言
<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>
Output:
119.087
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