从给定的周长和面积中找出长方体的最大体积

给定周长 P 和面积 A，任务是根据给定的周长和表面积计算可以制成长方体形式的最大体积。

例子：

Input: P = 24, A = 24
Output: 8

Input: P = 20, A = 14
Outpu: 3

方法：对于给定的长方体周长，我们有 P = 4(l+b+h) —(i)，
对于给定的长方体面积，我们有 A = 2 (lb+bh+lh) —(ii)。
长方体的体积是 V = lbh
体积取决于 3 个变量 l、b、h。让我们让它只依赖于长度。

as V = lbh,
=> V = l (A/2-(lb+lh)) {from equation (ii)}
=> V = lA/2 – l²(b+h)
=> V = lA/2 – l²(P/4-l) {from equation (i)}
=> V = lA/2 – l²P/4 + l³ —-(iii)
Now differentiate V w.r.t l for finding maximum of volume.
dV/dl = A/2 – lP/2 + 3l²
After solving the quadratic in l we have l = (P – (P²-24A)^1/2) / 12
Substituting value of l in (iii), we can easily find the maximum volume.

编程需要懂一点英语

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// function to return maximum volume
float maxVol(float P, float A)
{
    // calculate length
    float l = (P - sqrt(P * P - 24 * A)) / 12;
 
    // calculate volume
    float V = l * (A / 2.0 - l * (P / 4.0 - l));
 
    // return result
    return V;
}
 
// Driver code
int main()
{
    float P = 20, A = 16;
   
    // Function call
    cout << maxVol(P, A);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
class Geeks {
 
    // function to return maximum volume
    static float maxVol(float P, float A)
    {
        // calculate length
        float l
            = (float)(P - Math.sqrt(P * P - 24 * A)) / 12;
 
        // calculate volume
        float V
            = (float)(l * (A / 2.0 - l * (P / 4.0 - l)));
 
        // return result
        return V;
    }
 
    // Driver code
    public static void main(String args[])
    {
        float P = 20, A = 16;
       
        // Function call
        System.out.println(maxVol(P, A));
    }
}
 
// This code is contributed by Kirti_Mangal

Python3

# Python3 implementation of the
# above approach
from math import sqrt
 
# function to return maximum volume
 
 
def maxVol(P, A):
 
    # calculate length
    l = (P - sqrt(P * P - 24 * A)) / 12
 
    # calculate volume
    V = l * (A / 2.0 - l * (P / 4.0 - l))
 
    # return result
    return V
 
 
# Driver code
if __name__ == '__main__':
    P = 20
    A = 16
     
    # Function call
    print(maxVol(P, A))
 
# This code is contributed
# by Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
 
class GFG {
 
    // function to return maximum volume
    static float maxVol(float P, float A)
    {
        // calculate length
        float l
            = (float)(P - Math.Sqrt(P * P - 24 * A)) / 12;
 
        // calculate volume
        float V
            = (float)(l * (A / 2.0 - l * (P / 4.0 - l)));
 
        // return result
        return V;
    }
 
    // Driver code
    public static void Main()
    {
        float P = 20, A = 16;
        
        // Function call
        Console.WriteLine(maxVol(P, A));
    }
}
 
// This code is contributed
// by Akanksha Rai

PHP

Javascript

输出

4.14815