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Thread: Maximizing the cuboid volume with fixed perimeter

  1. #1
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    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?
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  2. #2
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    Re: Maximizing the cuboid volume with fixed perimeter

    Quote Originally Posted by andre.vignatti View Post
    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
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  3. #3
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    Re: Maximizing the cuboid volume with fixed perimeter

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