Maximize a value for a semicircle of given radius

• Difficulty Level : Medium
• Last Updated : 05 May, 2021

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 using namespace std; // Function to find the// maximum value of Fdouble maximumValueOfF (int R){    // using the formula derived for    // getting the maximum value of F    return 4 * R * R + 0.25;    }     // Drivers codeint main(){    int R = 3;    printf("%.2f", maximumValueOfF(R));    return 0;}

JAVA

 // JAVA program to find// the maximum value of Fimport 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 Fdef maximumValueOfF (R):         # using the formula derived for    # getting the maximum value of F    return 4 * R * R + 0.25       # Drivers codeR = 3print(maximumValueOfF(R)) # This code is contributed by Sam007.

C#

 // C# program to find the// maximum value of Fusing 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.



Javascript


Output :
36.25

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