# Maximizing the cuboid volume with fixed perimeter

• Aug 14th 2012, 11:09 AM
andre.vignatti
Maximizing the cuboid volume with fixed perimeter
First, the 2D case is easy...

Q0: In a rectangle, we know that the sum of the dimensions (lenght and width) is a fixed $C$. Which are the dimensions values that maximize the area?

A: One dimension is $x$, and thus the other is $C-x$. Therefore, the area is $x \cdot (C-x) = xC - x^2$. Then, setting the derivative equal to zero would provide us the answer ( $C/2$ on both dimensions)

Okay... now I want to generalize this to the 3D case. More precisely:

Q1: In a cuboid (see en.wikipedia.org/wiki/Cuboid), we know that the sum of the dimensions (lenght, width and depth) is a fixed $C$. Which are the dimensions values that maximize the volume?

Finally, is there a generalization of the answer to Q1 for higher dimensions?
• Aug 14th 2012, 12:25 PM
TheEmptySet
Re: Maximizing the cuboid volume with fixed perimeter
Quote:

Originally Posted by andre.vignatti
First, the 2D case is easy...

Q0: In a rectangle, we know that the sum of the dimensions (lenght and width) is a fixed $C$. Which are the dimensions values that maximize the area?

A: One dimension is $x$, and thus the other is $C-x$. Therefore, the area is $x \cdot (C-x) = xC - x^2$. Then, setting the derivative equal to zero would provide us the answer ( $C/2$ on both dimensions)

Okay... now I want to generalize this to the 3D case. More precisely:

Q1: In a cuboid (see en.wikipedia.org/wiki/Cuboid), we know that the sum of the dimensions (lenght, width and depth) is a fixed $C$. Which are the dimensions values that maximize the volume?

Finally, is there a generalization of the answer to Q1 for higher dimensions?

The easiest way is to use Lagrange Multipliers. So the problem would be we want to maximize the volume

$V(x,y,z)=xyz$

subject to the constraint that

$x+y+z=C \iff x+y+z-C=0 \implies F(x,y,z)=x+y+z-C$

Now we use the Lagrange multipliers to get

$\nabla V(x,y,z)=\lambda \nabla F(x,y,z)$

This gives

$yz \vec{i}+xz\vec{j}+xy\vec{k}=\lambda\vec{i}+\lambda \vec{j}+\lambda\vec{k}$

This gives a system of three equations with four unknowns. Using the original contraint gives the non linear system of equations

$yz=\lambda \\ xz=\lambda \\ xy=\lambda \\ x+y+z=C$

This gives that

$xyz=\lambda x =\lambda y =\lambda z$

or

$x=y=z= \frac{C}{3}$

You can use Lagrange multipliers in any dimention to solve this problem
• Aug 14th 2012, 01:17 PM
andre.vignatti
Re: Maximizing the cuboid volume with fixed perimeter
Hummm, thank you TheEmptySet! Now I need to learn about the Lagrange multipliers :-)