Maximize a value for a semicircle of given radius

We are given a semi circle with radius R. We can take any point on the circumference let it be P.Now, from that point P draw two lines to the two sides of diameter. Let the lines be PQ and PS.
The task is to find the maximum value of expression PS² + PQ for a given R.

Examples :

Input : R = 1 
Output : 4.25  
(4*1^2 + 0.25) = 4.25

Input : R = 2
Output : 16.25 
(4 * 2^2 + 0.25)= 16.25

Let F = PS^2 + PQ. We know QS = 2R (Diameter of the semicircle)
-> We also know that triangle PQS will always be right angled triangle irrespective of the position of point P on the circumference of circle 

1.)QS^2 = PQ^2 + PS^2 (Pythagorean Theorem)

2.) Adding and Subtracting PQ on the RHS
     QS^2 = PQ^2 + PS^2 + PQ - PQ

3.) Since QS = 2R
   4*R^2 + PQ - PQ^2 = PS^2 + PQ 
=> 4*R^2 + PQ - PQ^2 = F

4.) Using the concept of maxima and minima 
differentiating F with respect to PQ and equating 
it to 0 to get the point of maxima for F i.e.
   1 - 2 * PQ = 0
 => PQ = 0.5

5.) Now F will be maximum at F = 4*R^2 + 0.25 

C++

// C++ program to find
// the maximum value of F
#include <bits/stdc++.h>
using namespace std;

// Function to find the
// maximum value of F
double maximumValueOfF (int R)
{
    // using the formula derived for
    // getting the maximum value of F
    return 4 * R * R + 0.25;
}

// Drivers code
int main()
{
    int R = 3;
    printf("%.2f", maximumValueOfF(R));
    return 0;
}

JAVA

// JAVA program to find
// the maximum value of F
import java.io.*;

class GFG
{

    // Function to find the
    // maximum value of F
    static double maximumValueOfF (int R)
    {

        // using the formula derived for
        // getting the maximum value of F
        return 4 * R * R + 0.25;
    }

    // Driver code
    public static void main (String[] args)
    {
        int R = 3;
        System.out.println(maximumValueOfF(R));
    }

}

// This code is contributed
// by anuj_67.

Python3

# python program to find
# the maximum value of F

# Function to find the
# maximum value of F
def maximumValueOfF (R):

    # using the formula derived for
    # getting the maximum value of F
    return 4 * R * R + 0.25


# Drivers code
R = 3
print(maximumValueOfF(R))

# This code is contributed by Sam007.

C#

// C# program to find the
// maximum value of F
using System;

class GFG
{

    // Function to find the
    // maximum value of F
    static double maximumValueOfF (int R)
    {

        // using the formula derived for
        // getting the maximum value of F
        return 4 * R * R + 0.25;
    }

    // Driver code
    public static void Main ()
    {
        int R = 3;
        Console.WriteLine(maximumValueOfF(R));
    }

}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to find the
// maximum value of F

// Function to find the
// maximum value of F
function maximumValueOfF ($R)
{

    // using the formula derived
    // for getting the maximum
    // value of F
    return 4 * $R * $R + 0.25;
}

// Drivers code
$R = 3;
echo maximumValueOfF($R);

// This code is contributed
// by anuj_67.
?>

Javascript

<script>
// javascript program to find the
// maximum value of F

 // Function to find the
 // maximum value of F

 function maximumValueOfF(R) {

   // using the formula derived for
   // getting the maximum value of F

    return 4 * R * R + 0.25;
  }

  // Driver code

  var R = 3;
  document.write(maximumValueOfF(R));

// This code is contributed by bunnyram19.
</script>

Output :

36.25

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