可围绕大小为LxW的给定矩形外接的矩形的最大面积

给定一个尺寸为L和W的矩形。任务是找到可以围绕尺寸为L和W的给定矩形外接的矩形的最大面积。

例子：

Input: L = 10, W = 10
Output: 200

Input: L = 18, W = 12
Output: 450

方法：下面是给定的尺寸L和W的矩形EFGH 。我们必须找到包围ABCD矩形EFGH的ABCD矩形区域。

在上图中：
如果$\angle ABC = X$然后$\angle CGF = 90 - X$因为GCF是直角三角形。
所以，
$\angle HGD = 180 - \angle FGH - \angle CGF$
=> $\angle HGD = 180 - 90 - (90 - X)$
=> $\angle HGD = X$

相似地，
$\angle EHA = X$
$\angle FEB = X$

现在，矩形ABCD的面积由下式给出：

Area = AB * AD
Area = (AE + EB)*(AH + HD) …..(1)

According to the projection rule:
AE = L*sin(X)
EB = W*cos(X)
AH = L*cos(X)
HD = W*sin(X)

用上述等式(1)的值代替：

$Area = (AE + EB)*(AH + HD)$$Area = (L*sin(X)+ W*cox(X))*(L*cos(X) + W*sin(X))$$Area = ((L^{2} + W^{2})*sin(X)*cos(X) + L*W)$$Area = (\frac{(L^{2} + W^{2})*sin(2X)}{2} + L*W)$Now to maximize the area, the value of sin(2X) must be maximum i.e., 1.
Therefore after substituting sin(2X) as 1 we have,
$Area = (\frac{(L^{2} + W^{2})}{2} + L*W)$$Area = (\frac{(L+W)^{2}}{2})$

下面是上述方法的实现：

C++

// C++ program for the above approach 
#include  
using namespace std; 
  
// Function to find area of rectangle 
// inscribed another rectangle of 
// length L and width W 
double AreaofRectangle(int L, int W)
{
      
    // Area of rectangle 
    double area = (W + L) * (W + L) / 2;
      
    // Return the area 
    return area;
}
  
// Driver Code 
int main() 
{ 
      
    // Given dimensions 
    int L = 18;
    int W = 12;
      
    // Function call 
    cout << AreaofRectangle(L, W);
    return 0; 
} 
  
// This code is contributed by Princi Singh

Java

// Java program for the above approach 
import java.io.*;
import java.util.*; 
  
class GFG{
      
// Function to find area of rectangle 
// inscribed another rectangle of 
// length L and width W 
static double AreaofRectangle(int L, int W)
{
      
    // Area of rectangle 
    double area = (W + L) * (W + L) / 2;
      
    // Return the area 
    return area;
}
      
// Driver Code 
public static void main(String args[])
{ 
      
    // Given dimensions 
    int L = 18;
    int W = 12;
      
    // Function call 
    System.out.println(AreaofRectangle(L, W));
} 
} 
  
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
  
# Function to find area of rectangle 
# inscribed another rectangle of 
# length L and width W
def AreaofRectangle(L, W):
    
  # Area of rectangle
  area =(W + L)*(W + L)/2
  
# Return the area
  return area
  
# Driver Code
if __name__ == "__main__":
  
  # Given Dimensions
  L = 18
  W = 12
  
  # Function Call
  print(AreaofRectangle(L, W))

C#

// C# program for the above approach 
using System;
  
class GFG{
      
// Function to find area of rectangle 
// inscribed another rectangle of 
// length L and width W 
static double AreaofRectangle(int L, int W)
{
      
    // Area of rectangle 
    double area = (W + L) * (W + L) / 2;
      
    // Return the area 
    return area;
}
      
// Driver Code 
public static void Main(String []args)
{ 
      
    // Given dimensions 
    int L = 18;
    int W = 12;
      
    // Function call 
    Console.Write(AreaofRectangle(L, W));
} 
} 
  
// This code is contributed by shivanisinghss2110

输出：

450.0

时间复杂度： O(1)
辅助空间： O(1)