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# Maximum area of rectangle possible with given perimeter

• Last Updated : 30 May, 2021

Given the perimeter of a rectangle, the task is to find the maximum area of a rectangle which can use n-unit length as its perimeter.

Note: Length and Breadth must be an integral value.

Example:

```Input: perimeter = 15
Output: Maximum Area = 12

Input: perimeter = 16
Output: Maximum Area = 16```

Approach: For area to be maximum of any rectangle the difference of length and breadth must be minimal. So, in such case the length must be ceil (perimeter / 4) and breadth will be be floor(perimeter /4). Hence the maximum area of a rectangle with given perimeter is equal to ceil(perimeter/4) * floor(perimeter/4).

Below is the implementation of the above approach:

## C++

 `// C++ to find maximum area rectangle``#include ``using` `namespace` `std;` `// Function to find max area``int` `maxArea(``float` `perimeter)``{``    ``int` `length = (``int``)``ceil``(perimeter / 4);``    ``int` `breadth = (``int``)``floor``(perimeter / 4);` `    ``// return area``    ``return` `length * breadth;``}` `// Driver code``int` `main()``{``    ``float` `n = 38;``    ``cout << ``"Maximum Area = "` `<< maxArea(n);` `    ``return` `0;``}`

## Java

 `//Java to find maximum area rectangle` `import` `java.io.*;` `class` `GFG {``// Function to find max area``static` `int` `maxArea(``float` `perimeter)``{``    ``int` `length = (``int``)Math.ceil(perimeter / ``4``);``    ``int` `breadth = (``int``)Math.floor(perimeter / ``4``);` `// return area``return` `length * breadth;``}` `// Driver code``    ` `    ``public` `static` `void` `main (String[] args) {` `        ``float` `n = ``38``;``        ``System.out.println(``"Maximum Area = "` `+``                ``maxArea(n));``        ` `    ``}``}`

## Python3

 `# Python3 program to find``# maximum area rectangle``from` `math ``import` `ceil, floor` `# Function to find max area``def` `maxArea(perimeter):``    ``length ``=` `int``(ceil(perimeter ``/` `4``))``    ``breadth ``=` `int``(floor(perimeter ``/` `4``))` `    ``# return area``    ``return` `length ``*` `breadth` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `38``    ``print``(``"Maximum Area ="``, maxArea(n))`

## C#

 `// C# to find maximum area rectangle``using` `System;` `class` `GFG``{``// Function to find max area``static` `int` `maxArea(``float` `perimeter)``{``    ``int` `length = (``int``)Math.Ceiling(perimeter / 4);``    ``int` `breadth = (``int``)Math.Floor(perimeter / 4);` `    ``// return area``    ``return` `length * breadth;``}` `// Driver code``public` `static` `void` `Main()``{``    ``float` `n = 38;``    ``Console.WriteLine(``"Maximum Area = "` `+``                             ``maxArea(n));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## PHP

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## Javascript

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Output:
`Maximum Area = 90`

Time Complexity: O(1)
Auxiliary Space: O(1)

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