2
$\begingroup$

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

$\endgroup$
0

1 Answer 1

5
+50
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.