# Find maximum volume of a cuboid from the given perimeter and area

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 = 24Output:8Input:P = 20, A = 14Output: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 – l^{2}(b+h)

=> V = lA/2 – l^{2}(P/4-l) {from equation (i)}

=> V = lA/2 – l^{2}P/4 + l^{3}—-(iii)Now differentiate

Vw.r.tlfor finding maximum of volume.

dV/dl = A/2 – lP/2 + 3l^{2}

After solving the quadratic inlwe havel = (P – (P^{2}-24A)) / 12

Substituting value oflin (iii), we can easily find the maximum volume.

Below is the implementation of the above approach:

## C++

`// 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; ` ` ` `cout << maxVol(P, A); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// 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` `; ` ` ` `System.out.println(maxVol(P, A)); ` `} ` `} ` ` ` `// This code is contributed by Kirti_Mangal ` |

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## Python3

`# 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` ` ` `print` `(maxVol(P, A)) ` ` ` `# This code is contributed ` `# by Surendra_Gangwar ` |

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## C#

`// 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; ` ` ` `Console.WriteLine(maxVol(P, A)); ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

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## PHP

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