In the previous article, we have discussed Python Program for Volume of Pyramid

Given the sum of length, breadth, and height of a cuboid say **S** and the task is to get the maximum volume of a cuboid such that the sum of the side is **S**.

**Examples:**

**Example1:**

**Input:**

Given sum = 5

**Output:**

The maximum volume of a cuboid such that the sum of the side { 5 } = 4

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**Example2:**

**Input:**

Given sum = 11

**Output:**

The maximum volume of a cuboid such that the sum of the side { 11 } = 48

## Program for Maximize Volume of Cuboid with Given Sum of Sides in Python

Below are the ways to get the maximum volume of a cuboid such that the sum of the side is **S **in python:

### Method #1: Using For loop (Static Input)

**Approach:**

- Give the number (sum) as static input and store it in a variable.
- Create a function to say
**Maximum_volume()**which takes the given number as an argument and returns the maximum volume of a cuboid such that the sum of the side is the given sum. - Inside the function, take a variable to say
**maxim_vol**and initialize its value to 0. - Take another variable (for length) say
**a**and initialize its value to 1. - Loop till the given sum -1 using the for loop.
- Take another variable (for breadth) say
**b**and initialize its value to 1. - Loop till the given sum using another nested for loop.
- Subtract the above variables a, b values from the given sum and store it in a variable say
**c**(for height). - Multiply the above a, b, c values and store them in another variable.
- Get the maximum value from the above-initialized
**maxim_vol**and above multiplication result using the max() function and store it in another variable. - Return the above result which is the maximum volume of a cuboid.
- Pass the given number as an argument to the
**Maximum_volume****()**function and print it. - The Exit of the Program.

**Below is the implementation:**

# Create a function to say Maximum_volume() which takes the given number as an # argument and returns the maximum volume of a cuboid such that the sum of the side # is the given sum. def Maximum_volume(gvn_sum): # Inside the function, take a variable and initialize its value to 0. maxim_vol = 0 # Take another variable (for length) say a and initialize its value to 1. a = 1 # Loop till the given sum -1 using the for loop. for a in range(gvn_sum - 1): # Take another variable (for breadth) say b and initialize its value to 1. b = 1 # Loop till the given sum using the another nested for loop. for b in range(gvn_sum): # Subtract the above variables a, b values from the given sum and store # it in a variable say c. c = gvn_sum - a - b # Multiply the above a, b, c values and store them in another variable. mult = a * b * c # Get the maximum value from the above-initialized maxim_vol and above # multiplication result using the max() function and store it in # another variable. maxim_vol = max(maxim_vol, mult) # Return the above result which is the maximum volume of a cuboid. return maxim_vol # Give the number (sum) as static input and store it in a variable. gvn_sum = 5 # Pass the given number as an argument to the Maximum_volume() function # and print it. print("The maximum volume of a cuboid such that the sum of the side {", gvn_sum, "} = ", Maximum_volume( gvn_sum))

**Output:**

The maximum volume of a cuboid such that the sum of the side { 5 } = 4

### Method #2: Using For loop (User Input)

**Approach:**

- Give the number (sum) as user input using the int(input()) function and store it in a variable.
- Create a function to say
**Maximum_volume()**which takes the given number as an argument and returns the maximum volume of a cuboid such that the sum of the side is the given sum. - Inside the function, take a variable to say
**maxim_vol**and initialize its value to 0. - Take another variable (for length) say
**a**and initialize its value to 1. - Loop till the given sum -1 using the for loop.
- Take another variable (for breadth) say
**b**and initialize its value to 1. - Loop till the given sum using another nested for loop.
- Subtract the above variables a, b values from the given sum and store it in a variable say
**c**(for height). - Multiply the above a, b, c values and store them in another variable.
- Get the maximum value from the above-initialized
**maxim_vol**and above multiplication result using the max() function and store it in another variable. - Return the above result which is the maximum volume of a cuboid.
- Pass the given number as an argument to the
**Maximum_volume****()**function and print it. - The Exit of the Program.

**Below is the implementation:**

# Create a function to say Maximum_volume() which takes the given number as an # argument and returns the maximum volume of a cuboid such that the sum of the side # is the given sum. def Maximum_volume(gvn_sum): # Inside the function, take a variable and initialize its value to 0. maxim_vol = 0 # Take another variable (for length) say a and initialize its value to 1. a = 1 # Loop till the given sum -1 using the for loop. for a in range(gvn_sum - 1): # Take another variable (for breadth) say b and initialize its value to 1. b = 1 # Loop till the given sum using the another nested for loop. for b in range(gvn_sum): # Subtract the above variables a, b values from the given sum and store # it in a variable say c.(for height) c = gvn_sum - a - b # Multiply the above a, b, c values and store them in another variable. mult = a * b * c # Get the maximum value from the above-initialized maxim_vol and above # multiplication result using the max() function and store it in # another variable. maxim_vol = max(maxim_vol, mult) # Return the above result which is the maximum volume of a cuboid. return maxim_vol # Give the number (sum) as user input using the int(input()) function and # store it in a variable. gvn_sum = int(input("Enter some random number = ")) # Pass the given number as an argument to the Maximum_volume() function # and print it. print("The maximum volume of a cuboid such that the sum of the side {", gvn_sum, "} = ", Maximum_volume( gvn_sum))

**Output:**

Enter some random number = 11 The maximum volume of a cuboid such that the sum of the side { 11 } = 48