Maximize a value for a semicircle of given radius
Last Updated :
07 Jul, 2022
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++
#include <bits/stdc++.h>
using namespace std;
double maximumValueOfF (int R)
{
return 4 * R * R + 0.25;
}
int main()
{
int R = 3;
printf("%.2f", maximumValueOfF(R));
return 0;
}
|
JAVA
import java.io.*;
class GFG
{
static double maximumValueOfF (int R)
{
return 4 * R * R + 0.25;
}
public static void main (String[] args)
{
int R = 3;
System.out.println(maximumValueOfF(R));
}
}
|
Python3
def maximumValueOfF (R):
return 4 * R * R + 0.25
R = 3
print(maximumValueOfF(R))
|
C#
using System;
class GFG
{
static double maximumValueOfF (int R)
{
return 4 * R * R + 0.25;
}
public static void Main ()
{
int R = 3;
Console.WriteLine(maximumValueOfF(R));
}
}
|
PHP
<?php
function maximumValueOfF ($R)
{
return 4 * $R * $R + 0.25;
}
$R = 3;
echo maximumValueOfF($R);
?>
|
Javascript
<script>
function maximumValueOfF(R) {
return 4 * R * R + 0.25;
}
var R = 3;
document.write(maximumValueOfF(R));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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