Find Maximum Volume of a Cuboid

Given two integers p and a, where,

• p represents the perimeter
• a represents the surface area of a cuboid, 

Find the maximum possible volume of the cuboid and return it rounded to 2 decimal places.

Note: It is guaranteed that a valid cuboid always exists for the given values.

Examples:

Input: p = 22, a = 15
Output: 3.02
Explanation: The maximum attainable volume of the cuboid is 3.02

Input: p = 20, a = 5
Output: 0.33
Explanation: The maximum attainable volume of the cuboid is 0.33

[Expected Approach] Using Mathematical Formula - O(1) Time and O(1) Space

Let the dimensions of the cuboid be l, b, and h. Then:

2(lb + bh + lh) = a ...(i)
4(l + b + h) = p ...(ii)

Now Volume, V = lbh

From equation (i):

lb + bh + lh = a/2
l(b + h) + bh = a/2
bh = a/2 - l(b + h)

Therefore, V = l(a/2 - l(b + h))

From equation (ii):
l + b + h = p/4
b + h = p/4 - l

Hence,
V = la/2 - l²(b + h)
V = la/2 - l²(p/4 - l)
V = la/2 - l²p/4 + l³ ...(iii)

Now differentiate V with respect to l:
dV/dl = a/2 - lp/2 + 3l²

Setting it equal to 0:
a/2 - lp/2 + 3l² = 0
6l² - pl + a = 0

Thus,
l = (p ± √(p² - 24a)) / 12

Taking the valid value:
l = (p - √(p² - 24a)) / 12

Finally, substitute this value of l into equation (iii) to obtain the maximum volume.

#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;

double maxVolume(int p, int a) {
    
    // Calculate the two equal dimensions of the cuboid.
    double x = (p - sqrt(1.0 * p * p - 24.0 * a)) / 12.0;
    
    // Calculate the third dimension.
    double y = p / 4.0 - 2.0 * x;
    
    return x * x * y;
}

int main() {
    int p1 = 22;
    int a1 = 15;
    cout << fixed << setprecision(2) << maxVolume(p1, a1) << "\n";
    
    int p2 = 20;
    int a2 = 5;
    cout << fixed << setprecision(2) << maxVolume(p2, a2) << "\n";
    
    return 0;
}

import java.lang.Math;
import java.text.DecimalFormat;

public class Main {
    public static double maxVolume(int p, int a) {
        // Calculate the two equal dimensions of the cuboid.
        double x = (p - Math.sqrt(1.0 * p * p - 24.0 * a)) / 12.0;
        // Calculate the third dimension.
        double y = p / 4.0 - 2.0 * x;
        return x * x * y;
    }
    public static void main(String[] args) {
        int p1 = 22;
        int a1 = 15;
        DecimalFormat df = new DecimalFormat("#.##");
        System.out.println(df.format(maxVolume(p1, a1)));
        int p2 = 20;
        int a2 = 5;
        System.out.println(df.format(maxVolume(p2, a2)));
    }
}

import math


def maxVolume(p, a):
    # Calculate the two equal dimensions of the cuboid.
    x = (p - math.sqrt(1.0 * p * p - 24.0 * a)) / 12.0
    # Calculate the third dimension.
    y = p / 4.0 - 2.0 * x
    return x * x * y


p1 = 22
a1 = 15
print(f"{maxVolume(p1, a1):.2f}")

p2 = 20
a2 = 5
print(f"{maxVolume(p2, a2):.2f}")

using System;

class Program
{
    static double maxVolume(int p, int a)
    {
        // Calculate the two equal dimensions of the cuboid.
        double x = (p - Math.Sqrt(1.0 * p * p - 24.0 * a)) / 12.0;
        // Calculate the third dimension.
        double y = p / 4.0 - 2.0 * x;
        return x * x * y;
    }
    static void Main()
    {
        int p1 = 22;
        int a1 = 15;
        Console.WriteLine(maxVolume(p1, a1).ToString("0.00"));
        int p2 = 20;
        int a2 = 5;
        Console.WriteLine(maxVolume(p2, a2).ToString("0.00"));
    }
}

function maxVolume(p, a) {
    // Calculate the two equal dimensions of the cuboid.
    let x = (p - Math.sqrt(1.0 * p * p - 24.0 * a)) / 12.0;
    // Calculate the third dimension.
    let y = p / 4.0 - 2.0 * x;
    return x * x * y;
}

let p1 = 22;
let a1 = 15;
console.log(maxVolume(p1, a1).toFixed(2));

let p2 = 20;
let a2 = 5;
console.log(maxVolume(p2, a2).toFixed(2));

3.02
0.33