Given a perimeter P and area A, the task is to calculate the maximum volume that can be made in form of cuboid from the given perimeter and surface area.
Examples :
Input: P = 24, A = 24
Output: 8
Input: P = 20, A = 14
Output: 3
Approach: For a given perimeter of cuboid we have P = 4(l+b+h) —(i),
for given area of cuboid we have A = 2 (lb+bh+lh) —(ii).
Volume of cuboid is V = lbh
Volume is dependent on 3 variables l, b, h. Lets make it dependent on only length.
as V = lbh,
=> V = l (A/2-(lb+lh)) {from equation (ii)}
=> V = lA/2 – l2(b+h)
=> V = lA/2 – l2(P/4-l) {from equation (i)}
=> V = lA/2 – l2P/4 + l3 —-(iii)
Now differentiate V w.r.t l for finding maximum of volume.
dV/dl = A/2 – lP/2 + 3l2
After solving the quadratic in l we have l = (P – (P2-24A)1/2) / 12
Substituting value of l in (iii), we can easily find the maximum volume.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// function to return maximum volume float maxVol( float P, float A)
{ // calculate length
float l = (P - sqrt (P * P - 24 * A)) / 12;
// calculate volume
float V = l * (A / 2.0 - l * (P / 4.0 - l));
// return result
return V;
} // Driver code int main()
{ float P = 20, A = 16;
// Function call
cout << maxVol(P, A);
return 0;
} |
// Java implementation of the above approach import java.util.*;
class Geeks {
// function to return maximum volume
static float maxVol( float P, float A)
{
// calculate length
float l
= ( float )(P - Math.sqrt(P * P - 24 * A)) / 12 ;
// calculate volume
float V
= ( float )(l * (A / 2.0 - l * (P / 4.0 - l)));
// return result
return V;
}
// Driver code
public static void main(String args[])
{
float P = 20 , A = 16 ;
// Function call
System.out.println(maxVol(P, A));
}
} // This code is contributed by Kirti_Mangal |
# Python3 implementation of the # above approach from math import sqrt
# function to return maximum volume def maxVol(P, A):
# calculate length
l = (P - sqrt(P * P - 24 * A)) / 12
# calculate volume
V = l * (A / 2.0 - l * (P / 4.0 - l))
# return result
return V
# Driver code if __name__ = = '__main__' :
P = 20
A = 16
# Function call
print (maxVol(P, A))
# This code is contributed # by Surendra_Gangwar |
// C# implementation of the above approach using System;
class GFG {
// function to return maximum volume
static float maxVol( float P, float A)
{
// calculate length
float l
= ( float )(P - Math.Sqrt(P * P - 24 * A)) / 12;
// calculate volume
float V
= ( float )(l * (A / 2.0 - l * (P / 4.0 - l)));
// return result
return V;
}
// Driver code
public static void Main()
{
float P = 20, A = 16;
// Function call
Console.WriteLine(maxVol(P, A));
}
} // This code is contributed // by Akanksha Rai |
<script> // javascript implementation of the above approach // function to return maximum volume function maxVol( P, A)
{ // calculate length
let l = (P - Math.sqrt(P * P - 24 * A)) / 12;
// calculate volume
let V = l * (A / 2.0 - l * (P / 4.0 - l));
// return result
return V;
} // Driver code let P = 20, A = 16;
// Function call
document.write(maxVol(P, A).toFixed(5));
// This code is contributed by aashish1995 </script> |
<?php // PHP implementation of the above approach // function to return maximum volume function maxVol( $P , $A )
{ // calculate length
$l = ( $P - sqrt( $P * $P - 24 * $A )) / 12;
// calculate volume
$V = $l * ( $A / 2.0 - $l *
( $P / 4.0 - $l ));
// return result
return $V ;
} // Driver code $P = 20;
$A = 16;
// Function call echo maxVol( $P , $A );
// This code is contributed by mits ?> |
4.14815
Time Complexity: O(logn) as sqrt function is being used, time complexity of sqrt is logn
Auxiliary Space: O(1)