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Maximize volume of cuboid with given sum of sides
• Last Updated : 19 Mar, 2021

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.`

## 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

 ``

## Javascript

 ``

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

 ``

## Javascript

 ``

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

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