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.

semi-circle

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++

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// 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;
}

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JAVA

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

36.25


My Personal Notes arrow_drop_up

Maths is the language of nature

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Improved By : vt_m, Sam007



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