Question

Prove that of all the rectangular parallelepiped of the same volume the cube has the least surface

Answer 1

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Answer 2

Answer:

All the rectangular parallelepiped of the same volume the cube has the least surface (Proved)

Step-by-step explanation:

Don't try to perform a second derivative test in connection with Lagrange's method. The point you have found is clearly the maximum. Using the AGM inequality you have

$\sqrt[3]{{v}^{2} } = \sqrt[3]{ab.bc.ca} \leqslant \frac{ab + bc+ca }{3} \\ = \frac{1}{6} \times s$

with equality sign iff ab=bc=ca, i.e., iff a=b=ca=b=c. It follows that

$v \leqslant ( \frac{s}{6} )^{ \frac{3}{2} }$

with equality iff the parallelopiped is a cube with the given surface area.