Aumento porcentual en el volumen del cubo si un lado del cubo aumenta en un porcentaje dado

Aquí se presenta un cubo, cuyo lado se incrementa en un porcentaje dado. La tarea es encontrar el aumento porcentual en el volumen del cubo.
Ejemplos: 
 

Input: x = 10
Output: 33.1%

Input: x = 50
Output: 237.5%

Acercarse 
 

  • En un cubo todos los lados son iguales, entonces 
    largo = ancho = alto
  • sea ​​lado del cubo = a
  • incremento porcentual dado = x%
  • Entonces, el volumen antes del aumento = a^3
  • después del aumento, nuevo lado = a + ax/100
  • entonces, nuevo volumen = (a + ax/100)^3 = a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000
  • aumento de volumen = volumen nuevo – volumen antiguo = (a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000) – a^3 = (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000
  • entonces, aumento porcentual en volumen = (((ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000)/a^3) * 100 = ((x/100)^3 + 3x/ 100 + 3x^2/10000) * 100 = x^3/10000 + 3x + 3x^2/100


Below is the implementation of the above approach: 
 

C++

// C++ program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
 
#include <bits/stdc++.h>
using namespace std;
 
void newvol(double x)
{
    cout << "percentage increase "
         << "in the volume of the cube is "
         << pow(x, 3) / 10000 + 3 * x
                + (3 * pow(x, 2)) / 100
         << "%" << endl;
}
 
// Driver code
int main()
{
    double x = 10;
    newvol(x);
    return 0;
}

Java

// Java program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
import java.io.*;
 
class GFG
{
 
static void newvol(double x)
{
    System.out.print( "percentage increase "
    +"in the volume of the cube is "
        + (Math.pow(x, 3) / 10000 + 3 * x
                + (3 * Math.pow(x, 2)) / 100) );
                System.out.print("%");
}
 
// Driver code
public static void main (String[] args)
{
    double x = 10;
    newvol(x);
}
}
 
// This code is contributed by anuj_67..

Python3

# Python program to find percentage increase
# in the volume of the cube
# if a side of cube is increased
# by a given percentage
 
def newvol(x):
 
    print("percentage increase"
            "in the volume of the cube is ",
            ((x**(3)) / 10000 + 3 * x
                + (3 * (x**(2))) / 100),"%");
 
x = 10;
newvol(x);
 
# This code is contributed by PrinciRaj1992

C#

// C# program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
using System;
 
class GFG
{
 
static void newvol(double x)
{
    Console.Write( "percentage increase "
    +"in the volume of the cube is "
        + (Math.Pow(x, 3) / 10000 + 3 * x
                + (3 * Math.Pow(x, 2)) / 100) );
                Console.Write("%");
}
 
// Driver code
public static void Main ()
{
    double x = 10;
    newvol(x);
}
}
 
// This code is contributed by anuj_67..

Javascript

<script>
// javascript program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
function newvol( x)
{
    document.write("percentage increase "
         + "in the volume of the cube is "
         + (Math.pow(x, 3) / 10000 + 3 * x
                + (3 * Math.pow(x, 2)) / 100)
         + "%" );
}
 
// Driver code
    let x = 10;
    newvol(x);
     
// This code is contributed by gauravrajput1
</script>
Producción: 

percentage increase in the volume of the cube is 33.1%

 

Complejidad del tiempo: O(log(x))

Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.

Publicación traducida automáticamente

Artículo escrito por IshwarGupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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