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 $\displaystyle C$. Which are the dimensions values that maximize the area?

**A**: One dimension is $\displaystyle x$, and thus the other is $\displaystyle C-x$. Therefore, the area is $\displaystyle x \cdot (C-x) = xC - x^2$. Then, setting the derivative equal to zero would provide us the answer ($\displaystyle C/2$ 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 $\displaystyle C$. Which are the dimensions values that maximize the volume?

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

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 $\displaystyle C$. Which are the dimensions values that maximize the area?

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

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

subject to the constraint that

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

Now we use the Lagrange multipliers to get

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

This gives

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

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

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

This gives that

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

or

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

You can use Lagrange multipliers in any dimention to solve this problem

Re: Maximizing the cuboid volume with fixed perimeter

Hummm, thank you TheEmptySet! Now I need to learn about the Lagrange multipliers :-)