Question

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

Answer 1

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

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

3

v

2

=

3

ab.bc.ca

⩽

3

ab+bc+ca

=

6

1

×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} }v⩽(

6

s

)

2

3

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