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Maximize volume of cuboid with given sum of sides

We are given the sum of length, breadth and height, say S, of a cuboid. The task is to find the maximum volume that can be achieved so that sum of side is S.
Volume of a cuboid = length * breadth * height
Examples :

Input : s  = 4
Output : 2
Only possible dimensions are some combination of 1, 1, 2.

Input : s = 8
Output : 18
All possible edge dimensions:
[1, 1, 6], volume = 6
[1, 2, 5], volume = 10
[1, 3, 4], volume = 12
[2, 2, 4], volume = 16
[2, 3, 3], volume = 18



Method 1: (Brute Force)
The idea to run three nested, one for length, one for breadth and one for height. For each iteration, calculate the volume and compare with maximum volume.

Below is the implementation of this approach:

C++

#include <bits/stdc++.h>
using namespace std;
  
// Return the maximum volume.
int maxvolume(int s)
{
    int maxvalue = 0;
  
    // for length
    for (int i = 1; i <= s - 2; i++) {
  
        // for breadth
        for (int j = 1; j <= s - 1; j++) {
  
            // for height
            int k = s - i - j;
  
            // calculating maximum volume.
            maxvalue = max(maxvalue, i * j * k);
        }
    }
  
    return maxvalue;
}
  
// Driven Program
int main()
{
    int s = 8;
    cout << maxvolume(s) << endl;
    return 0;
}

Java

// Java code to Maximize volume of 
// cuboid with given sum of sides
  
class GFG
{
      
    // Return the maximum volume.
    static int maxvolume(int s)
    {
        int maxvalue = 0;
      
        // for length
        for (int i = 1; i <= s - 2; i++)
        {
      
            // for breadth
            for (int j = 1; j <= s - 1; j++)
            {
      
                // for height
                int k = s - i - j;
      
                // calculating maximum volume.
                maxvalue = Math.max(maxvalue, i * j * k);
            }
        }
      
        return maxvalue;
    }
    // Driver function
    public static void main (String[] args)
    {
        int s = 8;
        System.out.println(maxvolume(s));
    }
}
  
// This code is contributed by Anant Agarwal.

/div>

Python3

# Python3 code to Maximize volume of 
# cuboid with given sum of sides
  
# Return the maximum volume.
def maxvolume (s):
    maxvalue = 0
  
    # for length
    i = 1
    for i in range(s - 1):
        j = 1
          
        # for breadth
        for j in range(s):
              
            # for height
            k = s - i - j
              
            # calculating maximum volume.
            maxvalue = max(maxvalue, i * j * k)
              
    return maxvalue
      
# Driven Program
s = 8
print(maxvolume(s))
  
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# code to Maximize volume of 
// cuboid with given sum of sides
using System;
  
class GFG
{
      
    // Return the maximum volume.
    static int maxvolume(int s)
    {
        int maxvalue = 0;
      
        // for length
        for (int i = 1; i <= s - 2; i++)
        {
      
            // for breadth
            for (int j = 1; j <= s - 1; j++)
            {
      
                // for height
                int k = s - i - j;
      
                // calculating maximum volume.
                maxvalue = Math.Max(maxvalue, i * j * k);
            }
        }
      
        return maxvalue;
    }
      
      
    // Driver function
    public static void Main ()
    {
        int s = 8;
        Console.WriteLine(maxvolume(s));
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP code to Maximize volume of 
// cuboid with given sum of sides
  
// Return the maximum volume.
function maxvolume( $s)
{
    $maxvalue = 0;
  
    // for length
    for ( $i = 1; $i <= $s - 2; $i++) 
    {
  
        // for breadth
        for ( $j = 1; $j <= $s - 1; $j++)
        {
  
            // for height
            $k = $s - $i - $j;
  
            // calculating maximum volume.
            $maxvalue = max($maxvalue
                            $i * $j * $k);
        }
    }
  
    return $maxvalue;
}
  
// Driver Code
$s = 8;
echo(maxvolume($s));
  
// This code is contributed by vt_m.
?>


Output :

18

Time Complexity: O(n2)

 

Method 2: (Efficient approach)
The idea is to divide edges as equally as possible.
So,
length = floor(s/3)
width = floor((s – length)/2) = floor((s – floor(s/3)/2)
height = s – length – width = s – floor(s/3) – floor((s – floor(s/3))/2)

Below is the implementation of this approach:

C++

#include <bits/stdc++.h>
using namespace std;
  
// Return the maximum volume.
int maxvolume(int s)
{
    // finding length
    int length = s / 3;
  
    s -= length;
  
    // finding breadth
    int breadth = s / 2;
  
    // finding height
    int height = s - breadth;
  
    return length * breadth * height;
}
  
// Driven Program
int main()
{
    int s = 8;
    cout << maxvolume(s) << endl;
    return 0;
}

Java

// Java code to Maximize volume of
// cuboid with given sum of sides
import java.io.*;
  
class GFG 
{
    // Return the maximum volume.
    static int maxvolume(int s)
    {
        // finding length
        int length = s / 3;
      
        s -= length;
      
        // finding breadth
        int breadth = s / 2;
      
        // finding height
        int height = s - breadth;
      
        return length * breadth * height;
    }
      
    // Driven Program
    public static void main (String[] args) 
    {
        int s = 8;
        System.out.println ( maxvolume(s));
                  
    }
}
  
// This code is contributed by vt_m.

Python3

# Python3 code to Maximize volume of
# cuboid with given sum of sides
  
# Return the maximum volume.
def maxvolume( s ):
  
    # finding length
    length = int(s / 3)
      
    s -= length
      
    # finding breadth
    breadth = s / 2
      
    # finding height
    height = s - breadth
      
    return int(length * breadth * height)
      
# Driven Program
s = 8
print( maxvolume(s) )
  
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# code to Maximize volume of
// cuboid with given sum of sides
using System;
  
class GFG 
{
    // Return the maximum volume.
    static int maxvolume(int s)
    {
        // finding length
        int length = s / 3;
      
        s -= length;
      
        // finding breadth
        int breadth = s / 2;
      
        // finding height
        int height = s - breadth;
      
        return length * breadth * height;
    }
      
    // Driven Program
    public static void Main () 
    {
        int s = 8;
        Console.WriteLine( maxvolume(s));
                  
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// Return the maximum volume.
function maxvolume($s)
{
    // finding length
    $length = (int)($s / 3);
  
    $s -= $length;
  
    // finding breadth
    $breadth = (int)($s / 2);
  
    // finding height
    $height = $s - $breadth;
  
    return $length * $breadth * $height;
}
  
// Driven Code
$s = 8;
echo(maxvolume($s));
  
// This code is contributed by Ajit.
?>


Output :

18

Time Complexity: O(1)

How does this work?

We basically need to maximize product of
three numbers, x, y and z whose sum is given.

Given s = x + y + z
Maximize P = x * y * z
= x * y * (s – x – y)
= x*y*s – x*x*s – x*y*y

We get dp/dx = sy – 2xy – y*y
and dp/dy = sx – 2xy – x*x

We get dp/dx = 0 and dp/dy = 0 when
x = s/3, y = s/3

So z = s – x – y = s/3



This article is attributed to GeeksforGeeks.org

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