最大化给定半径的半圆的值

我们得到了一个机会 半圆 半径 R 我们可以取圆周上的任意一点 P 从那时起 P 在直径的两边画两条线。别管台词了 PQ 附言 . 任务是找到表达式的最大值 附言 2. +PQ 在一定的时间内 R

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semi-circle

例如:

Input : R = 1 Output : 4.25  (4*1^2 + 0.25) = 4.25Input : R = 2Output : 16.25 (4 * 2^2 + 0.25)= 16.25

设F=PS^2+PQ。我们知道QS=2R(半圆的直径) -> 我们也知道 三角形PQS 永远都是 直角三角形 不考虑点P在圆周上的位置

1.)QS^2 = PQ^2 + PS^2 (Pythagorean Theorem)2.) Adding and Subtracting PQ on the RHS     QS^2 = PQ^2 + PS^2 + PQ - PQ3.) Since QS = 2R   4*R^2 + PQ - PQ^2 = PS^2 + PQ => 4*R^2 + PQ - PQ^2 = F4.) 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 = 0.55.) 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>


输出:

36.25

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