Thread: 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 . Which are the dimensions values that maximize the area?


A: One dimension is , and thus the other is . Therefore, the area is . Then, setting the derivative equal to zero would provide us the answer ( on both dimensions)

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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 . Which are the dimensions values that maximize the volume?




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

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 . Which are the dimensions values that maximize the area?


A: One dimension is , and thus the other is . Therefore, the area is . Then, setting the derivative equal to zero would provide us the answer ( 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 . 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



subject to the constraint that



Now we use the Lagrange multipliers to get



This gives



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



This gives that



or



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

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