Among all the retangular parallelpipeds of volume $V$, find one whose total surface área is minimum

Using the Lagrange Multipliers method, I've found that it is a cube with dimensions $ \sqrt[3]{V} $. But I don't know how to prove that it is, indeed, a cube with those dimensions, since I couldn't prove that the function $S_A(x,y,z)=2xy + 2xz + 2yz$ (surface total area) have a minimum. Can you help me with it?

Thanks in advance


1 Answer 1


By the arithmetic-geometric mean inequality, surface area of any rectangular parallelpiped satisfies $$2(xy+yz+zx)\geq 2(3)\left(\sqrt[3]{x^2y^2z^2}\right) = 6V^{\frac{2}{3}}.$$

On the other hand, a cube with side lengths $V^{\frac{1}{3}}$ has surface area $$2(V^{\frac{1}{3}}V^{\frac{1}{3}}+V^{\frac{1}{3}}V^{\frac{1}{3}}+V^{\frac{1}{3}}V^{\frac{1}{3}}) = 6V^{\frac{2}{3}}.$$

Hence the cube does indeed have minimum surface area.


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