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 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.

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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



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 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



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